Tao, §1.1 introduction, and §1.2.3.

There exists a set \(A\subset {\mathbb{R}}\) such that no measure can be assigned to \(A\) consistently with (a)–(c).

Rather than work on \({\mathbb{R}}\), we will work on the half-open unit interval \([0,1)\) with the addition operation taken modulo \(1\). This is ok, since if there is a measure \(m\) on all subsets of \({\mathbb{R}}\), then by properties (b) and (c), \(m\) restricts to a measure on subsets of \([0,1)\) which satisfies property (b) with respect to addition modulo \(1\).

Now let \({\mathbb{Q}}_1\) denote the rationals of \([0,1)\), that is, \({\mathbb{Q}}_1={\mathbb{Q}}\cap [0,1)\), and consider the collection of additive cosets of \({\mathbb{Q}}_1\) inside \([0,1)\). The cosets are of the form \(a+{\mathbb{Q}}_1\) where again addition is interpreted modulo \(1\). We now let \(A\subset [0,1)\) denote a system of coset representatives for this collection.

Now every number in \([0,1)\) can be written uniquely as \(a+q\) for \(a\in A\) and \(q\in {\mathbb{Q}}_1\). This means that the collection of translates of \(A\) by elements \(q\in {\mathbb{Q}}_1\) covers all of \([0,1)\). In particular, by (a) the measure of \(\bigcup _{q\in {\mathbb{Q}}_1}(A+q)\) is exactly \(1\).

On the other hand, by (b) and (c) we have that

By the previous paragraph, the left-hand side of the above equation is \(1\). On the other hand the right-hand side is an infinite sum of some nonnegative constant, and hence must be either \(0\) or \(\infty \). This is a contradiction!

The proof of Vitali’s theorem requires the Axiom of Choice. Specifically, it is needed to find a system of coset representatives for an uncountable collection. Solovay showed that the use of AC is essential, and that it is consistent with \(\neg \)AC that there is a measure function \(m\) on all subsets of \({\mathbb{R}}\).

Show that the properties (a)–(c) of a measure imply finite addivitity: If \(A\) and \(B\) are disjoint then \(m(A\cup B)=m(A)+m(B)\).

Show that the properties (a)–(c) of a measure imply the inclusion–exclusion principle: For any sets \(A,B\) we have \(m(A\cup B)+m(A\cap B)=m(A)+m(B)\).

Complete the details of the proof that if there is a measure on \({\mathbb{R}}\) satisfying properties (a)–(c), then there is a measure on \([0,1)\) (with additional modulo \(1\)) satisfying properties (a)–(c).

If \(A\) is a bounded set of real numbers, the supremum \(\sup (A)\) is the least upper bound of \(A\), and the infemum \(\inf (A)\) is the greatest lower bound of \(A\). Show that \(s=\sup (A)\) if and only if:

\(\circ \)

for all \(a\in A\) we have \(a\leq s\), and;

\(\circ \)

for all \(\epsilon >0\), there exists \(a\in A\) such that \(s-a<\epsilon \).

Formulate and prove the analogous statement for the infemum.

Tao, §1.1.1

A subset \(E\subset {\mathbb{R}}^n\) is elementary if it can be expressed as a union of finitely many boxes.

Any elementary set \(E\) can be expressed as a finite union of disjoint boxes.

First assume that \(E\subset {\mathbb{R}}^1\) and that \(E=\bigcup I_i\). Then by considering all endpoints of the \(I_i\) in increasing order \(a_1,\ldots ,a_m\) it is easy to write \(E\) as the union of sets of the form \((a_i,a_{i+1})\) together with sets of the form \([a_i,a_i]\) (single points). Such a union is clearly disjoint.

In general if \(E\subset {\mathbb{R}}^n\) and \(E=\bigcup B_i\) then for each dimension \(d\leq n\) consider in turn the \(d\)th sides of the boxes \(I_i^d\). Again consider the endpoints of these intervals in increasing order \(a_i^d,\ldots ,a_{m_d}^d\). Then we can write \(E\) as a union of small boxes which are products of sets of the form \((a_i^d,a_{i+1}^d)\) or of the form \([a_i^d,a_{i+1}^d]\). Such boxes are again disjoint.

Suppose the elementary set \(E\) can be expressed in two ways a a finite union of disjoint boxes: \(E=\bigsqcup B_i=\bigsqcup C_j\). Then \(\sum \operatorname {vol}(B_i)=\sum \operatorname {vol}(C_j)\).

We first note that \(I\) is an interval with endpoints \(a,b\), and if \(a=a_1,a_2,\ldots ,a_m=b\) is an increasing sequence then \(\operatorname {len}(I)=\sum \operatorname {len}(a_i,a_{i+1})\). This is simply because the latter summation telescopes.

Next if \(B\) is a box whose \(d\)th side has endpoints \(a^d,b^d\), and if \(a^d=a_1^d,a_2^d,\ldots ,a_{m_d}^d=b^d\) then \(\operatorname {vol}(B)=\) the sum of all small boxes of the form \(\prod (a_{i_d}^d,a_{i_d+1}^d)\). We will call the set of such small boxes a perfect grid. Intuitively, if you break a box into a perfect grid of sub-boxes, then the volume of the box is the sum of the volumes of the sub-boxes.

Now if \(B\) is a box and one expresses it as a disjoint union of sub-boxes \(B=\bigsqcup B_i\), then \(\operatorname {vol}(B)=\sum \operatorname {vol}(B_i)\). This is because it is possible to find a refinement of the given disjoint union which is a perfect grid as in the previous paragraph. That is, it is possible to write \(B=\bigsqcup D_i\) where \(\{D_i\}\) is a perfect grid, and each \(B_i\) is the union of a perfect grid of sets taken from the collection \(\{D_i\}\). Then one can simply apply the argument of the previous paragraph to \(B\) and to each \(B_i\).

Finally given \(E\), \(B_i\), and \(C_j\) as in the problem statement, one can find a third expression \(E=\bigsqcup D_k\) where \(\{D_k\}\) is a refinement of both \(\{B_i\}\) and of \(\{C_j\}\). That is, each \(B_i\) and each \(C_j\) is a disjoint union of elements of \(\{E_k\}\). It follows from the previous paragraph that \(\sum \operatorname {vol}(B_i)=\sum \operatorname {vol}(D_k)\) and analogously that \(\sum \operatorname {vol}(C_j)=\sum \operatorname {vol}(D_k)\). This completes the proof.

The elementary measure function \(m\) satisfies

(a)

(normality) \(m(B)=\operatorname {vol}(B)\) for any box \(B\);

(b)

(translation-invariance) \(m(x+E)=m(E)\) for any elementary set \(E\); and

(c)

(finite additivity) \(m(E\cup F)=m(E)+m(F)\) for any disjoint elementary sets \(E,F\).

The elementary measure function \(m\) satisfies

\(\circ \)

(monotonicity) \(m(E)\leq m(F)\) for elementary sets \(E\subset F\); and

\(\circ \)

(finite subadditivity) \(m(E\cup F)\leq m(E)+m(F)\) for elementary \(E,F\).

Show that the class of elementary sets is closed under the operations: union, intersection, set difference, symmetric difference, and translation.

Prove Proposition 2.5: The elementary measure satisfies the monotonicity and finite subadditivity properties.

Show that if \(E\) is an elementary subset of \({\mathbb{R}}^m\) and \(F\) is an elementary subset of \({\mathbb{R}}^n\) then \(E\times F\) is an elementary subset of \({\mathbb{R}}^{m+n}\). Furthermore show that \(m_e(E\times F)=m_e(E)m_e(F)\).

Tao, §1.1.2.

Let \(A\) be a bounded subset of \({\mathbb{R}}^n\). First define the inner and outer Jordan measures (sometimes called lower and upper):

If \(m_{*j}(A)=m^{*j}(A)\) we say that \(A\) is Jordan measurable, we call the common quantity the Jordan measuref of \(A\), and we denote it by \(m(A)\).

The set \(A\) is Jordan measurable if and only if either of the following holds:

\(\circ \)

For all \(\epsilon >0\) there are elementary sets \(E,F\) such that \(E\subset A\subset F\) such that \(m(F\smallsetminus E)<\epsilon \).

\(\circ \)

For all \(\epsilon >0\) there is an elementary set \(E\) such that \(m^{*j}(E\bigtriangleup A)<\epsilon \).

We establish only the equivalence of Jordan measurability with the first item. To begin, assume that \(A\) is Jordan measurable and let \(\epsilon >0\) be given. By the \(m_{*j}\) definition of Jordan measure, we can find an elementary set \(E\subset A\) such that \(m(A)-m(E)<\epsilon /2\). By the \(m^{*j}\) definition of jordan measure we can find an elementary set \(F\) such that \(A\subset F\) and \(m(F)-m(A)<\epsilon /2\). It follows that

as desired.

For the converse, assume that the first bullet holds true, and let \(\epsilon >0\) be arbitrary. Then we can find elementary sets \(E,F\) such that \(E\subset A\subset F\) and \(m(F)-m(E)<\epsilon \). From the definitions of inner and outer Jordan measure, we have that \(m(E)\leq m_{*j}(A)\leq m^{*j}(A)\leq m(F)\). It follows that \(m^{*j}(A)-m_{*j}(A)<\epsilon \). Since \(\epsilon \) was arbitrary, we may conclude that \(m_{*j}(A)=m^{*j}(A)\) and therefore that \(A\) is Jordan measurable.

If \(A,B\) are Jordan measurable, then so are \(A\cup B\), \(A\cap B\), and \(A\smallsetminus B\).

We prove only the case of \(A\cup B\). Suppose that \(A,B\) are Jordan measurable. By the previous lemma, we can find elementary sets \(E,F,E',F'\) such that \(E\subset A\subset F\), and \(E'\subset B\subset F'\), and \(m(F\smallsetminus E),m(F'\smallsetminus E')<\epsilon /2\). Then we have \(E\cup E'\subset A\cup B\subset F\cup F'\) and using some algebra together with the finite subadditivity of elementary measure, \(m(F\cup F'\smallsetminus (E\cup E'))\leq m(F\smallsetminus E)+m(F'\smallsetminus E')<\epsilon \). Again by the previous lemma, this shows that \(A\cup B\) is Jordan measurable.

The Jordan measure satisfies finite additivity, that is, if \(A,B\) are Jordan measurable and disjoint, then \(m(A\cup B)=m(A)+m(B)\).

We first show subbaditivity, that is, that \(m(A\cup B)\leq m(A)+m(B)\). Let \(\epsilon >0\) be given. Using the fact that \(m=m^{*j}\) we can find elementary sets \(F,F'\) such that \(A\subset F\), \(B\subset F'\), \(m(F)-m(A)<\epsilon /2\), and \(m(F')-m(A')<\epsilon /2\). Using the monotonicity and subadditivity properties of the elementary measure, together with the definition of Jordan measurability, we now have:

Since \(\epsilon \) was arbitrary, we achieve the desired inequality.

Now additionally assume that \(A,B\) are disjoint, and again let \(\epsilon >0\). This time using \(m=m_{*j}\), we can find elementary sets \(E,E'\) such that \(E\subset A\), \(E'\subset B\), \(m(A)-m(E)<\epsilon /2\), and \(m(B)-m(E')<\epsilon /2\). Using the fact that \(E,E'\) are disjoint, the finite additivity of elementary measure, and the definition of Jordan measurability, we now have:

Again letting \(\epsilon \) tend to \(0\), we achieve that \(m(A\cup B)\geq m(A)+m(B)\).

Let \(f\) be a continuous function defined on a closed, bounded interval. Then the graph of \(f\), considered as a subset of \({\mathbb{R}}^2\), has Jordan measure \(0\).

Let \(I\) denote the domain of \(f\). Recall that since \(I\) is closed and bounded, it is compact. Recall also that a continuous function with a compact domain is uniformly continuous: for any \(\epsilon >0\) there exists a \(\delta >0\) such that for any interval \(J\), \(\operatorname {len}(J)<\delta \) implies \(\operatorname {len}(f(J))<\epsilon \).

So let \(\epsilon >0\) be given, and choose \(\delta >0\) as above. Shrinking \(\delta \) if necessary, we can suppose that \(\operatorname {len}(I)/\delta \) is an integer \(k\). Partitioning \(I\) into intervals \(J_1,\ldots ,J_k\) each of lengh \(\delta \), we have that the graph of \(f\) is contained in the set

Note that the min and max values in the definition of \(A\) exist by the extreme value theorem. Now \(A\) is a union of \(k\) many rectangles each of size at most \(\delta \epsilon \). Thus \(A\) is elementary and its measure is at most \(k\delta \epsilon \). This latter value is \(\operatorname {len}(I)\epsilon \), so the upper measure \(m^{*j}\) of the graph of \(f\) is at most \(\operatorname {len}(I)\epsilon \). Taking \(\epsilon \to 0\), we conclude that \(f\) is Jordan measurable with measure \(0\).

Verify that Jordan measure agrees with the elementary measure on elementary sets (thus satisfies the normality property). Verify that Jordan measure satisfies the translation-invariance property.

Complete the proof of Lemma 3.2: \(A\) is Jordan measurable iff for all \(\epsilon >0\) there is an elementary set \(E\) such that \(m^{*j}(E\bigtriangleup A)<\epsilon \).

Complete the proof of Proposition 3.3: If \(A,B\) are Jordan measurable, then so are \(A\cap B\) and \(A\smallsetminus B\).

Say that \(A\) is Jordan null if \(A\) is Jordan measurable and \(m(A)=0\). Show that any subset of a Jordan null set is a Jordan null set.

Show that the outer Jordan measure \(m^{*j}(A)\) is equal to:

Let \(A\) be an arbitrary bounded set, and let \(E\) be an elementary set. Show that

Show that \(A\) is Jordan measurable if and only if for all \(\epsilon >0\) there exists an elementary set \(E\) such that \(A\subset E\) and \(m^{*j}(E\smallsetminus E)<\epsilon \).

Tao, §1.1.3.

Let \(f\) be a real-valued function defined on \([a,b]\). Then \(f\) is said to be piecewise constant if there exists a partition \(\mathcal{P}\) of \([a,b]\) into finitely many subintervals \(I_j\) such that \(f\) takes a constant value \(c_j\) on each interval \(I_j\).

If \(f=\sum _1^kc_j\chi _{I_j}\) then the pc integral of \(f\) is defined to be \(\sum _1^kc_j\operatorname {len}(I_j)\).

Let \(f\) be a bounded function on \([a,b]\). First define the lower and upper Riemann forms:

Then if \(\underline{\int }f=\overline{\int }f\) we say that \(f\) is Riemann integrable, and denote the common value simply by \(\int f\).

The Riemann integral satisfies the three properties:

\(\circ \)

(normality) If \(A\) is a Jordan measurable subset of \([a,b]\), then \(\chi _A\) is Riemann integrable over \([a,b]\) and \(\int \chi _A=m(A)\).

\(\circ \)

(linearity) If \(f,g\) are Riemann integrable then so are \(cf\) and \(f+g\) and we have \(\int cf=c\int f\), and \(\int (f+g)=\int f+\int g\).

\(\circ \)

(monotonicity) If \(f,g\) are Riemann integrable and \(f\leq g\) then \(\int f\leq \int g\).

We establish only the normality property. By Lemma 3.2, for any \(\epsilon \) we can find disjoint intervals \(I_j\) and disjoint intervals \(J_k\) such that \(\bigcup I_j\subset A\subset \bigcup J_k\) and \(m(\bigcup J_k\smallsetminus \bigcup I_j)<\epsilon \). It is easy to see from the definition of the pc integral that \({\text{pc}}\int \chi _{\bigcup I_j}=m(\bigcup I_i)\), and similarly \({\text{pc}}\int \chi _{\bigcup J_k}=m(\bigcup J_k)\). We now have

Since the left and right-hand sides differ by \(<\epsilon \), it follows that the lower and upper integrals differ by \(<\epsilon \) as well. Since \(\epsilon \) was arbitrary, it follows that \(\chi _A\) is integrable. And since we also have

we may conclude that \(\int \chi _A\) is equal to \(m(A)\).

If \(f\) is a nonnegative, bounded function on \([a,b]\), then \(f\) is Riemann integrable if and only if the region \(A=\left\{\,(x,y)\mid 0\leq y\leq f(x)\,\right\}\) is Jordan measurable. Moreover, in this case we have \(\int f=m(A)\).

First suppose that \(f\) is Riemann integrable and let \(\epsilon >0\) be given. Choose pc functions \(g,h\) such that \(g\leq f\leq h\) and \({\text{pc}}\int (h-g)<\epsilon \). Let \(E\) be the region under the graph of \(g\) and let \(F\) be the region under the graph of \(h\). It is clear that \(E,F\) are elementary, \(E\subset A\subset F\), and \(m(F\smallsetminus E)<\epsilon \).

Conversely if \(A\) is Jordan measurable we can find an elementary \(E\) such that \(E\subset A\) and \(m(A\smallsetminus E)<\epsilon \). Using our usual grid argument, we can suppose that there is a sequence of disjoint intervals \(I_j\) such that \(E\) is a union of boxes with horizontal sides selected from the \(I_j\). Pairing each \(I_j\) with the constant \(c_j=\) the maximum of the vertical coordinates of all of the boxes with horizontal side \(I_j\), we obtain a pc function \(g\). It is easy to see that \(m(E)\leq {\text{pc}}\int g\leq m(A)\). This shows that the lower Riemann integral of \(f\) is \(m(A)\). We can proceed similarly using an outer approximation \(B\) to show that the upper Riemann integral of \(f\) is \(m(A)\) too.

With \(f\), \(\mathcal{P}\), \(\delta x_i\), \(x_i^*\) as above, the corresponding Riemann sum is:

The Riemann integral of \(f\) on \([a,b]\) is then defined by

provided this limit exists. Here the limit “exists” and equals \(L\) if for all \(\epsilon >0\) there exists \(\delta >0\) such that for all \(\mathcal{P}\) and \(x_i^*\) we have \(\|\mathcal{P}\|<\delta \) implies \(|R(f,\mathcal{P},x_i^*)-L|<\epsilon \).

Show that the pc integral is well-defined, and satisfies the normality, linearity, and monotonicity properties.

Let \(f\) be a bounded function on the interval \([a,b]\). Then \(f\) is integrable in the Darboux sense if and only if \(f\) is integrable in the classical Riemann sense, and in this case the two values agree.

Let \(f\colon [a,b]\to {\mathbb{R}}\). Show that if \(f\) is continuous, then \(f\) is Riemann integrable. Show that if \(f\) is bounded and piecewise continuous, then \(f\) is Riemann integrable.

Complete the proof of Proposition 4.4: Show that the Riemann integral satisfies the linearity and monotonicity properties. (Hint: first establish these properties for the pc integral.)

Tao, §1.2, first few pages

Let \(A\) be any subset of \({\mathbb{R}}^n\). The Lebesgue outer measure of \(A\) is

Let \(A\) be any subset of \({\mathbb{R}}^n\). We say that \(A\) is Lebesgue measurable if for every \(\epsilon >0\) there exists a sequence of boxes \(B_i\) such that \(A\subset \bigcup B_i\) and \(m^*(\bigcup B_i\smallsetminus A)<\epsilon \). When this is the case, we define \(m(A)=m^*(A)\) to be the Lebesgue measure of \(A\).

The Lebesgue outer measure satisfies the outer measure axioms (a)–(c).

The axioms (a) and (b) are both trivial, so it remains to prove only axiom (c). Let \(E_n\) be arbitrary sets and let \(\epsilon >0\) be given. From the definition of Lebegue outer measure, for each \(n\) we can find a sequence of boxes \(B_i^n\) such that \(A_n\subset \bigcup _iB_i^n\) and \(\sum _im(B_i^n)-m^*(A_n)<\epsilon /2^n\).

Taking unions, we have \(\bigcup A_n\subset \bigcup _n\bigcup _iB_i^n\), and moreover:

Taking \(\epsilon \to 0\), we obtain the desired inequality \(m^*(\bigcup A_n)\leq \sum m^*(A_n)\).

Show that the countable union of Jordan measurable sets need not be Jordan measurable, even when bounded. Show that the countable intersection of Jordan measurable sets need not be Jordan measurable.

Give an example of a sequence of uniformly bounded, Riemann integrable functions on \([0,1]\) which converges pointwise to a function that is not Riemann integrable. Is it possible to give an example which converges uniformly?

Show that \(m^*(A)\leq m^{*j}(A)\). Give an example of a set \(A\) such that \(m^*(A)<m^{*j}(A)\).

Show that Jordan outer measure does not satisfy countable subadditivity.

Show that if \(A\) is Lebesgue null, that is, \(m^*(A)=0\), then \(A\) is Lebesgue measurable.

Tao, §1.2.1.

Suppose that \(A,B\) are positively separated, that is, that \(d(A,B)=\inf \left\{\,d(x,y)\mid x\in A, y\in B\,\right\}>0\). Then \(m^*(A\cup B)=m^*(A)+m^*(B)\).

Subbaditivity implies that \(m^*(A\cup B)\leq m^*(A)+m^*(B)\), so it remains only to show \(m^*(A\cup B)\geq m^*(A)+m^*(B)\). Applying the definition of \(m^*(A\cup B)\), given any \(\epsilon >0\) we can find boxes \(C_i\) such that \(A\cup B\subset \bigcup C_i\) and \(\sum \operatorname {vol}(C_i)-m^*(A\cup B)<\epsilon \).

Let us first consider an easy case when each \(C_i\) meets at most one of the sets \(A,B\). Then we can rewrite the sequence \(\{C_i\}\) as \(\{D_i\}\cup \{E_i\}\), where the \(D_i\)’s meet only \(A\) and the \(E_i\)’s meet only \(B\). Now

Taking \(\epsilon \to 0\), we are done in this case.

In the general case, we can reduce to the easy one by partitioning each \(B_i\) into smaller boxes, each with diameter smaller than \(d(A,B)\). Once this is done, each new box meets at most one of \(A,B\) and we may proceed as above.

If \(E\) is an elementary subset of \({\mathbb{R}}^n\), then \(m^*(E)\) agrees with the elementary measure \(m_e(E)\).

It is clear that \(m^*(E)\leq m_e(E)\), since \(E\) is itself a union of boxes whose volumes sum to \(m(E)\). Thus it remains only to show \(m^*(E)\geq m_e(E)\). Appealing to the definition of \(m^*(E)\), given any \(\epsilon >0\) we can find boxes \(B_i\) such that \(E\subset \bigcup B_i\) and \(\sum \operatorname {vol}(B_i)-m^*(E)<\epsilon \). Rearranging, this says \(m^*(E)>\sum \operatorname {vol}(B_i)-\epsilon \). Now we would like to say that \(\sum \operatorname {vol}(B_i)\geq m_e(E)\), but unfortunately the elementary measure is only finitely subadditive.

In order to proceed, let us temporarily assume that \(E\) is closed and the \(B_i\) are open. We recall that any closed and bounded set is compact, and that any covering of a compact set by open sets has a finite subcovering. Thus under these assumptions, we have that just finitely many of the \(B_i\) are needed to cover \(E\). Thus the argument of the previous paragraph works in this case!

In order to assume that the \(B_i\) are open, we can enlarge each slightly and find an open box \(B'_i\) such that \(B_i\subset B'_i\) and \(\operatorname {vol}(B'_i)-\operatorname {vol}(B_i)<\epsilon /2^i\).

In order to assume that \(E\) is closed, first write it as a finite union of disjoint boxes \(C_1,\ldots ,C_k\). Shrinking each \(C_i\) slightly, we can find a closed box \(C'_i\subset C_i\) such that \(m_e(C_i\smallsetminus C'_i)<\epsilon /k\). Replacing \(E\) with \(\bigcup C'_i\) we obtain a closed set as desired.

Suppose \(B_i\) is a sequence of pairwise almost disjoint boxes. Then \(m^*(\bigcup B_i)=\sum \operatorname {vol}(B_i)\).

By subadditivity together with the previous theorem, we have \(m^*(\bigcup B_i)\leq \sum m^*(B_i)=\sum \operatorname {vol}(B_i)\). Hence it remains only to show \(m^*(\bigcup B_i)\geq \sum \operatorname {vol}(B_i)\). For this, let \(N\in {\mathbb{N}}\) be given, and note that \(\bigcup _1^NB_i\) is an elementary set. Thus by monotonicity together with the previous theorem, we have

Taking \(N\to \infty \), we obtain the desired inequality.

Any open set \(O\) can be written as a union of a sequence of pairwise almost disjoint boxes.

Consider the family \(\mathcal{Q}\) of dyadic cubes, that is, cubes with each side of the form \([m/2^n,(m+1)/2^n]\) where \(n\geq 0\). The family \(\mathcal{Q}\) has a nesting property: for any two cubes in \(\mathcal{Q}\), either one is contained in the other or else the two cubes are almost disjoint.

It is not difficult to observe that \(\mathcal{Q}\) is a basis for the topology of \({\mathbb{R}}^n\). In particular, for any \(x\in O\) there exists a cube \(B_x\in \mathcal{Q}\) such that \(x\in B_x\subset O\). Thu union of all \(B_x\) for \(x\in O\) is a covering of \(O\) by dyadic cubes. Now we eliminate duplicates from this covering, that is, remove any cube in the covering that is contained in some other cube of the covering. Since any nested chain of cubes has a maximal element, the cubes which remain will still cover \(O\). And by the nesting property, the cubes which remain will also be almost disjoint.

Let \(A\) be any subset of \({\mathbb{R}}^n\). Then \(m^*(A)=\inf \left\{\,m^*(O)\mid {\text{$O$ is open and }}A\subset O\,\right\}\).

It is clear from monotonicity of \(m^*\) that \(\leq \) holds. Thus it remains only to show \(\geq \). Applying the definition of \(m^*(A)\) we can find boxes \(B_i\) such that \(A\subset \bigcup B_i\) and \(\sum \operatorname {vol}(B_i)-m^*(A)<\epsilon \). Arguing as in an earlier proof, we can enlarge the \(B_i\) slightly to assume without loss of generality that they are open. Then

Taking \(\epsilon \to 0\), we obtain the desired result.

Suppose \(A\) is expressible as a countable union of pairwise almost disjoint boxes. Show that \(m^*(A)=m_{*j}(A)\).

Show that it is not true in general that

Tao, §1.2.2.

Open and closed sets are Lebesgue measurable. Complements, countable unions, and countable intersections of measurable sets are measurable.

Since the boxes form a base for the topology of \({\mathbb{R}}^n\), any open set can be written as a union of boxes. (Or see Proposition 6.4.) Thus by the remark above, open sets are Lebesgue measurable.

For countable unions, suppose that \(A_n\) are Lebesgue measurable. Given \(\epsilon >0\), find for each \(n\) a countable union of boxes \(U_n\) such that \(A_n\subset U_n\) and \(m^*(U_n\smallsetminus A_n)<\epsilon /2^n\). Then we have

This shows that \(\bigcup A_n\) is measurable.

For closed sets, assume first that \(A\) is closed and bounded, and thus compact. Using the outer regularity lemma we can find an open set \(O\) such that \(A\subset O\) and \(m^*(O)-m^*(A)<\epsilon \). We wish to show that \(m^*(O\smallsetminus A)<\epsilon \) too. Since \(O\smallsetminus A\) is open, we can use Proposition 6.4 to write \(O\smallsetminus A\) as an almost disjoint union of closed dyadic cubes \(C_n\). Then \(\bigcup _1^N C_n\) is compact and thus positively separated from the compact set \(A\). By Lemma 6.1, additivity holds for positively separated sets, so we have:

It follows that \(\sum _1^N\operatorname {vol}(C_i)=m^*(\bigcup _1^NC_i)<\epsilon \), and taking \(N\to \infty \) we have \(\sum \operatorname {vol}(C_i)\leq \epsilon \). This shows that \(m^*(O\smallsetminus A)\leq \epsilon \), as desired.

In general a closed set can be written as a countable union of compact sets, and we have already handled the case of countable unions.

For complements, let \(A\) be measurable and for each \(n\) find a union of boxes \(U_n\) such that \(A\subset U_n\) and \(m^*(U_n\smallsetminus A)<1/n\). We can enlarge the constituent boxes of each \(U_n\) slightly to find an open set \(O_n\) such that \(U_n\subset O_n\) and \(m^*(U_n\smallsetminus O_n)<1/n\). (Here we are using normality and subadditivity to achieve this estimate.) Taking the intersection of the \(O_n\) we now have \(A\subset \cap O_n\) and \(m^*(\bigcap O_n\smallsetminus A)=0\). Writing these two expressions in complement, they become \(\bigcup O_n^c\subset A^c\) and \(m^*(A^c\smallsetminus \bigcup O_n^c)=0\). Now \(A^c\) can be expressed as a union of two sets: \(\bigcup O_n^c\) and \(A^c\smallsetminus \bigcup O_n^c\). The first is Lebesgue measurable because it is a countable union of closed sets. The second is Lebesgue measurable because it is null (see an earlier exercise). Appealing again to the closure under unions, we conclude that \(A^c\) is measurable too.

For countable intersections, we can simply apply Demorgan’s laws to reduce it to complements and countable unions. Whew!

The collection of Lebesgue measurable sets is the least \(\sigma \)-algebra containing both the open sets and the Lebesgue null sets.

It is clear that the Lebesgue measurable sets are a \(\sigma \)-algebra containing the open sets and the Lebesgue null sets. On the other hand suppose that \(E\) is a Lebesgue measurable set. By the previous lemma for all \(n\) we can find open sets \(O_n\) such that \(E\subset O_n\) and \(m^*(O_n\smallsetminus E)<1/n\). It follows that \(N=\bigcap O_n\smallsetminus E\) is Lebesgue null. Now have that

and thus \(E\) lies in the \(\sigma \)-algebra generated by the open sets and the Lebesgue null sets.

A set \(A\) is Lebesgue measurable if and only if for all \(\epsilon >0\) there exists an open set \(O\) such that \(m^*(O\bigtriangleup A)<\epsilon \).

Our original definition of Lebesgue measurability automatically gives an open set \(O\) such that \(m^*(O\bigtriangleup A)<\epsilon \). Conversely, let \(A\) be any set and suppose the condition holds. Then for any \(\epsilon \) we can find an open set \(O_\epsilon \) such that \(m^*(O_\epsilon \bigtriangleup E)<\epsilon \). Let \(U_\epsilon =\bigcup O_{\epsilon /2^k}\). Then it is not difficult to check that \(m^*(U_\epsilon \smallsetminus A)\leq \epsilon \), and \(m^*(A\smallsetminus U_\epsilon )=0\). Finally letting \(B=\bigcap _nU_{1/n}\) we have that \(B\) is a measurable set and both \(m^*(A\smallsetminus B)=0\) and \(m^*(B\smallsetminus A)=0\). We have thus shown that \(A\) differs from a measurable set by a null set, and we leave it as an exercise to check that this implies \(A\) is measurable too.

A set \(A\) is Lebesgue measurable with finite Lebesgue measure if and only if for all \(\epsilon >0\) there exists an elementary set \(E\) such that \(m^*(E\bigtriangleup A)<\epsilon \).

Suppose that \(A\) is Lebesgue measurable and let \(\epsilon >0\) be given. Let \(O\) be an open set such that \(A\subset O\) and \(m^*(O\smallsetminus A)<\epsilon \). Then \(O\) can be written as a union of almost disjoint boxes \(O=\bigcup B_i\), and we know that \(m(O)=\sum \operatorname {vol}(B_i)\).

Now \(m(O)<m(A)+\epsilon \) and the right-hand side is finite, so the sum \(\sum \operatorname {vol}(B_i)\) converges. Thus there exists some \(N\) such that \(\sum _{N+1}^\infty \operatorname {vol}(B_i)<\epsilon \). Letting \(E=\bigcup _1^N B_i\), we have that \(E\) is elementary and \(m(O\smallsetminus E)<\epsilon \). Thus we have

which is sufficient to prove the implication. The converse implication is similar to the previous lemma.

(a)

Show that \(A\) is measurable iff for all \(\epsilon >0\) there exists a closed set \(F\subset A\) such that \(m^*(A\smallsetminus F)<\epsilon \).

(b)

Show that \(A\) is measurable iff for all \(\epsilon >0\) there exists a measurable set \(B\) such that \(m^*(A\bigtriangleup B)<\epsilon \).

Show that any set \(A\) is contained in a Lebesgue measurable set \(B\) such that \(m(B)=m^*(A)\).

Show the inner regularity property: If \(A\) is Lebesgue measurable, then

Tao, §1.2.2

The Lebesge measure satisfies the axioms

(a)

(normality) if \(B\) is a box then \(m(B)=\operatorname {vol}(B)\);

(b)

(translation-invariance) \(m(x+A)=m(A)\) for every \(x\in {\mathbb{R}}^n\) and measurable set \(A\);

(c)

(countable additivity) \(m(\bigcup A_n)=\sum m(A_n)\) for every sequence of pairwise disjoint measurable sets \(A_n\).

We have already established normality for \(m^*\) and hence for \(m\). Translation-invariance of \(m^*\) is clear from the definition since it is clear for boxes.

For countable additivity, first recall that we always have subadditivity so we need only show \(m(\bigcup A_n)\geq \sum m(A_n)\). Suppose first that the \(A_n\) are compact. Then they are pairwise positively separated, so using Lemma 6.1 inductively we can establish that \(m(\bigcup _1^N A_n)=\sum _1^N m(A_n)\). It follows that \(m(\bigcup A_n)\geq \sum _1^Nm(A_n)\). Taking \(N\to \infty \) we have \(m(\bigcup A_n)\geq \sum m(A_n)\) as desired.

Next assume that the \(A_n\) are bounded but not necessarily closed. By the measurability of \(A_n^c\) we can find open sets \(O_n\) such that \(A_n^c\subset O_n\) and \(m^*(O_n\smallsetminus A_n^c)<\epsilon /2^n\). Taking complements we thus have compact sets \(K_n\subset E_n\) such that \(m^*(A_n\smallsetminus K_n)<\epsilon /2^n\). Now using the additivity for compact sets,

Taking \(\epsilon \to 0\), we are finished in this case.

Finally for general \(A_n\), decompose \({\mathbb{R}}^d\) into disjoint bounded cells \(C_m\). Then \(A_n=\bigcup _m A_n\cap C_m\). Now the sets \(A_n\cap C_n\) are bounded, so applying the result for bounded sets twice we have:

and the proof is complete.

\(\circ \)

(upwards monotone convergence theorem) If \(A_n\) are measurable and \(A_{n+1}\supset A_n\) then \(m(\bigcup A_n)=\lim m(A_n)\).

\(\circ \)

(downwards monotone convergence theorem) If \(A_n\) are measurable and \(A_{n+1}\subset A_n\) then \(m(\bigcap A_n)=\lim m(A_n)\), provided some \(A_n\) has finite measure.

For the upwards MCT, let \(A'_n=A_n\smallsetminus A_{n-1}\) and note that the \(A'_n\) are disjoint and have the same union as before: \(\bigcup A'_n=\bigcup A_n\). Note that we implicitly set the value \(A_0=\emptyset \). Applying countable additivity, we now have

The last equality holds simply by telescoping cancellation.

For the downwards MCT, we can suppose without loss of generality that \(A_1\) has finite measure. We thus take complements inside \(A_1\) to obtain the sequence \(B_n=A_1\smallsetminus A_n\). Then the \(B_n\) form an increasing sequence and \(\bigcup B_n=A_1\smallsetminus \bigcap A_n\). Using this and the upwards MCT, we now have

Cancelling the \(m(A_1)\) from the first and last expression, we obtain that \(0=m(\bigcap A_n)-\lim m(A_n)\), which implies the desired result.

Give a counterexample showing that the hypothesis that some \(A_n\) has finite measure is necessary for the downwards MCT.

Suppose you know that the domain of \(m\) is a \(\sigma \)-algebra, and \(m\) satisfies \(m(\emptyset )=0\) and the countable additivity property. Show that \(m\) satisfies the monotonicity property and the countable subadditivity property.

Let us say that a sequence of sets \(A_n\) converges to \(A\) if the characteristic functions \(\chi _{A_n}\) converge pointwise to \(\chi _A\).

(a)

Show that if \(A_n\) are Lebesgue measurable and \(A_n\) converges to \(A\) then \(A\) is Lebesgue measurable. [Hint: Show that if \(A_n\) converges to \(A\) then \(A=\bigcup _n\bigcap _{m>n}A_m\) and also \(A=\bigcap _n\bigcup _{m>n}A_m\).]

(b)

Suppose that if \(A_n\) are all contained in a set of finite measure and \(A_n\) converges to \(A\), then \(m(A_n)\to m(A)\). This is an example of the dominated convergence theorem.

(c)

Give a counterexample showing that the hypothesis that \(A_n\) are all contained in a set of finite measure cannot be replaced with the hypothesis that the values \(m(A_n)\) are bounded.

Tao, §1.3 introduction

A function \(f\) mapping \({\mathbb{R}}^d\) into the extended real numbers \([0,\infty ]\) (or sometimes into \({\mathbb{C}}\)) is called simple if there exists a partition of \({\mathbb{R}}^d\) into finitely many Lebesgue measurable subsets \(A_1,\ldots ,A_k\) such that \(f\) takes a constant value \(c_i\) on each \(A_i\).

If \(f=\sum _1^kc_i\chi _{A_i}\) is a simple function and \(f\geq 0\), then the simple integral of \(f\) is defined to be \({\text{s}}\kern -3pt\int f=\sum _1^kc_im(A_i)\).

If \(f=\sum _1^lc_i\chi _{A_i}\) and \(f=\sum _1^md_j\chi _{B_j}\) then we have \(\sum _1^l c_i m(A_i)=\sum _1^m d_j m(B_j)\).

We use a common refinement approach. By considering intersections of all \(k+l\) sets, we can find a sequence of nonempty disjoint sets \(C_1,\ldots ,C_n\) such that each of the \(A_i\) and \(B_j\) can be written as a union of some of the \(C_k\)’s. Note that since the \(A_i\) and \(B_j\) are measurable, the \(C_k\) are measurable too. Now let \(x_k\) be an arbitrary point in \(A_k\). We can calculate

This is what we desired.

Let \(f,g\) be simple functions.

(a)

(equivalence) if \(f=g\) almost everywhere then \({\text{s}}\kern -3pt\int f={\text{s}}\kern -3pt\int g\)

(b)

(monotonicity) if \(f\leq g\) almost everywhere then \({\text{s}}\kern -3pt\int f\leq {\text{s}}\kern -3pt\int g\).

(c)

(linearity) \({\text{s}}\kern -3pt\int (cf)=c\cdot {\text{s}}\kern -3pt\int f\), and \({\text{s}}\kern -3pt\int (f+g)={\text{s}}\kern -3pt\int f+{\text{s}}\kern -3pt\int g\).

We prove the first property (i), since after that properties (ii) and (iii) are very similar to the analogous properties of the Riemann integral. Given simple functions \(f\) and \(g\), we can refine their expressions to find measurable sets \(A_1,\ldots ,A_n\) and a null set \(N\) such that \(f(x)=\sum c_i\chi _{A_i}(x)\) for all \(x\notin N\) and \(g(x)=\sum c_i\chi _{A_i}(x)\) for all \(x\notin N\). Then clearly the simple integral of both \(f\) and \(g\) evaluates to \(\sum c_im(A_i)+0\).

Show that a function \(f\) is simple if and only if it can be expressed as \(f=\sum _1^kc_i\chi _{A_i}\), where \(A_i\) are (not necessarily disjoint) Lebesgue measurable sets.

Show that the simple integral satisfies the properties:

(a)

(finiteness) \({\text{s}}\kern -3pt\int f<\infty \) if and only if \(f\) is finite almost everywhere and supported on a set of finite measure

(b)

(vanishing) \({\text{s}}\kern -3pt\int f=0\) if and only if \(f=0\) almost everywhere

(c)

(normality) \({\text{s}}\kern -3pt\int \chi _A=m(A)\) for any Lebesgue measurable set \(A\)

Tao, §1.3.2

A nonnegative function \(f\) on \({\mathbb{R}}^n\) is said to be a measurable function if \(f\) is the pointwise limit of nonnegative simple functions.

A nonnegative function \(f\) is measurable if and only if either of the following holds.

(a)

there is a sequence \(f_n\) of simple functions such that the \(f_n\) are bounded and have bounded support, the \(f_n\) are increasing \(f_n\leq f_{n+1}\), and \(f=\sup f_n\);

(b)

for any open set \(S\) (respectively: closed set, interval, ray, etc) the preimage \(f^{-1}(S)\) is Lebesgue measurable.

We first show that if \(f\) is measurable, then (b) holds. So let \(f_n\) be simple functions such that \(f=\lim f_n\) pointwise. Note that by the above discussion, we have that \(f=\limsup f_n\) pointwise. Now suppose that \(S=(\lambda ,\infty )\) is an open ray. Then we want to say that

Since the \(f_n\) are simple, it is clear that the set in the last line is measurable and therefore \(f^{-1}(S)\) would be measurable. However the argument isn’t right, since for instance it isn’t quite true that \((\forall n)z_n>\lambda \) implies \(\inf _n z_n>\lambda \). The correct calculation introduces a couple additional steps but ultimately accomplishes the same thing:

Once again, this establishes that \(f^{-1}(S)\) is measurable. Now an analogous argument will allow us to handle the case when \(S=(-\infty ,\mu )\). Since any open interval is an intersection of two open rays, and any open set is a countable union of intervals, we can conclude that for any open \(S\) the set \(f^{-1}(S)\) is measurable. This establishes (b).

Next we argue that (b) implies (a). Suppose that \(f\) satisfies condition (b). Given any \(n\), we will define a nonnegative simple function \(f_n\leq f\) as follows:

The above prescription clearly ensures that \(f_n\leq f\), that \(f_n\) is bounded above by \(n\), and that \(f_n\) has bounded support \([-n,n]^d\). Moreover it is not difficult to check that \(f_n\leq f_{n+1}\) and that \(f_n\to f\). It remains only to show that \(f_n\) is simple, and since \(f_n\) clearly takes just finitely many values we really only have to check that it takes each of its values on a measurable set. For this, for example we have \(f_n(x)=\frac{i}{2^n}\) if and only if \(f(x)\) lies in the interval \([\frac{i}{2^n},\frac{i+1}{2^n})\). It follows from property (b) that \(f_n^{-1}(\frac{i}{2^n})\) is a measurable set, and therefore we have that \(f_n\) is simple and \(f\) satisfies property (a).

Finally it is trivial that (a) implies \(f\) is measurable, so we have completed the proof.

If \(f\) is an almost-everywhere defined complex-valued function on \({\mathbb{R}}^n\) then \(f\) is a measurable function if and only if the positive and negative parts of its real and imaginary parts are measurable functions.

If \(f\) is a complex-valued function on \({\mathbb{R}}^d\) then \(f\) is measurable if and only if \(f\) is a pointwise limit of complex-valued simple functions.

Show that \(\lim x_n=x\) if and only if \(\liminf x_n=x=\limsup x_n\).

…

Show that if \(f\) is a bounded, nonnegative measurable function on \({\mathbb{R}}^d\), then there is a sequence of bounded simple functions \(f_n\) which converges uniformly to \(f\) (not just pointwise).

Let \(f\) be a nonnegative function on \({\mathbb{R}}^d\). Show that \(f\) is simple if and only if \(f\) is measurable and takes on at most finitely many values.

If \(f\) is a nonnegative measurable function, show that the region under \(f\) is a measurable set.

Tao, §1.3.3

Let \(f\) be a nonnegative function on \({\mathbb{R}}^d\). We define the lower Lebesgue integral of \(f\) by

If \(f\) is measurable, we define the Lebesgue integral of \(f\) to be \(\int f=\underline{\int }f\).

The lower Lebesgue integral satisfies the properties:

(a)

(equivalence) if \(f=g\) almost everywhere then \(\underline{\int }f=\underline{\int }g\);

(b)

(monotonicity) if \(f\leq g\) almost everywhere then \(\underline{\int }f\leq \underline{\int }g\); and

(c)

(superadditivity) \(\underline{\int }(f+g)\geq \underline{\int }f+\underline{\int }g\).

The equivalence and monotonicity properties are clear from the analogous properties of simple integrals. For superadditivity, let \(\epsilon >0\) be given and find simple functions \(h\) and \(k\) such that \(h\leq f\), \(k\leq g\), \(\underline{\int }f-{\text{s}}\kern -3pt\int h<\epsilon \), and \(\underline{\int }g-{\text{s}}\kern -3pt\int k<\epsilon \). Then we have \(h+k\leq f+g\), and using monotinicity plus additivity for simple integrals:

Letting \(\epsilon \to 0\), we obtain the desired result.

Let \(f\) be a nonnegative function on \({\mathbb{R}}^d\). The lower Lebesgue integral satisfies the following identities.

(a)

(range truncation) If \(f^N=\min (f,N)\) then \(\underline{\int }f^N\to \underline{\int }f\).

(b)

(support trunctation) If \(f_N=f\chi _{[-N,N]^d}\) then \(\underline{\int }f_N\to \underline{\int }f\).

(a) Let us first assume that \(\underline{\int }f<\infty \). Given \(\epsilon >0\) we can find a simple function \(g\) such that \(g\leq f\) and \(\underline{\int }f-{\text{s}}\kern -3pt\int g<\epsilon \). By our assumption \(g\) must be bounded almost everywhere, which implies that for \(N\) large enough we have \(g\leq f^N\) too. Now by monotonicity \(\underline{\int }f-\underline{\int }f^N\leq \underline{\int }f-{\text{s}}\kern -3pt\int g<\epsilon \), which shows the desired result. The argument is similar in the case \(\underline{\int }f=\infty \).

(b) Again let \(\epsilon \) be given and find a simple function \(g\) such that \(g\leq f\) and \(\underline{\int }f-{\text{s}}\kern -3pt\int g<\epsilon \). Write \(g=\sum _1^kc_i\chi _{A_i}\). Now we look at the simple integral of \(g_N\):

Where here \(N\to \infty \) and we are applying the upwards monotone convergence theorem. Thus we can find \(N\) large enough that \({\text{s}}\kern -3pt\int g-{\text{s}}\kern -3pt\int g_N<\epsilon \). Again using monotonicity we conclude that \(\underline{\int }f-\underline{\int }f_N\leq \underline{\int }f-{\text{s}}\kern -3pt\int g_N<2\epsilon \).

If \(f,g\) are nonnegative measurable functions, then \(\int (f+g)=\int f+\int g\).

First suppose that \(f,g\) are bounded functions with bounded supports. For such functions, it is useful to define the upper Lebesgue integral in the obvious way:

We claim that under our hypotheses, we in fact have \(\underline{\int }f=\overline{\int }f\) (and similarly for \(g\) and \(f+g\)).

To see this recall that since \(f\) is measurable, we can find simple functions \(f_n\) such that \(f_n\leq f_{n+1}\) and \(f_n\to f\). Note also that since \(f\) is bounded, the construction of the \(f_n\) from Theorem 10.2 in fact showed that the \(f_n\) converge uniformly to \(f\). Thus given \(\epsilon >0\) we can find \(n\) such that

where \(S\) is a support for \(f\). Taking \(\underline{\int }\) of the first inequality and \(\overline{\int }\) of the second, we obtain

Letting \(\epsilon \to 0\) we obtain the desired result.

Now we have already shown that \(\underline{\int }\) satisfies the superadditivity property. Using a parallel argument, it is easy to show that \(\overline{\int }\) satisfies the analogous subadditivity property. And since \(\underline{\int }=\overline{\int }\) on the functions \(f\), \(g\), and \(f+g\), we can put the two together to conclude the additivity property!

Finally for general measurable functions \(f\) and \(g\), we can always apply the truncation lemmas to replace \(f\) and \(g\) with bounded functions with bounded supports. Each truncation costs us an \(\epsilon \) in the additivity property, but afterwards we can let \(\epsilon \to 0\) and obtain the desired result.

Let \(f\) be a nonnegative measurable function on \({\mathbb{R}}\). Show that \(\int f\) is equal to the \(2\)-dimensional Lebesgue measure of the region

Let \(f\) be an nonnegative measurable function on \({\mathbb{R}}^d\).

(a)

Show that if \(\int f<\infty \) then \(f\) is finite almost everywhere. Give a counterexample to show that the converse is false.

(b)

Show that \(\int f=0\) if and only if \(f=0\) almost everywhere.

Give an example of a nonnegative function which is measurable, but has different lower and upper Lebesgue integrals.

Tao, §1.3.4

Let \(f\) be a complex-valued measurable function on \({\mathbb{R}}^d\) (it need only be defined almost everywhere). We say that \(f\) is absolutely integrable, or a member of \(L^1\), if \(\int |f|<\infty \).

If \(f\) is absolutely integrable and real-valued, let \(\int f=\int f^+-\int f^-\). If \(f\) is absolutely integrable and complex-valued, let \(\int f=\int \Re f+i\int \Im f\).

Let \(f\colon [a,b]\to {\mathbb{R}}\) be a Riemann integrable function. Then intepreting \(f\) as a function defined on all of \({\mathbb{R}}\) which is zero outside of \([a,b]\), we have that \(f\) is Lebesgue absolutely integrable with the same value.

Let us first assume that \(f\) is nonnegative. Then since pc functions are simple, we clearly have that the lower Darboux integral of \(f\) is less than or equal to the lower Lebesgue integral of \(f\). On the other hand by monotonicity the lower Lebesgue integral of \(f\) is less than or equal to the upper Darboux integral of \(f\). Since the lower and upper Darboux integrals are equal, we must have that it agrees with the Lebesgue integral of \(f\).

If \(f\) is real-valued, then we can write \(\int f=\int f^+-\int f^-\), and this expression is valid for both the Darboux and absolutely convergent Lebesgue integrals. Applying the previous argument to both \(f^+\) and \(f^-\), we have the desired result.

(a)

(linearity and conjugation) \(\int (f+g)=\int f+\int g\), \(\int cf=c\int f\), and \(\int \bar{f}=\overline{\int f}\);

(b)

(triangle inequality for integrals) \(\left|\int f\right|\leq \int |f|\).

We should first check that if \(f,g\) are absolutely integrable, then \(f+g\) and \(cf\) are absolutely integrable. For the first we can simply use the classical triangle inequality \(|f+g|\leq |f|+|g|\). Then monotonicity implies that \(\int |f+g|\leq \int |f|+\int |g|\) is finite. For the second, simply note that \(\int |cf|=\int |c||f|=|c|\int |f|\) is again finite.

Now linearity is easily proved using the analogous property for the nonnegative integral. For example, if \(f,g\) are real-valued and \(h=f+g\), then \(h^+-h^-=f^+-f^-+g^+-g^-\). Rearranging terms we have \(f^-+g^-+h^+=f^++g^++h^-\). From the nonnegative linearity we know \(\int f^-+\int g^-+\int h^+=\int f^++\int g^++\int h^-\). Rearranging back, we obtain \(\int h=\int f+\int g\).

For the triangle inequality for integrals, first assume that \(f\) is real-valued. Then we can write \(f=f^+-f^-\) and \(|f|=f^++f^-\). Thus using linearity together with the triangle inequality, we have

Next if \(f\) is complex-valued we can find an angle \(\theta \) such that \(e^{i\theta }\int f=\left|\int f\right|\). Again using linearity, together with the definition of the complex-valued integral, we have

The last inequality following from monotonicity, and gives the desired result.

The collection of absolutely integrable functions forms a vector space. In fact it is a seminormed vector space with the seminorm \(\|f\|=\int |f|\).

We argued in the previous proof that \(\int |f+g|\leq \int |f|+\int |g|\) and also that \(\int |cf|=|c|\int |f|\). These two identities imply that the space of absolutely integrable functions is closed under linear combinations and moreover that the two properties of the seminorm hold.

The following are all dense subsets of the space of absolutely integrable functions.

(a)

absolutely integrable simple functions;

(b)

absolutely integrable simple functions \(\sum _1^kc_i\chi _{B_i}\) where \(B_i\) are all boxes; and

(c)

continuous, compactly supported functions.

(a) First assume that \(f\) is nonnegative. Then by the definition of the integral, we can find a simple function \(g\) such that \(g\leq f\) and \(\int f-\int g<\epsilon \). It follows that \(\int (f-g)<\epsilon \), and since \(f-g\) is nonnegative, clearly \(\int |f-g|<\epsilon \). It is easy to extend this argument to complex-valued functions using the standard technique.

(b) We now know from (a) that it is sufficient to approximate any simple function by a function of this type. Using linearity, it is enough to approximate a single term \(\chi _A\), with \(m(A)<\infty \), by a function of this type. We have already seen that for any such \(A\) there exists an elementary set \(E\) such that \(m(A\bigtriangleup E)<\epsilon \). This means that \(\int |\chi _A-\chi _E|<\epsilon \), so the result follows.

(c) We now know from (b) that it is sufficient to approximate any \(\chi _B\), \(B\) a box, by a function of this type. It is possible to do this explicity. For example in one dimension we have \(B=I\) is an interval, and the step function \(I\) can easily be approximated by a continuous function which looks like a trapezoid.

Let \(f\) be absolutely integrable. Show that for any \(\epsilon >0\) there exists a bounded measurable set \(A\) such that \(\int |f|\chi _{A^c}<\epsilon \).

Let \(f\) be a nonnegative measurable function, and assume \(f\) is finite almost everywhere. Show that for any \(\epsilon >0\), there exists a measurable set \(A\) such that \(m(A)<\epsilon \) and \(f\) is locally bounded outside of \(A\). In other words, for every \(n\) there exists \(M\) such that for all \(x\in [-n,n]^d\smallsetminus A\) we have \(f(x)\leq M\).

Show that the space of absolutely integrable functions is separable, that is, has a countable dense subset.

Tao, §1.4.5, though assume all functions are defined on \({\mathbb{R}}^d\).

Let \(f_n=\chi _{[n,n+1]}\) and \(f=0\). That is, \(f_n\) is a sequence of moving unit bumps. Then \(f_n\to f\) pointwise (not uniformly), but we have \(\int f_n=1\) for all \(n\), and \(\int f=0\).

Let \(f_n=\frac{1}{n}\chi _{[0,n]}\) and \(f=0\). That is, \(f_n\) is a sequence of widening and shortening bumps. Then \(f_n\to f\) uniformly, but once again we have \(\int f_n=1\) for all \(n\), and \(\int f=0\).

Let \(f_n=n\chi _{[1/n,2/n]}\) and \(f=0\). That is, \(f_n\) is a sequence of narrowing and tallening bumps. Then \(f_n\to f\) pointwise (not uniformly), but once again we have \(\int f_n=1\) for all \(n\), and \(\int f=0\).

Let \(f_n\) be nonnegative measurable functions. Then

In particular, if \(f_n\to f\) then we have \(\int f\leq \liminf \int f_n\).

Let \(f_n\) be a sequence of measurable complex-valued functions and suppose that \(f_n\to f\). Suppose that there exists a nonnegative function \(G\) such that \(\int |G|<\infty \) and for all \(n\), we have \(|f_n|\leq G\). Then \(\int f_n\to \int f\).

By separating \(f_n\) and \(f\) into their real and imaginary parts, we may assume that they are all real-valued. Thus our hypothesis says that \(-G\leq f_n\leq G\). Since \(f_n\to f\) we also have that \(-G\leq f\leq G\). Now \(f_n+G\) is nonnegative, so we can use Fatou’s lemma to obtain:

Similarly \(G-f_n\) is nonnegative so we can again use Fatou’s lemma to obtain:

Putting the two equations together, we conclude that

It follows that \(\int f=\lim \int f_n\).

Let \(f_n\) be a sequence of nonnegative measurable functions, and suppose that \(f_n\leq f_{n+1}\) for all \(n\). Let \(f=\sup f_n\), so that we automatically have \(f_n\to f\). Then \(\int f_n\to \int f\).

We first observe that if the \(f_n\) were all characteristic functions, then this theorem would follow directly from the upwards monotone convergence theorem for Lebesgue measurable sets. Thus our strategy is to reduce to a situation in which we can apply the upwards monotone convergence theorem.

First, since \(f_n\leq f_{n+1}\), by the montonicity of the integral we know that \(\int f_n\leq \int f_{n+1}\). Similarly since \(f_n\leq f\) we know that \(\int f_n\leq \int f\). It follows that the values \(\int f_n\) converge and that \(\lim \int f_n\leq \int f\).

To show that \(\int f\leq \lim \int f_n\), it is sufficient to show that \(\int g\leq \lim \int f_n\) for any simple function \(g\) such that \(g\leq f\). Given such a \(g\), we can express it as \(g=\sum c_i\chi _{A_i}\) where the \(A_i\) are disjoint measurable sets. Using the range truncation lemma, we can suppose that \(c_i\neq \infty \) for all \(i\).

Now fix just one of the sets \(A_i\) and let \(\epsilon >0\) be given. We define the sets

About this defintion, we first note that it immediately implies \(f_n>\sum (1-\epsilon )c_i\chi _{A_{i,n}}\). We second note that for all \(x\in A_i\), for \(n\) large enough, we have \(f_n(x)\geq (1-\epsilon )g(x)=(1-\epsilon )c_i\). This means that \(A_i\) is the union of the \(A_{i,n}\). Thus the upwards monotone convergence theorem for sets implies that \(m(A_{i,n})\to m(A_i)\) as \(n\to \infty \).

Putting this all together, we have

Taking \(\epsilon \to 0\) we therefore conclude that \(\lim \int f_n\geq \int g\), as desired.

If \(f_n\) is a sequence of nonnegative measurable functions, then \(\int \sum f_n=\sum \int f_n\).

Using the monotone convergence theorem and then finite linearity, we have:

as desired.

We will need the general fact that \(\int \inf g_n\leq \inf \int g_n\). This holds simply because \(\int \inf g_n\leq \int g_n\) for any particular \(n\), and then one can take the \(\inf _n\) over both sides.

Now recall that \(\liminf f_n=\lim _N\inf _{n\geq N}f_n\). The functions \(\inf _{n\geq N}f_n\) are increasing in \(N\), so using the monotone convergence theorem together with the above we have

as desired.

(a)

Let \(A_n\) be measurable sets and assume that \(\sum m(A_n)<\infty \). Show that almost every \(x\in {\mathbb{R}}^d\) is contained in at most finitely many of the \(A_n\). [Hint: Use Tonelli’s theorem on the functions \(\chi _{A_n}\).] This is the Borel–Cantelli lemma.

(b)

Give a counterexample to the above conclusion, showing that the hypothesis \(\sum m(A_n)<\infty \) cannot be replaced by the weaker condition \(\lim m(A_n)=0\).

Use the dominated convergence theorem to show that the harmonic series \(\sum \frac{1}{n}\) diverges. [Hint: Let \(f_n=\frac{1}{n}\chi _{[0,n]}\), show that \(\sum \frac{1}{n}<\infty \) plus the dominated convergence theorem implies \(\int f_n\to 0\), and obtain a contradiction from this.]

Tao, §1.4.1–1.4.3

A measurable space is a pair \((X,\mathcal{B})\) where \(X\) is any set and \(\mathcal{B}\) is a Boolean \(\sigma \)-algebra on \(X\): a collection of subsets of \(X\) which contains the sets \(\emptyset \) and \(X\), and is closed under countable intersections, countable unions, and complements.

Suppose \(X\) is a set and \(\mathcal{B}\) is a \(\sigma \)-algebra on \(X\). A function \(\mu \colon \mathcal{B}\to [0,\infty ]\) is said to be a measure if it satisfies

(a)

(empty set) \(\mu (\emptyset )=0\); and

(b)

(countable additivity) for every sequence of pairwise disjoint sets \(A_n\in \mathcal{B}\), we have \(\mu (\bigcup A_n)=\sum (\mu (A_n))\).

Suppose that \(\mathcal{B}\) is a \(\sigma \)-algebra on \(X\), and \(\mu \) is a measure on \(\mathcal{B}\). Let \(A,B\), and \(A_n\) denote elements of \(\mathcal{B}\). Then we have:

\(\circ \)

(monotonicity) if \(A\subset B\) then \(\mu (A)\leq \mu (B)\);

\(\circ \)

(inclusion–exclusion) \(\mu (A\cup B)+\mu (A\cap B)=\mu (A)+\mu (B)\).

\(\circ \)

(countable subadditivity) \(\mu (\bigcup A_n)\leq \sum \mu (A_n)\);

\(\circ \)

(upwards monotone convergence) if \(A_{n+1}\supset A_n\) then \(\mu (\bigcup A_n)=\lim \mu (A_n)\); and

\(\circ \)

(downwards monotone convergence) if \(A_{n+1}\subset A_n\) and \(\mu (A_1)<\infty \) then \(\mu (\bigcap A_n)=\lim \mu (A_n)\).

A Boolean algebra on \(X\) is a collection of subsets of \(X\) which contains the sets \(\emptyset \), and \(X\), and is closed under pairwise intersections, pairwise unions, and complements.

Suppose \(X\) is a set and \(\mathcal{A}\) is a Boolean algebra on \(X\). A function \(\mu \colon \mathcal{A}\to [0,\infty ]\) is said to be a finitely additive measure if it satisfies

(a)

(empty set) \(\mu (\emptyset )=0\); and

(b)

(finite additivity) for all disjoint sets \(A,B\in \mathcal{A}\), we have \(\mu (A\cup B)=\mu (A)+\mu (B)\).

Let \(X\) be any set, \(\mathcal{A}\) a Boolean algebra on \(X\), and \(\mu _0\) a finitely additive measure on \(\mathcal{A}\). Then \(\mu _0\) is said to be a premeasure if it satisfies the additional axiom:

\(\circ \)

for every sequence of pairwise disjoint sets \(A_n\in \mathcal{A}\), if \(\bigcup A_n\in \mathcal{A}\), then \(\mu _0(\bigcup A_n)=\sum \mu (A_n)\).

Let \(X\) be a set, \(\mathcal{A}\) an algebra on \(X\), and \(\mu _0\) a premeasure on \(\mathcal{A}\). Let \(\mathcal{B}\) be the \(\sigma \)-algebra generated by \(\mathcal{A}\). Then \(\mu _0\) extends to a measure \(\mu \) on \(\mathcal{B}\).

Let \(X\) be any set and \(\mu ^*\) a function on all subsets of \(X\). Then \(\mu ^*\) is said to be a outer measure if it satisfies

(a)

(empty set) \(\mu ^*(\emptyset )=0\);

(b)

(monotonicity) if \(A\subset B\) then \(\mu ^*(A)\leq \mu ^*(B)\); and

(c)

(countable subadditivity) \(\mu ^*(\bigcup A_n)\leq \sum \mu ^*(A_n)\).

If \(\mu ^*\) is constructed as in Equation (14.e1), then \(\mu ^*\) is an outer measure.

The empty set and monotonicity axioms are clear from the definition. For countable subadditivity, let \(E_i\) be given and \(\epsilon >0\). Then for every \(i\) there exists a sequence of \(A_i^n\) such that \(E_i\subset A_i^n\) and \(\sum _n\mu _0(A_i^n)-\mu ^*(E_i)<\epsilon /2^i\). It follows that \(\bigcup E_i\subset \bigcup _{i,n} A_i^n\) and \(\sum _{n.i}\mu (A_i^n)-\sum \mu ^*(E_i)\leq \epsilon \). Therefore we have \(\mu ^*(\bigcup E_i)-\sum (\mu ^*(E_i))\leq \epsilon \). Taking \(\epsilon \to 0\) the proof is complete.

Let \(\mu \) be a measure on \((X,\mathcal{B})\). Show that \(\mathcal{B}\) can be extended to \(\sigma \)-algebra \(\hat{\mathcal{B}}\) and \(\mu \) to a measure \(\hat{\mu }\) on \((X,\hat{\mathcal{B}})\) in such a way that \(\hat{\mu }\) is complete.

Let \(f\) be a nonnegative Lebesgue measurable function. Show that \(\mu (A)=\int f\chi _A\) is a measure.

Give an example of a finitely additive measure that is not a premeasure. [Hint: work on the measurable space \(({\mathbb{N}},\mathcal{P}({\mathbb{N}}))\) and define \(\mu _0\) separately for finite and infinite sets.]

Tao, §1.7

Suppose that \(X\) is any set and \(\mu ^*\) is an outer measure on \(X\). Then a subset \(A\subset X\) is said to be \(\mu ^*\)-measurable if for every subset \(S\subset X\) we have

If \(\mu ^*\) is an outer measure on \(X\), then the \(\mu ^*\)-measurable subsets of \(X\) form a \(\sigma \)-algebra, and \(\mu ^*\) is a measure when restricted to the \(\mu ^*\)-measurable sets.

The first part of the proof will be to show that the collection of \(\mu ^*\)-measurable sets is a Boolean algebra and \(\mu ^*\) is finitely additive on the \(\mu ^*\)-measurable sets. To begin, it is clear that a set \(A\) is \(\mu ^*\)-measurable if and only if \(A^c\) is \(\mu ^*\)-measurable.

We now show that the \(\mu ^*\)-measurable sets are closed under pairwise unions. Suppose that \(A,B\) are \(\mu ^*\)-measurable. Since we have already proved \(\mu ^*\) is subadditive for arbitrary sets, it suffices to prove that for any set \(S\) we have

To achieve this, we expand the right-hand side and then apply the measurability of \(B\) followed by the measurability of \(A\):

Next we show that \(\mu ^*\) is finitely additive on the \(\mu ^*\)-measurable sets. Suppose that \(A,B\) are disjoint and \(\mu ^*\)-measurable. Then using the measurability of \(A\), we have

The second part of the proof is to show that the \(\mu ^*\)-measurable sets form a \(\sigma \)-algebra and that \(\mu ^*\) is countably additive on the \(\mu ^*\)-measurable sets. Let \(A_n\) be a sequence of \(\mu ^*\)-measurable sets. Since the \(\mu ^*\)-measurable sets form an algebra, we can assume without loss of generality that the \(A_n\) are pairwise disjoint. Then using the argument of the previous paragraph, with \(S\cap \) inserted and then induction, for any set \(S\) we have \(\mu ^*(S\cap \bigcup _1^N A_n)=\sum _1^N\mu ^*(S\cap A_n)\). It follows that

Taking \(N\to \infty \), we obtain

Thus we have shown that \(\bigcup A_n\) is \(\mu ^*\)-measurable.

Finally if we take \(S=\bigcup A_n\) in the above, we obtain that \(\mu ^*(\bigcup A_n)\geq \sum \mu ^*(A_n)\), which shows that \(\mu ^*\) is countably additive as desired.

Let \(\mathcal{A}\) be a Boolean algebra on \(X\), and \(\mu _0\) a premeasure on \(\mathcal{A}\). Let \(\mu ^*\) be the outer measure induced by \(\mu _0\) according to Equation (14.e1) and \(\mathcal{B}\) the \(\sigma \)-algebra of \(\mu ^*\)-measurable sets. Then \(\mathcal{B}\) is an extension of \(\mathcal{A}\), and \(\mu ^*\) is an extension of \(\mu _0\).

We first show that if \(A\) lies in \(\mathcal{A}\) then \(A\) is \(\mu ^*\)-measurable. For this, let \(S\) be an arbitrary set and find sets \(B_n\in \mathcal{A}\) such that \(S\subset \bigcup B_n\) and \(\sum \mu _0(B_n)-\mu ^*(S)<\epsilon \). Then

Taking \(\epsilon \to 0\), we conclude that \(A\) is \(\mu ^*\)-measurable.

Next we show that if \(A\) lies in \(\mathcal{A}\) then \(\mu ^*(A)=\mu _0(A)\). It is clear from the definition that \(\mu ^*(A)\leq \mu _0(A)\). To show that \(\mu _0(A)\leq \mu ^*(A)\), let \(B_n\in \mathcal{A}\) be such that \(A\subset \bigcup B_n\). We can assume without loss of generality that the \(B_n\) are pairwise disjoint. Then using the axiom of a premeasure, we obtain

Taking the infemum over all such sequences \(B_n\), we conclude that \(\mu _0(A)\leq \mu ^*(A)\), as desired.

An interval is said to be half-open if it is of the form \((a,b]\) where \(a\in [-\infty ,\infty )\) and \(b\in {\mathbb{R}}\), or else of the form \((a,\infty )\). We let \(\mathcal{H}\) denote the Boolean algebra generated by the half-open intervals.

For any increasing, right-continuous function \(F\), \(\mu _F\) is a premeasure on \(\mathcal{H}\).

It is not difficult to see that \(\mu _F\) is well-defined on \(\mathcal{H}\). Moreover \(\mu _F\) is finitely additive by construction.

Thus it remains only to show that \(\mu _F\) satisfies the axiom of a premeasure. Suppose that \(A_n\in \mathcal{H}\) and that \(\bigcup A_n\in \mathcal{H}\) too. We can assume without loss of generality that \(A_n\) are each single half-open intervals, and that \(\bigcup A_n\) is a single half-open interval \(I\).

It is clear from monotonicity that \(\mu _F(\bigcup A_n)\geq \sum \mu _F(A_n)\), so it suffices to show that \(\mu _F(\bigcup A_n)\leq \sum \mu _F(A_n)\). In order to achieve this, let us define \(\mu _F\) on open intervals by letting \(\mu _F((a,b))=\lim _{x\to b^-}F(x)-F(a)\), and similarly for closed intervals. Then using the right-continuity of \(F\), we can pay \(\epsilon \) to replace each \(A_n\) with an open \(U_n\) such that \(A_n\subset U_n\). Similarly we can pay \(\epsilon \) to replace \(I\) with a closed (bounded) interval \(K\) such that \(K\subset I\).

Now by the Heine–Borel theorem, just finitely many of the \(U_n\) suffice to cover \(K\), and hence \(\mu _F(K)\leq \sum \mu _F(U_n)\). It follows that

Thus the result follows by taking \(\epsilon \to 0\).

Let \(\mu _0\) be a premeasure on \((X,\mathcal{A})\) and let \(\mu \) on \((X,\mathcal{B})\) be the Carathéodory extension. Show that for any \(B\in \mathcal{B}\) [such that \(\mu (B)<\infty \)?] there exists \(C\) in the \(\sigma \)-algebra generated by \(\mathcal{A}\) such that \(B\subset C\) and \(\mu (C\smallsetminus B)=0\).

Let \(\mu _0\) be a premeasure on \((X,\mathcal{A})\), \(\mu ^*\) the corresponding outer measure, and \(\mu \) on \((X,\mathcal{B})\) the Carathéodory extension. Show that for any set \(B\) such that \(\mu ^*(B)<\infty \), we have \(B\in \mathcal{B}\) if and only if for all \(\epsilon >0\) there exists \(A\in \mathcal{A}\) such that \(\mu ^*(A\bigtriangleup B)<\epsilon \).

Suppose \(F\) is a monotone and non-decreasing function. Show that \(F\) is continuous if and only if \(\mu _F(\{x\})=0\) for every \(x\in {\mathbb{R}}\).

Tao, §1.4.4

Suppose that \((X,\mathcal{B})\) is a measurable space. A function \(f\) which maps \(X\) to \([0,\infty ]\), \({\mathbb{R}}\), or \({\mathbb{C}}\) is called measurable if the pre-image of any open set is in \(\mathcal{B}\), that is, for every open subset \(U\) of the codomain, \(f^{-1}(U)\in \mathcal{B}\).

The measurable functions on \((X,\mathcal{B})\) are closed under sums, products, continuous compositions, and pointwise limits.

We address only the last point, that measurable functions are closed under pointwise limits. For this we refer to the method of proof of Theorem 10.2. The idea is that if \(f\) is the limit of the functions \(f_n\), then it is also the \(\limsup \) of the functions \(f_n\). This allows one to write \(f^{-1}(U)\) as a countable Boolean combination of the sets \(f_n^{-1}(U)\).

Let \((X,\mathcal{B})\) be a measurable space. A function \(f\colon X\to [0,\infty ]\) is said to be simple if it is measurable and takes finitely many values, that is, if \(f=\sum _1^kc_i\chi _{A_i}\) where \(A_i\in \mathcal{B}\).

If \(f\) is the simple function above, and \(\mu \) is a measure on \((X,\mathcal{B})\), then we define the simple integral of \(f\) with respect to \(\mu \) by

Suppose \((X,\mathcal{B})\) is a measurable space, and \(\mu \) is a measure on it. If \(f\colon X\to [0,\infty ]\) is measurable, then we define the integral of \(f\) with respect to \(\mu \) by

The integral of nonnegative measurable functions satisfies the following properties

(a)

(equivalence) if \(f=g\) almost everywhere then \(\int f\,d\mu =\int g\,d\mu \);

(b)

(monotonicity) if \(f\leq g\) almost everywhere then \(\int f\,d\mu \leq \int g\,d\mu \);

(c)

(range truncation) if \(f^N=\min (f,N)\) then \(\int f^N\,d\mu \to \int f\,d\mu \); and

(d)

(support truncation) if \(A_n\) is a sequence of measurable sets such that \(A_n\subset A_{n+1}\) and \(\bigcup A_n=X\), then \(\int f\chi _{A_n}\,d\mu \to \int f\,d\mu \).

(e)

(Markov’s inequality) if \(f\) is measurable, then \(\mu (\{x\mid f(x)\geq \lambda \})\leq \frac{1}{\lambda }\int f\,d\mu \);

(f)

(additivity) \(\int (f+g)\,d\mu =\int f\,d\mu +\int g\,d\mu \);

All of the properties (a)–(d) can be proved in a fashion similar to the case of the Lebesgue integral.

The same is true of property (e), but we neglected to state it earlier so let us provide the proof now. Let \(g=\lambda \chi _{\{x\mid f(x)\geq \lambda \}}\). Then clearly \(g\) is a simple function and \(g\leq f\). It follows that the simple integral of \(g\) is a lower bound for the integral of \(f\), so \(\lambda \mu (\{x\mid f(x)\geq \lambda \})\leq \int f\,d\mu \), as desired.

The proof of additivity is similar to the Lebesgue case, with one wrinkle. Recall that in the Lebesgue proof we applied the truncation lemma to the sets \([-N,N]^d\) to assume without loss of generality that \(f,g\) have finite measure support. In general we cannot assume that \(X\) is a union of countably many sets of finite measure. Instead we let \(A_n=\{x\mid f(x)>\frac{1}{n}{\text{ or }}g(x)>\frac{1}{n}\}\). This family of sets is increasing and the union contains the supports of \(f\) and \(g\). By Markov’s inequality, the sets \(A_n\) have finite measure, and the proof can now proceed by truncating to \(A_n\).

Suppose that \((X,\mathcal{B})\) is a measurable space and \(\mu \) is a measure on it. If \(f\colon X\to {\mathbb{C}}\) is measurable and defined \(\mu \)-almost everywhere, we say that \(f\) is absolutely integrable with respect to \(\mu \) if \(\int |f|\,d\mu <\infty \).

In this case, we define \(\int f\,d\mu =\int \Re f\,d\mu +i\int \Im f\,d\mu \), where for real-valued functions \(f\) we define \(\int f\,d\mu =\int f^+\,d\mu -\int f^-\,d\mu \).

The space \(L^1(\mu )\) together with the norm \(\|\cdot \|\) is a normed vector space.

\(\circ \)

Show that \(f\) is measurable if and only if \(f^+\) and \(f^-\) are measurable.

\(\circ \)

Show that sums and products of measurable functions are measurable.

Establish the support truncation property: if \(A_n\) is a sequence of measurable sets such that \(A_n\subset A_{n+1}\) and \(\bigcup A_n=X\), then \(\int f\chi _{A_n}\,d\mu \to \int f\,d\mu \).

BBT §12.1, §12.3

A normed vector space consists of a (real or complex) vector space \(X\) together with a mapping \(\|\cdot \|\colon X\to [0,\infty )\) satisfying:

\(\circ \)

(homogeneity) \(\|ax\|=|a|\cdot \|x\|\);

\(\circ \)

(triangle inequality) \(\|x+y\|\leq \|x\|+\|y\|\); and

\(\circ \)

(non-vanishing) \(\|x\|=0\) implies \(x=0\).

The mapping \(\|\cdot \|\) is called a norm on \(X\).

A normed vector space is called a Banach space if the associated metric \(d(x,y)=\|x-y\|\) is complete.

The ordinary finite-dimensional vector space \({\mathbb{R}}^d\), together with its usual Euclidean norm \(\|x\|=\left(\sum _1^dx_i^2\right)^{1/2}\), is a Banach space. It is a classical annoying exercise to establish the triangle inequality. One can also verify that the completeness property holds, using the classical fact/construction that it is true for \({\mathbb{R}}\).

The space \({\mathbb{R}}^{\mathbb{N}}\) consisting of all real sequences will not be a Banach space in any reasonable way. However it does contain many classical Banach spaces. For example we can consider the space of all square summable sequences

This is a Banach space with its generalized Euclidean norm \(\|x\|_2=\left(\sum x_i^2\right)^{1/2}\). It is not particularly easy to verify that \(\ell ^2\) is a Banach space and \(\|\cdot \|_2\) is a complete norm on it. We will do this later!

As we have mentioned, spaces of integrable functions may also form Banach spaces. Let \((X,\mathcal{B})\) be a measurable space and \(\mu \) a measure on it. We have already described the space \(L^1(\mu )\) of all absolutely integrable functions on \(X\). It is also possible to use other powers, for example, let

We may then define a norm on \(L^2\) be \(\|f\|_2=\left(\int |f|^2\right)^{1/2}\). The spaces \(L^1(\mu )\) and \(L^2(\mu )\) are not quite Banach spaces, but only because they fail to satisfy the non-vanishing property. This can easily be remedied by identifying two functions if they agree almost everywhere.

As a final series of examples, if \([a,b]\) is any closed, bounded interval then let \(B[a,b]\) be the space of all bounded functions \(f\colon [a,b]\to {\mathbb{R}}\) together with the supremum norm \(\|f\|=\sup \left\{\,f(x)\mid x\in [a,b]\,\right\}\). This is a Banach space which contains many other spaces of independent interest. For example it contains the space \(C[a,b]\) of continuous functions on \([a,b]\), the space \(D[a,b]\) of all differentiable functions on \([a,b]\) with continuous derivative, and the space \(P[a,b]\) of all polynomial functions on \([a,b]\). The last two turn out to be incomplete.

Let \(X,Y\) be normed vector spaces and let \(T\colon X\to Y\) be an operator. If \(T\) is continuous at one point \(x\in X\), then \(T\) is uniformly continuous.

Using the translation-invariance property, we can assume that \(x=0\). By definition this means that for all \(\epsilon >0\) there exists \(\delta >0\) such that for all \(x\in X\) we have \(\|x\|<\delta \implies \|Tx\|<\epsilon \). Given two points \(x,x'\) we can apply the latter property to \(x-x'\) and use the linearity of \(T\) to conclude that \(\|x-x'\|<\delta \implies \|Tx-Tx'\|<\epsilon \). In other words, \(T\) is uniformly continuous.

Let \(X,Y\) be normed vector spaces and let \(T\colon X\to Y\) be an operator. Then \(T\) is continuous if and only if \(T\) is bounded.

First suppose that \(T\) is continous. Apply the continuity with \(\epsilon =1\) to obtain \(\delta >0\) such that \(\|x\|<\delta \implies \|Tx\|<1\). The for any \(x\neq 0\) we have

Conversely suppose that \(T\) is bounded, and let \(M\) be such that \(\|Tx\|\leq M\|x\|\) for \(x\in X\). Given any \(\epsilon >0\) we let \(\delta =\epsilon /M\). Then we have

This shows that \(T\) is continuous at the point \(0\), and hence by the previous lemma \(T\) is continuous.

Show that the addition and constant multiple operations are continuous on a normed vector space.

Show that the unit ball of a normed vector space is convex. That is, for \(x,y\) in the ball and \(\lambda \in (0,1)\) we have \(\lambda x+(1-\lambda )y\) is also in the ball.

Consider the operators \(D(f)=f'\), \((Sf)(x)=\int _a^x fd\mu \), and \(I(f)=\int fd\mu \). What are appropriate domains and codomains of each operator? Show that \(S\) and \(I\) are continuous, and \(D\) is not continuous.

Let \(D[0,1]\) denote the space of differentiable functions on \([0,1]\) with continuous derivative, equipped with the supremum norm of \(B[0,1]\). Show that \(D[a,b]\) is not complete.

BBT §1.5

If \(X\) is a normed vector space, a linear functional on \(X\) is an operator \(\phi \colon X\to {\mathbb{R}}\). A bounded linear functional on \(X\) is a bounded, which is to say continuous, linear functional \(X\).

Let \(X\) be a vector space. A sublinear functional on \(X\) is a function \(p\colon X\to {\mathbb{R}}\) that satisfies:

(a)

(positive homogeneity) \(p(cx)=cp(x)\) for all \(c\geq 0\); and

(b)

(subadditivity) \(p(x+y)\leq p(x)+p(y)\).

Let \(X\) be a vector space, \(Y\leq X\) a subspace of \(X\), and \(p\) a sublinear functional on \(X\). Then any linear functional \(\phi _0\) on \(Y\) such that \(\phi _0\leq p\) extends to a linear functional \(\phi \) on \(X\) such that \(\phi \leq p\).

Let \(X\) be a normed vector space, and \(Y\leq X\) a subspace of \(X\).

(a)

Any bounded linear functional \(\phi _0\) on \(Y\) extends to a bounded linear functional \(\phi \) on \(X\).

(b)

If \(Y\) is closed and \(z\notin Y\), then there exists a bounded linear functional \(\phi \) on \(X\) such that \(\phi (Y)=0\) and \(\phi (z)\neq 0\).

(c)

The bounded linear functionals separate points: for all \(x,x'\in X\), if \(x\neq x'\) then there is a bounded linear functional \(\phi \) on \(X\) such that \(\phi (x)\neq \phi (x')\).

(a) Since \(\phi _0\) is a bounded linear functional on \(Y\), there exists some \(M\) such that \(\phi _0(y)\leq M\|y\|\) for all \(y\in Y\). We may therefore apply the Hahn–Banach theorem with the sublinear functional \(p(x)=M\|x\|\). Thus \(\phi _0\) extends to a linear functional \(\phi \) on \(X\) such that \(\phi (x)\leq M\|x\|\). In particular, \(\phi \) is bounded too.

(b) We first define a function \(\phi _0\) on the space \(Y+{\mathbb{R}}z\) which is bounded by \(p=\) the norm. For this we will let \(\phi _0(y+cz)=c\phi _0(z)\) where \(\phi _0(z)\) remains to be determined. In order to satisfy \(\phi _0(y+cz)\leq \|y+cz\|\) we require that \(c\phi _0(z)\leq \|y+cz\|\) for all \(y\in Y\).

Substituting \(y\) with \(-cy\), we see that we must choose \(\phi _0(z)\leq \|-y+z\|\) for all \(y\in Y\). Since \(Y\) is closed we must have \(\inf _{y\in Y}\|y+z\|\neq 0\) (otherwise \(z\) would be a limit of elements of \(Y\) and hence in \(Y\)). It follows that we can choose a nonzero value of \(\phi _0(z)\), and we may then use part (a) to extend \(\phi _0\) to a bounded linear functional \(\phi \) on \(X\) that meets our requirements.

(c) If \(x\neq x'\) then \(x-x'\neq 0\). Applying part (b) with \(Y=\{0\}\) we can find a bounded linear functional \(\phi \) such that \(\phi (x-x')\neq 0\). It follows that \(\phi (x)\neq \phi (x')\).

We begin by showing that we can find a proper extension of \(\phi _0\). Specifically, given any \(z\in X\smallsetminus Y\) we will find an extension of \(\phi _0\) to a linear functional \(\phi _1\) on \(Y\oplus {\mathbb{R}}z\) satisfying \(\phi _1\leq p\). For this we will define

where \(\phi _1(z)\) will be determined a little bit later. When we do choose \(\phi _1(z)\), it will have to satisfy the requirement that for all \(y\in Y\) and all \(c\in {\mathbb{R}}\):

To isolate \(\phi _1(z)\) we must consider the cases of negative and positive values of the coefficient \(c\) separately. Thus assume \(c>0\) and split the last equation into two conditions:

Solving each for \(\phi '(z)\) and substituting \(y\) with \(cy\) in each gives us the two new conditions:

In order for the constraints to be satisfiable, it is sufficient to have for all \(y,y'\in Y\) that \(-p(y-z)+\phi _0(y)\leq p(y'+z)-\phi _0(y')\). And this is indeed the case, since

Thus we can find a suitable value for \(\phi _1(z)\) and successfully extend \(\phi \) to \(\phi '\) as required.

To complete the proof we wish to apply the above step repeatedly. Since the number of steps will be uncountable in general, it is necessary to phrase our construction using the standard Zorn’s lemma: if \(P\) is a partially ordered set and every chain of \(P\) has an upper bound, then \(P\) has a maximal element.

Now let \(P\) be the collection of all linear functionals \(\phi \) such that the domain of \(\phi \) is a subspace of \(X\), \(\phi \) extends \(\phi _0\), and \(\phi \leq p\). We partially order \(P\) by function extension. A chain \(\mathcal{C}\) in \(P\) always has an upper bound, namely the set-theoretic union \(\bigcup \mathcal{C}\) of the members of the chain. Moreover the union will be \(\leq p\) because each member of the chain is \(\leq p\).

Therefore we can apply Zorn’s lemma to find an element \(\phi \) of \(P\) which is maximal with respect to function extension. We claim moreover that the domain of \(\phi \) must be all of \(X\). Indeed, otherwise we can use the argument above to properly extend the domain of \(\phi \) to find a larger element \(\phi '\) of \(P\). This contradicts the maximality of \(\phi \), and completes the proof.

Let \(f\) be a bounded real-valued function on \([0,1]\), and let \(U(f)\) denote the Upper Lebesgue integral of \(f\). Show that \(U\) is a sublinear functional. What can you conclude from the Hahn–Banach theorem?

Let \(\ell ^\infty \) denote the space of bounded real sequences with the supremum norm, and let \(c\) denote the subspace of convergent real sequences. Define \(p\) on \(\ell ^\infty \) by

Verify that \(p\) is a sublinear functional such that \(\lim x\leq p(x)\). If we apply the Hahn–Banach theorem to obtain a bounded linear functional \(L\) extending \(\lim \), show that \(\liminf x\leq L(x)\leq \limsup x\) and calculate the value of \(L(0,1,0,1,\ldots )\).

Let \({\mathbb{R}}^d\) be equipped with any norm that makes it into a normed vector space. Show that every linear functional on \({\mathbb{R}}^d\) is continuous.

BBT §12.3, 12.7

Let \(X,Y\) be normed vector spaces. Then \(B(X,Y)\) denotes the space of bounded linear operators \(T\colon X\to Y\).

(a)

\(B(X,Y)\) is a normed vector space with the operations of pointwise addition and scaling, and with the operator norm.

(b)

If \(Y\) is a Banach space then so is \(B(X,Y)\).

(a) We first show that the operator norm is indeed a norm. The homogeneity and non-vanishing properties are easy to check. For the triangle inequality let \(T,T'\in B(X,Y)\), and calculate \(\|(T+T')(x)\|=\|Tx+T'x\|\leq \|Tx\|+\|T'x\|\leq \|T\|\|x\|+\|T'\|\|x\|\). It follows that \(\|T+T'\|\leq \|T\|+\|T'\|\), as desired.

It also follows from homogeneity and the triangle inequality that \(B(X,Y)\) is closed under scalar multiplication and addition. Thus \(B(X,Y)\) is a normed vector space.

(b) It remains only to show that \(B(X,Y)\) is complete. For this let \(T_n\) be a sequence of elements of \(B(X,Y)\) and assume that it is Cauchy in the operator norm. This means that for all \(\epsilon >0\), there exists \(N\) such that for all \(m,n\geq N\) we have \(\|T_m-T_n\|<\epsilon \).

Now for any fixed \(x\in X\), it follows from the last equation that \(\|T_mx-T_nx\|<\epsilon \|x\|\). In particular, the sequence \(T_nx\) is a Cauchy sequence in the space \(Y\). Since we are assuming that \(Y\) is complete, the sequence \(T_nx\) converges and we define \(Tx=\lim T_nx\).

Now \(T\) is a well-defined function from \(X\) to \(Y\), and by definition \(T\) is the pointwise limit of the \(T_n\). We need to check that \(T\in B(X,Y)\) and moreover that \(T_n\to T\) in the operator norm.

To see that \(T\in B(X,Y)\), first note that it is easy to check \(T\) is a linear map. For example, \(T(x+y)=\lim T_n(x+y)=\lim T_n(x)+T_n(y)=T(x)+T(y)\). To see that \(T\) is bounded, we first claim that the sequence of operator norms \(\|T_n\|\) is itself bounded. For this claim, recall from the reverse triangle inequality that \(|\|T_n\|-\|T_m\||\leq \|T_n-T_m\|\). Thus the fact that \(T_n\) is Cauchy implies that \(\|T_n\|\) is Cauchy in \({\mathbb{R}}\), and any Cauchy sequence in \({\mathbb{R}}\) is bounded. This completes the claim.

Now let \(M\) be a bound for the sequence \(\|T_n\|\). Then given \(x\in X\), for all \(n\) we have \(\|T_nx\|\leq M\|x\|\). Taking the limit of both sides, we conclude that \(\|Tx\|\leq M\|x\|\), and this means that \(T\) is a bounded operator. Thus we have shown that \(T\in B(X,Y)\).

Last we establish that \(T_n\to T\) in operator norm. For this, given \(\epsilon >0\) we have already argued that we can find \(N\) such that \(m,n\geq N\) implies \(\|T_mx-T_nx\|<\epsilon \|x\|\) for all \(x\in X\). Now let \(n\geq N\) be fixed and take the limit as \(m\to \infty \). This gives us \(\|Tx-T_nx\|\leq \epsilon \|x\|\) for all \(x\). We thus have that \(\|T-T_n\|\leq \epsilon \), which means that \(T_n\to T\) in operator norm.

If \(X\) is a normed (real) vector space then the dual of \(X\) is \(X^*=B(X,{\mathbb{R}})\).

Let \(X\) be a normed vector space, \(Y\leq X\) a subspace of \(X\).

(a)

Any \(\phi _0\in Y^*\) extends to an element \(\phi \in X^*\) such that \(\|\phi \|=\|\phi _0\|\).

(b)

If \(Y\) is closed and \(z\notin Y\), then there exists \(\phi \in X^*\) such that \(\phi (Y)=0\) and \(\phi (z)\neq 0\). Moreover one may take \(\|\phi \|=1\) and \(\phi (z)=\inf _{y\in Y}\|y+z\|\).

(a) Since \(\phi \) is an extension of \(\phi _0\), we always have \(\|\phi \|\geq \|\phi _0\|\). And in the proof Corollary 18.4(a), it is apparent that \(\|\phi \|\leq \|\phi _0\|\).

(b) Recall that in the proof of Corollary 18.4(b), we showed one may define \(\phi _0\) on \(Y+{\mathbb{R}}z\) in such a way that \(\phi (Y)=0\) and \(\phi (z)=\inf _{y\in Y}\|y+z\|\) and \(\phi _0\leq \|\cdot \|\). It is not too difficult to argue that with this definition, \(\|\phi _0\|=1\). Therefore by part (a) we can extend \(\phi _0\) to \(\phi \) with \(\|\phi \|=1\) as desired.

Let \(X\) be a normed vector space. Then there is a norm-preserving operator from \(X\) into its double dual \(X^{**}\).

We define an embedding \(x\mapsto \hat{x}\) from \(X\) to \(X^{**}\) as follows. Given an element \(x\in X\), we define the corresponding element \(\hat{x}\in X^{**}\) by the formula:

So \(\hat{x}\) is a function from \(X^*\) to \({\mathbb{R}}\). And \(\hat{x}\) is linear because \(\hat{x}(\phi _1+\phi _2)=(\phi _1+\phi _2)(x)=\phi _1(x)+\phi _2(x)=\hat{x}(\phi _1)+\hat{x}(\phi _2)\). Next \(\hat{x}\) is bounded because \(|\hat{x}(\phi )|=|\phi (x)|\leq \|\phi \|\cdot \|x\|\). Thus we have verified that \(\hat{x}\in X^{**}\).

Now we verify that the map \(x\mapsto \hat{x}\) is itself linear. Indeed, we have \(\widehat{x_1+x_2}(\phi )=\phi (x_1+x_2)=\phi (x_1)+\phi (x_2)=\hat{x}_1(\phi )+\hat{x}_2(\phi )\). And similarly for scalar multiplication.

Finally to show that \(x\mapsto \hat{x}\) is norm-preserving, note that the calculation in the first paragraph shows that \(\|\hat{x}\|\leq \|x\|\). To show that \(\|\hat{x}\|\geq \|x\|\), we use Corollary 19.4(a) to obtain \(\phi \in X^*\) such that \(\phi (x)=\|x\|\) and \(\|\phi \|=1\). Thus we have that \(\hat{x}(\phi )=\phi (x)=\|x\|=\|x\|\|\phi \|\), which witnesses that \(\|\hat{x}\|\geq \|x\|\).

Show that \(\|T\|\) is equal to \(\sup \left\{\,\|Tx\|:\|x\|\leq 1\,\right\}\).

Show that \(\|x\|=\sup \left\{\,|\phi (x)|:\phi \in X^*{\text{ and }}\|\phi \|=1\,\right\}\).

If \(X,Y\) are Banach spaces and \(T\in B(X,Y)\), show that \((T^*\phi )(x)=\phi (Tx)\) defines an element of \(B(Y^*,X^*)\) such that \(\|T^*\|=\|T\|\).

Show that a Banach space \(X\) is reflexive if and only if \(X^*\) is reflexive.

Complete the proof of Corollary 19.4(b).

BBT §12.11, 12.13, 12.14

Let \(X,Y\) be Banach spaces and \(\mathcal{F}\) a family of bounded operators from \(X\) to \(Y\). Suppose that for all \(x\) there exists a constant \(M_x\) such that for all \(T\in \mathcal{F}\) we have \(\|Tx\|\leq M_x\). Then there exists a constant \(M\) such that for all \(T\in \mathcal{F}\) we have \(\|T\|\leq M\).

Let \(X,Y\) be Banach spaces and let \(T_n\colon X\to Y\) be a sequence of bounded operators. If \(T_n\to T\) pointwise, then \(T\) is a bounded operator too.

We have already observed that a pointwise limit of operators is an operator. Hence it remains only to check that \(T\) is bounded. Now given any \(x\in X\), since \(\{T_nx\}\) is a convergent sequence of \(Y\), it is necessarily a bounded sequence of \(Y\). In other words, the sequence \(\{\|T_nx\|\}\) is bounded. By the uniform boundedness principle, there exists a constant \(M\) such that \(\|T_n\|\leq M\) for all \(n\). Thus for any \(x\in X\) we have

In particular, \(T\) is bounded and \(\|T\|\leq M\).

Let \(X,Y\) be Banach spaces and \(T\colon X\to Y\) be a bounded operator. If \(T\) is onto, then \(T\) is open.

Suppose \(X\) is a Banach space with two complete norms \(\|\cdot \|_a\) and \(\|\cdot \|_b\). Then if there is a constant \(c\) such that \(\|x\|_a\leq c\|x\|_b\) for all \(x\in X\), then \(\|\cdot \|_a\) and \(\|\cdot \|_b\) are equivalent.

Let \(\operatorname {id}\colon X\to X\) denote the identity mapping. If we consider \(\operatorname {id}\) as an operator from \((X,\|\cdot \|_b)\) to \((X,\|\cdot \|_a)\), then the hypothesis implies that \(\operatorname {id}\) is bounded and hence continuous. It follows from the open mapping theorem that \(\operatorname {id}\) is open, that is, it maps open sets to open sets. Since \(\operatorname {id}\) is a bijection, this simply means that \(\operatorname {id}^{-1}\) is continuous and hence bounded. Thus we conclude that there exists a constant \(d\) suh that \(\|x\|_b\leq d\|x\|_d\) for all \(x\in X\).

Let \(X,Y\) be Banach spaces and \(T\colon X\to Y\) be an operator. If the graph of \(T\) is a closed subset of \(X\times Y\), then \(T\) is bounded.

Give an example of a function from \({\mathbb{R}}\) to \({\mathbb{R}}\) which has a closed graph but is not continuous. Give an example of function from \({\mathbb{R}}\) to \({\mathbb{R}}\) which is continuous and surjective but not open. Is it possible to give a bijective example?

Equip the space \(C[0,1]\) with both the \(L^1\) norm and the supremum norm. Show that the \(L^1\) norm is bounded by a constant times the supremum norm. Show that the reverse is not true. Explain why the two results do not contradict Corollary 20.4.

BBT §13.1, 13.2

Let \((X,\mathcal{B})\) be a measurable space and let \(\mu \) be a measure on it. For any measurable \(f\) defined on \(X\) we let

We then define the space

with the understanding that \(f,g\) are identified when \(f-g=0\) almost everywhere.

Let \(p,q\geq 1\) be real numbers such that \(1/p+1/q=1\). If \(f\in L^p(\mu )\) and \(g\in L^q(\mu )\), then \(fg\) is absolutely integrable and

The theorem follows from a classical inequality which we will call Hölder’s inequality for real numbers:

There are many proofs and one is left as an exercise.

To begin the proof, note that given \(f,g\) as in the theorem statement, we can rescale to assume that \(\|f\|_p=\|g\|_q=1\). This is because both \(\|\cdot \|_p\) and \(\|\cdot \|_q\) satisfy the positive homogeneity property.

Now our objective is to show that \(\int |fg|\,d\mu \leq 1\). For this we plug \(a=|f(x)|\) and \(b=|g(x)|\) into Hölder’s inequality for real numbers to obtain

Taking the integral of both sides we have

as desired.

Suppose that \(p\geq 1\). If \(f,g\in L^p(\mu )\), then \(\|f+g\|_p\leq \|f\|_p+\|g\|_p\).

We can assume without loss of generality that \(f,g\) never take the value \(\infty \). We begin by writing

We now integrate both sides of this inequality, and then apply Hölder’s inequality to each of the resulting terms. In the following calculation, we also note that our hypothesis implies that the value \(q\) used in Hölder’s inequality is equal to \(p/(p-1)\). Here is the computation:

We may now divide both sides by \((\|f+g\|_p)^{p-1}\) to obtain the desired conclusion.

The space \(L^p(\mu )\) with the norm \(\|\cdot \|_p\) is a Banach space.

It is clear that the norm is homogeneous and non-vanishing, and we have just shown it satisfies the triangle inequality. This also implies that \(L^p(\mu )\) is closed under linear combinations and therefore it is a vector space. So it only remains to show that the norm \(\|\cdot \|_p\) is complete.

For this let \(f_n\) be a sequence of elements of \(L^p(\mu )\) which is Cauchy in the \(\|\cdot \|_p\) norm. Passing to a subsequence if necessary, we can suppose without loss of generality that for all \(n\) we have \(\|f_{n+1}-f_n\|_p<1/2^n\). We first wish to show that this implies \(f_n\) has a pointwise limit \(f\).

Let \(g_k=\sum _1^k|f_{n+1}-f_n|\) and \(g=\sum _1^\infty |f_{n+1}-f_n|\). So \(g\) is the limit of the \(g_k\). While the function \(g\) may take the value \(+\infty \), we claim that this cannot happen too often. Indeed by Minkowski’s inequality we have

It then follows from Fatou’s lemma that \(\int |g|^p\leq \liminf \int |g_k|^p\leq 1\). Thus we can conclude that \(g\) is finite \(\mu \)-almost everywhere.

Now using a simple telescoping we can write:

We have just observed that for \(\mu \)-almost every \(x\), the latter series is absolutely convergent. Thus the series is convergent, and we can define the function \(f(x)=\lim f_n(x)\).

It remains only to show that \(f\) lies in \(L^p\) and that \(f_n\to f\) in the norm \(\|\cdot \|_p\). For this, let \(\epsilon \) be given and choose \(N\) large enough that \(m,n\geq N\) implies \(\|f_m-f_n\|_p<\epsilon \). Fixing \(n\) and applying Fatou’s Lemma to the resulting \(m\)-sequence we obtain

This shows that \(\|f-f_n\|_p\to 0\), or in other words that \(f_n\to f\) in the norm of \(L^p\). Finally if \(n\geq N\) then we have \(\|f\|_p\leq \|f-f_n\|_p+\|f_n\|_p<\infty \), so \(f\in L^p\) too.

Let \((X,\mathcal{B})\) be a measurable space and let \(\mu \) be a measure on it. For any measurable \(f\) defined on \(X\) we let

We then define the space

Let \((X,\mathcal{B})\) be a measurable space and \(\mu \) a measure on it.

\(\circ \)

If \(f\in L^1(\mu )\) and \(g\in L^\infty (\mu )\) then \(fg\) is absolutely integrable and Hölder’s inequality is true: \(\int |fg|\,d\mu \leq \|f\|_1\cdot \|g\|_\infty \).

\(\circ \)

The space \(L^\infty (\mu )\) is a Banach space with the norm \(\|\cdot \|_\infty \).

Show that the inequality \(\|f+g\|_1\leq \|f\|_1+\|g\|_1\) is strict precisely when there exists a nonnegative measurable function \(h\) such that \(g=fh\) for almost every element \(x\) of the set where \(f,g\neq 0\).

Recall the argument from Theorem 12.6 that the absolutely integrable simple functions are dense in \(L^1\). Show that the absolutely integrable simple functions are dense in \(L^p\).

Prove Theorem 21.6.

Prove Hölder’s inequality for real numbers. (Should probably give a hint.)

BBT §13.6, Tao “an \(\epsilon \) of room” §1.2

Let \((X,\mathcal{B})\) be a measure space, \(\mu \) a measure on it, and \(L^p=L^p(\mu )\). Assume that \(\mu \) is \(\sigma \)-finite, that is, \(X=\bigcup A_n\) where \(\mu (A_n)<\infty \). Let \(1\leq p<\infty \) and let \(q\) be the conjugate exponent, that is, \(1/p+1/q=1\). Then for every \(\phi \in (L^p)^*\) there exists a unique \(g\in L^q\) such that

Moreover \(\|\phi \|=\|g\|_q\), and \((L^p)^*\cong L^q\).

Let \((X,\mathcal{B}\)) be a measurable space. A signed measure on \((X,\mathcal{B})\) is a function \(\nu \colon \mathcal{B}\to [-\infty ,\infty ]\) with the properties:

\(\circ \)

\(\nu (\emptyset )=0\);

\(\circ \)

\(\nu \) does not take both the values \(\infty \) and \(-\infty \); and

\(\circ \)

If \(A_n\) are disjoint then \(\sum \nu (A_n)\) converges to \(\nu (\bigcup A_n)\).

Let \((X,\mathcal{B})\) be a measurable space and let \(\mu \) be an unsigned measure and \(\nu \) be a signed measure on it. Assume that \(\mu ,\nu \) are \(\sigma \)-finite. Then if \(\mu (A)=0\implies \nu (A)=0\), then there exists \(g\in L^1(\mu )\) such that \(\nu =\mu _g\).

In this proof, we will sketch only the case when \(1<p<\infty \) and \(\mu (X)<\infty \). It is not essentially more difficult to complete the proof from this simplified version.

Given a functional \(\phi \in (L^p)^*\), we first define the mapping \(\nu (A)=\phi (\chi _A)\). We have already checked that \(\nu \) is finitely additive. We claim that in fact \(\nu \) is a signed measure. Indeed, if \(A_n\) is a given sequence of pairwise disjoint sets, then using the finiteness of \(\mu (X)\) and the dominated convergence theorem we have:

In other words, we have that \(\chi _{\bigcup _1^kE_n}\to \chi _{\bigcup E_n}\) in the \(L^p\)-norm. Using the fact that \(\phi \) is a continuous function on \(L^p\), it follows that

and so \(\nu \) is countably additive.

Now by the Radon–Nikodym theorem, there exists a function \(g\in L^1\) such that \(\phi (\chi _A)=\int \chi _A g\,d\mu \) for all sets \(E\). It therefore follows from linearity that \(\phi (f)=\int fg\,d\mu \) for all simple functions \(f\).

We next claim that \(\phi (f)=\int fg\,d\mu \) for all functions \(f\) in \(L^\infty \). For this recall that any bounded measurable function is a uniform limit of simple functions. So given \(f\in L^\infty \) let \(f_n\) be a sequence of simple functions such that \(f_n\to f\) uniformly. Using the uniform convergence on a finite measure space, we can easily argue that \(\int f_ng\,d\mu \to \int fg\,d\mu \). For the same reason, we can also argue that \(f_n\to f\) in \(L^p\). Since \(\phi \) is continuous on \(L^p\) we therefore have that:

as desired.

While our next goal is of course to show that \(\phi (f)=\int fg\,d\mu \) for all functions \(f\in L^p\), we first take a break and show that \(g\) lies in \(L^q\). In fact we will show that \(\|g\|_q\leq \|\phi \|\). First we can use the truncation lemma to suppose that \(g\) is bounded. Then \(|g|^q/g\) is bounded too, and so by our work for functions in \(L^\infty \) we can calculate:

This inequality implies that \(\|g\|_q\leq \|\phi \|\), as desired. We remark that this implies the functional \(f\mapsto \int fg\,d\mu \) is continuous, since Hölder’s inequality states that \(\left|\int fg\,d\mu \right|\leq \|f\|_p\cdot \|g\|_q\).

We now claim that we have \(\phi (f)=\int fg\,d\mu \) for any \(f\in L^p\). For this, recall that we have previously shown that the simple functions are dense in \(L^1\), that is any \(L^1\) function is an \(L^1\)-limit of simple functions. The same argument can be used to show that the simple functions are dense in \(L^p\). We have shown above that the two functionals, \(f\mapsto \phi (f)\) and \(f\mapsto \int fg\,d\mu \), agree on the simple functions. Our hypothesis states that \(\phi \) continuous, and the previous paragraph implies that \(f\mapsto \int fg\,d\mu \) is continuous too. Since two continuous functions that agree on a dense set must agree on their domain, we can conclude that \(\phi (f)=\int fg\,d\mu \) for all \(f\in L^p(\mu )\).

Finally we claim that \(\|g\|_q=\|\phi \|\). Indeed we now know that \(|\phi (f)|=\left|\int fg\,d\mu \right|\leq \|f\|_p\cdot \|g\|_q\), and hence that \(\|\phi \|\leq \|g\|_q\). We have also shown two paragraphs previously that \(\|g\|_q\leq \|\phi \|\). This concludes the proof.

The Radon–Nikodym theorem has a generalization called the Lebesgue–Radon–Nykodym theorem which states that given \(\mu \), any signed measure \(\nu \) can be decomposed \(\nu =\mu _g+\delta \) where \(\mu ,\delta \) have disjoint supports.

We have stated that the dual of \(L^1\) is equal to \(L^\infty \) when \(\mu \) is \(\sigma \)-finite. However the reverse is usually not true. Instead the dual of \(L^\infty (\mu )\) is the space of all finitely additive signed measures \(\nu \) such that \(\mu (E)=0\implies \nu (E)=0\).

Let \(g\in L^1[0,1]\). Show that the map \(f\mapsto \int fg\) is a bounded linear functional on \(L^\infty [0,1]\).

Show that there is a nonzero bounded linear functional on \(L^\infty [0,1]\) that vanishes on the (closed) subspace of continuous functions.

Show that there is a bounded linear functional on \(L^\infty [0,1]\) that is not of the form \(f\mapsto \int fg\) for any \(g\in L^1[0,1]\).

BBT §13.5, 14.1, 14.2, 14.3

Let \(X\) be a complex vector space. A function \(\langle \cdot ,\cdot \rangle \colon X\times X\to {\mathbb{C}}\) is called an inner product if it satisfies

(a)

(positivity/nonvanishing) for \(x\in X\) we have \(\langle x,x\rangle \geq 0\), and \(\langle x,x\rangle =0\) iff \(x=0\);

(b)

(conjugate symmetry) for \(x,y\in X\) we have \(\langle x,y\rangle =\overline{\langle y,x\rangle }\); and

(c)

(linearity in the first coordinate) \(\langle c_1x_1+c_2x_2,y\rangle =c_1\langle x_1,y\rangle +c_2\langle x_2,y\rangle \)

If \(X\) admits an inner product then it automatically admits a norm \(\|x\|=\sqrt {\langle x,x\rangle }\), and \(X\) is called a Hilbert space if this norm is complete.

Let \(X\) be an inner product space. Then \(|\langle x,y\rangle |\leq \|x\|\cdot \|y\|\).

The proof is a simpler version of the proof of Hölder’s inequality. First, by multiplying \(x\) by a scalar of the form \(e^{i\theta }\), we may assume that \(\langle x,y\rangle \) is real. Next given \(x,y\) we define a real function \(p(\alpha )=\langle \alpha x+y,\alpha x+y\rangle \). Then by axiom (a) we have that \(p(\alpha )\geq 0\). And by axiom (c) we have

Thus \(p\) is a quadratic and \(p\geq 0\), which implies \(p\) has at most one real root. This means that the discriminant is non-positive, that is, \(4|\langle x,y\rangle |^2-4\|x\|^2\cdot \|y\|^2\leq 0\). This last equation is plainly equivalent to the desired result.

Let \(X\) be an inner product space. Then \(\|x\|=\sqrt {\langle x,x\rangle }\) makes \(X\) into a normed vector space.

Since the nonvanishing and homogeneity properties are automatic from the axioms, it remains only to verify the triangle inequality. For this we simply calculate:

Here, the first inequality uses axiom (b) and the ordinary triangle inequality, and the second inequality uses the Schwarz inequality. Taking the square root of both sides, we achieve the desired result.

Let \(X\) be a Hilbert space. We say that vectors \(x,y\in X\) are orthogonal if \(\langle x,y\rangle =0\). Given a vector subspace \(Y\subset X\) we define its orthogonal complement \(Y^\perp =\left\{\,x\in X\mid (\forall y\in Y)\;\langle x,y\rangle =0\,\right\}\).

Let \(X\) be a Hilbert space and let \(Y\leq X\) be a closed subspace. Then \(X=Y\oplus Y^\perp \) in the sense that every \(x\in X\) can be uniqely expressed as \(x=y+y'\) where \(y\in Y\) and \(y'\in Y^\perp \).

If \(X\) is a Hilbert space, then \(X\) is self-dual. That is, for any \(\phi \in X^*\) there exists \(y\in X\) such that \(\phi (x)=\langle x,y\rangle \). Moreover the correspondence \(\phi \mapsto y\) is a conjugate-linear isomorphism \(X^*\cong X\).

Given \(\phi \), we let \(Y=\left\{\,x\in X\mid \phi (x)=0\,\right\}\). Assuming \(Y\neq X\), we may choose \(z\in Y^\perp \) such that \(\|z\|=1\). We then let \(y=\overline{\phi (z)}z\). Then we have

Here the last equality follows from the fact that \(\phi (x)z-\phi (z)x\) lies in \(Y\).

Use the discussion in BBT, §14.2 to prove Proposition 23.5.

BBT, §14.4

Let \(X\) be a Hilbert space. A subset \(\{e_\alpha \}_{\alpha \in A}\) of \(X\) is called an orthonormal basis if it satisfies the following properties:

(a)

(normality) for all \(\alpha \), \(\|e_\alpha \|=1\);

(b)

(orthogonality) for all \(\alpha \neq \beta \), \(\langle e_\alpha ,e_\beta \rangle =0\); and

(c)

(maximality) for any \(x\in X\), if \(\langle x,e_\alpha \rangle =0\) for all \(\alpha \in A\), then \(x=0\).

Let \(X\) be a Hilbert space and \(\{e_\alpha \}_{\alpha \in A}\) an orthonormal basis for \(X\). Then for any \(x\in X\), we have \(x=\sum \langle x,e_\alpha \rangle e_\alpha \), with the convergence being in norm. Moreover, we have Parseval’s identity, which states that \(\|x\|^2=\sum _{\alpha \in A}|\langle x,e_\alpha \rangle |^2\).

In the proof, we will need the Pythagorean theorem, which states that if \(e_1,\ldots ,e_n\) are orthognal, then \(\|e_1+\cdots +e_n\|^2=\|e_1\|^2+\cdots +\|e_n\|^2\). The calculation is the same as the classical version, and is obtained by distributing out the expression \(\langle e_1+\cdots +e_n,e_1+\cdots +e_n\rangle \).

We first show one half of Parseval’s identity, namely that \(\sum |\langle x,e_\alpha \rangle |^2\leq \|x\|^2\). (This is called Bessel’s inequality.) For this we let \(A_0\subset A\) be a finite subset. Then an elementary calculation together with the Pythagorean theorem gives

Since the left-hand side is nonnegative, we have \(\sum _{\alpha \in A_0}|\langle x,e_\alpha \rangle |^2\leq \|x\|^2\). Since \(A_0\) was arbitrary, this completes the claim.

Now we know that \(\sum |\langle x,e_\alpha \rangle |^2\) converges. It follows that there are just countably many nonzero terms, let us enumerate them \(|\langle x,e_n\rangle |^2\). Then the sequence of partial sums of \(\sum |\langle x,e_n\rangle |^2\) is Cauchy. By the Pythagorean theorem,

Thus the sequence of partial sums of \(\sum \langle x,e_n\rangle e_n\) is Cauchy too. Since \(X\) is complete, we conclude that there exists an element \(y\in X\) defined by \(y=\sum \langle x,e_\alpha \rangle e_\alpha \).

Next we claim that in fact \(x=y\). Indeed, it is easy to see that \(\langle x-y,e_\alpha \rangle =0\) for all \(\alpha \), so the completeness of the orthonormal set implies that \(x-y=0\).

Finally, we conclude the proof of Parseval’s identity by returning to Equation (24.e1). Since we now know that the left-hand side converges to \(0\) as \(A_0\to A\), it follows that the right-hand side does as well, establishing the equality.

For any index set \(A\), we let

Let \(X\) be a Hilbert space. Then \(X\) is isomorphic to \(\ell ^2(A)\) for some \(A\) by a linear bijection that preserves inner products.

Let \(\{e_\alpha \}_{\alpha \in A}\) be an orthonormal basis for \(X\). The index set \(A\) will be the same set we use to form \(\ell ^2(A)\). We define a function \(\phi \colon X\to \ell ^2(A)\) by \(\phi (x)(\alpha )=\langle x,e_\alpha \rangle \).

It is easy to see that \(\phi \) is linear, and Parseval’s identity implies that \(\phi \) preserves the norm. In particular \(\phi \) is injective. To see that \(\phi \) preserves the inner product, it suffices to note that the inner product can be recovered from the norm by the polarization identity

Finally to see that \(\phi \) is surjective, let \(f\in \ell ^2(A)\) be given. Since \(\sum _{\alpha \in A}|f(\alpha )|^2<\infty \), the series has just countably many nonzero terms and its partial sums are Cauchy. By the Pythagorean theorem, the partial sums of \(\sum _{\alpha \in A}f(\alpha )e_\alpha \) are Cauchy too. It follows that there exists an element \(x\in X\) defined by \(x=\sum _{\alpha \in A}f(\alpha )e_\alpha \). Clearly \(\phi (x)=f\), as desired.

Prove the Pythagorean theorem in a Hilbert space.

Prove the polarization identity ina Hilbert space.