# 8.12: Integration and Differentiation

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I. We shall now link RN-derivatives (§11) to those of Chapter 7, §12.

Below, we use the notation of Definition 3 in Chapter 7, §10 and Definition 1 of Chapter 7, §12. (Review them!) In particular,

$m : \mathcal{M}^{*} \rightarrow E^{*}$

is Lebesgue measure in $$E^{n}$$ (presupposed in such terms as "a.e.," etc.); $$s$$ is an arbitrary set function. For convenience, we set

$s^{\prime}(\overline{p})=0$

and

$\int_{X} f dm=0,$

unless defined otherwise; thus $$s^{\prime}$$ and $$\int_{X} f$$ exist always.

## Lemma $$\PageIndex{1}$$

With the notation of Definition 3 of Chapter 7, §10, the functions

$\overline{D} s, \underline{D} s, \text { and } s^{\prime}$

are Lebesgue measurable on $$E^{n}$$ for any set function

$s : \mathcal{M}^{\prime} \rightarrow E^{*} \quad\left(\mathcal{M}^{\prime} \supseteq \overline{\mathcal{K}}\right).$

Proof

By definition,

$\overline{D} s(\overline{p})=\inf _{r} h_{r}(\overline{p}),$

where

$h_{r}(\overline{p})=\sup \left\{\frac{s I}{m I} | I \in \mathcal{K}_{\overline{p}}^{r}\right\}$

and

$\mathcal{K}_{\overline{p}}^{r}=\left\{I \in \overline{\mathcal{K}} | \overline{p} \in I, d I<\frac{1}{r}\right\}, \quad r=1,2, \ldots.$

As is easily seen (verify!),

$E^{n}\left(h_{r}>a\right)=\bigcup\left\{I \in \overline{\mathcal{K}} | a<\frac{s I}{m I}, d I<\frac{1}{r}\right\}, \quad a \in E^{*}.$

The right-side union is Lebesgue measurable by Problem 2 in Chapter 7, §10. Thus by Theorem 1 of §2, the function $$h_{r}$$ is measurable on $$E^{n}$$ for $$r=1,2, \ldots$$ and so is

$\overline{D} s=\inf _{r} h_{r}$

by Lemma 1 of §2 and Definition 3 in Chapter 7, §10. Similarly for $$\underline{D} s$$.

Hence by Corollary 1 in §2, the set

$A=E^{n}(\underline{D} s=\overline{D} s)$

is measurable. As $$s^{\prime}=\overline{D} s$$ on $$A, s^{\prime}$$ is measurable on $$A$$ and also on $$-A$$ (by convention, $$s^{\prime}=0$$ on $$-A ),$$ hence on all of $$E^{n}. \quad \square$$

## Lemma $$\PageIndex{2}$$

With the same notation, let $$s : \mathcal{M}^{\prime} \rightarrow E^{*}\left(\mathcal{M}^{\prime} \supseteq \overline{\mathcal{K}}\right)$$ be a regular measure in $$E^{n}.$$ Let $$A \in \mathcal{M}^{*}$$ and $$B \in \mathcal{M}^{\prime}$$ with $$A \subseteq B,$$ and $$a \in E^{1}$$.

If

$\overline{D} s>a \quad \text { on } A,$

then

$a \cdot m A \leq s B.$

Proof

Fix $$\varepsilon>0.$$ By regularity (Definition 4 in Chapter 7, §7), there is an open set $$G \supseteq B,$$ with

$s B+\varepsilon \geq s G.$

Now let

$\mathcal{K}^{\varepsilon}=\{I \in \overline{\mathcal{K}} | I \subseteq G, s I \geq(a-\varepsilon) m I\}.$

As $$\overline{D} s>a,$$ the definition of $$\overline{D} s$$ implies that $$\mathcal{K}^{\varepsilon}$$ is a Vitali covering of $$A$$. (Verify!)

Thus Theorem 1 in Chapter 7, §10, yields a disjoint sequence $$\left\{I_{k}\right\} \subseteq \mathcal{K}^{\varepsilon}$$, with

$m\left(A-\bigcup_{k} I_{k}\right)=0$

and

$m A \leq m\left(A-\bigcup I_{k}\right)+m \bigcup I_{k}=0+m \bigcup I_{k}=\sum_{k} m I_{k}.$

As

$\bigcup I_{k} \subseteq G \text { and } s B+\varepsilon \geq s G$

(by our choice of $$\mathcal{K}^{\varepsilon}$$ and $$G$$, we obtain

$s B+\varepsilon \geq s \bigcup_{k} I_{k}=\sum_{k} s I_{k} \geq(a-\varepsilon) \sum_{k} m I_{k} \geq(a-\varepsilon) m A.$

Thus

$(a-\varepsilon) m A \leq s B+\varepsilon.$

Making $$\varepsilon \rightarrow 0,$$ we obtain the result.$$\quad \square$$

## Lemma $$\PageIndex{3}$$

If

$t=s \pm u,$

with $$s, t, u : \mathcal{M}^{\prime} \rightarrow E^{*}$$ and $$\mathcal{M}^{\prime} \supseteq \overline{\mathcal{K}},$$ and if $$u$$ is differentiable at a point $$\overline{p} \in E^{n}$$, then

$\overline{D} t=\overline{D} s \pm u^{\prime} \text { and } \underline{D} t=\underline{D} s \pm u^{\prime} \text { at } \overline{p}.$

Proof

The proof, from definitions, is left to the reader (Chapter 7, §12, Problem 7).

## Lemma $$\PageIndex{4}$$

Any $$m$$-continuous measure $$s : \mathcal{M}^{*} \rightarrow E^{1}$$ is strongly regular.

Proof

By Corollary 3 of Chapter 7, §11, $$v_{s}=s<\infty$$ ($$s$$ is finite!). Thus $$v_{s}$$ is certainly $$m$$-finite.

Hence by Theorem 2 in Chapter 7, §11, $$s$$ is absolutely $$m$$-continuous. So given $$\varepsilon>0,$$ there is $$\delta>0$$ such that

$\left(\forall X \in \mathcal{M}^{*} | m X<\delta\right) \quad s X<\varepsilon.$

Now, let $$A \in \mathcal{M}^{*}.$$ By the strong regularity of Lebesgue measure $$m$$ (Chapter 7, §8, Theorem 3(b)), there is an open set $$G \supseteq A$$ and a closed $$F \subseteq A$$ such that

$m(A-F)<\delta \text { and } m(G-A)<\delta.$

Thus by our choice of $$\delta$$,

$s(A-F)<\varepsilon \text { and } s(G-A)<\varepsilon,$

as required.$$\quad \square$$

## Lemma $$\PageIndex{5}$$

Let $$s, s_{k}(k=1,2, \ldots)$$ be finite $$m$$-continuous measures, with $$s_{k} \nearrow s$$ or $$s_{k} \searrow s$$ on $$\mathcal{M}^{*}.$$

If the $$s_{k}$$ are a.e. differentiable, then

$\overline{D} s=\underline{D} s=\lim _{k \rightarrow \infty} s_{k}^{\prime} \text{ a.e.}$

Proof

Let first $$s_{k} \nearrow s.$$ Set

$t_{k}=s-s_{k}.$

By Corollary 2 in Chapter 7, §11, all $$t_{k}$$ are $$m$$-continuous, hence strongly regular (Lemma 4). Also, $$t_{k} \searrow 0$$ (since $$s_{k} \nearrow s$$). Hence

$t_{k} I \geq t_{k+1} I \geq 0$

for each cube $$I;$$ and the definition of $$\overline{D} t_{k}$$ implies that

$\overline{D} t_{k} \geq \overline{D} t_{k+1} \geq \underline{D} t_{k+1} \geq 0.$

As $$\{\overline{D} t_{k}\} \downarrow,$$ $$\lim_{k \rightarrow \infty} \overline{D} t_{k}$$ exists (pointwise). Now set

$A_{r}=E^{n}\left(\lim _{k \rightarrow \infty} \overline{D} t_{k} \geq \frac{1}{r}\right), \quad r=1,2, \ldots.$

By Lemma 1 (and Lemma 1 in §2), $$A_{r} \in \mathcal{M}^{*}.$$ Since

$\overline{D} t_{k} \geq \lim _{i \rightarrow \infty} \overline{D} t_{i} \geq \frac{1}{r}$

on $$A_{r},$$ Lemma 2 yields

$\frac{1}{r} m A_{r} \leq t_{k} A_{r}.$

As $$t_{k} \searrow 0,$$ we have

$\frac{1}{r} m A_{r} \leq \lim _{k \rightarrow \infty} t_{k} A_{r}=0.$

Thus

$m A_{r}=0, \quad r=1,2, \ldots.$

Also, as is easily seen

$E^{n}\left(\lim _{k \rightarrow \infty} \overline{D} t_{k}>0\right)=\bigcup_{r=1}^{\infty} E^{n}\left(\lim _{k \rightarrow \infty} \overline{D} t_{k} \geq \frac{1}{r}\right)=\bigcup_{r=1}^{\infty} A_{r}$

and

$m \bigcup_{r=1}^{\infty} A_{r}=0.$

Hence

$\lim _{k \rightarrow \infty} \overline{D} t_{k} \leq 0 \quad \text {a.e.}$

As

$\overline{D} t_{k} \geq \underline{D} t_{k} \geq 0$

(see above), we get

$\lim _{k \rightarrow \infty} \overline{D} t_{k}=0=\lim _{k \rightarrow \infty} D t_{k} \quad \text { a.e. on } E^{n}.$

Now, as $$t_{k}=s-s_{k}$$ and as the $$s_{k}$$ are differentiable, Lemma 3 yields

$\overline{D} t_{k}=\overline{D} s-s_{k}^{\prime} \text { and } \underline{D} t_{k}=\underline{D} s-s_{k}^{\prime} \quad \text {a.e.}$

Thus

$\lim _{k \rightarrow \infty}\left(\overline{D} s-s_{k}^{\prime}\right)=0=\lim \left(\underline{D} s-s_{k}^{\prime}\right),$

i.e.,

$\overline{D} s=\lim _{k \rightarrow \infty} s_{k}^{\prime}=\underline{D} s \quad \text {a.e.}$

This settles the case $$s_{k} \nearrow s$$.

In the case $$s_{k} \searrow s,$$ one only has to set $$t_{k}=s_{k}-s$$ and proceed as before. (Verify!)$$\quad \square$$

## Lemma $$\PageIndex{6}$$

Given $$A \in \mathcal{M}^{*}, m A<\infty,$$ let

$s=\int C_{A} dm$

on $$\mathcal{M}^{*}.$$ Then $$s$$ is a.e. differentiable, and

$s^{\prime}=C_{A} \text { a.e. on } E^{n}.$

$$\left(C_{A}=\text { characteristic function of } A.\right)$$

Proof

First, let $$A$$ be open and let $$\overline{p} \in A$$.

Then $$A$$ contains some $$G_{\overline{p}}(\delta)$$ and hence also all cubes $$I \in \overline{\mathcal{K}}$$ with $$d I<\delta$$ and $$\overline{p} \in I.$$

Thus for such $$I \in \overline{\mathcal{K}}$$,

$s I=\int_{I} C_{A} d m=\int_{I}(1) d m=m I;$

i.e.,

$\frac{s I}{m I}=1=C_{A}(\overline{p}), \quad \overline{p} \in A.$

Hence by Definition 1 of Chapter 7, §12,

$s^{\prime}(\overline{p})=1=C_{A}(\overline{p})$

if $$\overline{p} \in A;$$ i.e., $$s^{\prime}=C_{A}$$ on $$A$$.

We calim that

$\overline{D} s=s^{\prime}=0 \quad \text {a.e. on } -A.$

To prove it, note that

$s=\int C_{A} dm$

is a finite (why?) $$m$$-continuous measure on $$\mathcal{M}^{*}$$. By Lemma 4, $$s$$ is strongly regular. Also, as $$s I \geq 0$$ for any $$I \in \overline{\mathcal{K}},$$ we certainly have

$\overline{D} s \geq \underline{D} s \geq 0.$

(Why?) Now let

$B=E^{n}(\overline{D} s>0)=\bigcup_{r=1}^{\infty} B_{r},$

where

$B_{r}=E^{n}\left(\overline{D} s \geq \frac{1}{r}\right), \quad r=1,2, \ldots.$

We have to show that $$m(B-A)=0.$$

Suppose

$m(B-A)>0.$

Then by (2), we must have $$m\left(B_{r}-A\right)>0$$ for at least one $$B_{r};$$ we fix this $$B_{r}$$ Also, by (3),

$\overline{D} s \geq \frac{1}{r} \text { on } B_{r}-A$

(even on all of $$B_{r}$$). Thus by Lemma 2,

$0<\frac{1}{r} m\left(B_{r}-A\right) \leq s\left(B_{r}-A\right)=\int_{B_{r}-A} C_{A} dm.$

But this is impossible. Indeed, as $$C_{A}=0$$ on $$-A$$ (hence on $$B_{r}-A$$),the integral in (4) cannot be $$>0.$$ This refutes the assumption $$m(B-A)>0;$$ so by (2),

$m\left(E^{n}(\overline{D} s>0)-A\right)=0;$

i.e.,

$\overline{D} s=0=\underline{D} s \quad \text { a.e. on } -A.$

We see that

$s^{\prime}=0=C_{A} \quad \text { a.e. on } -A,$

and

$s^{\prime}=1=C_{A} \quad \text { on } A,$

proving the lemma for open sets $$A.$$

Now take any $$A \in \mathcal{M}^{*}, m A<\infty.$$ As Lebesgue measure is regular (Chapter 7, §8, Theorem 3(b)), we find for each $$k \in N$$ an open set $$G_{k} \supseteq A,$$ with

$m\left(G_{k}-A\right)<\frac{1}{k} \text { and } G_{k} \supseteq G_{k+1}.$

Let

$s_{k}=\int C_{G_{k}} dm.$

Then $$s_{k} \searrow s$$ on $$\mathcal{M}^{*}$$ (see Problem 5 (ii) in §6). Also, by what was shown above, the $$s_{k}$$ are differentiable, with $$s_{k}^{\prime}=C_{G_{k}}$$ a.e.

Hence by Lemma 5,

$\overline{D} s=\underline{D} s=\lim _{k \rightarrow \infty} C_{G_{k}}=C_{A} \text { (a.e.).}$

The lemma is proved.$$\quad \square$$

## Theorem $$\PageIndex{1}$$

Let $$f : E^{n} \rightarrow E^{*}\left(E^{r}, C^{r}\right)$$ be $$m$$-integrable, at least on each cube in $$E^{n}.$$ Then the set function

$s=\int f dm$

is differentiable, with $$s^{\prime}=f,$$ a.e. on $$E^{n}.$$

Thus $$s^{\prime}$$ is the $$RN$$-derivative of $$s$$ with respect to Lebesgue measure $$m$$ (Theorem 1 in §11).

Proof

As $$E^{n}$$ is a countable union of cubes (Lemma 2 in Chapter 7, §2), it suffices to show that $$s^{\prime}=f$$ a.e. on each open cube $$J,$$ with $$s$$ differentiable a.e. on $$J.$$

Thus fix such a $$J \neq \emptyset$$ and restrict $$s$$ and $$m$$ to

$\mathcal{M}_{0}=\left\{X \in \mathcal{M}^{*} | X \subseteq J\right\}.$

This does not affect $$s^{\prime}$$ on $$J;$$ for as $$J$$ is open, any sequence of cubes

$I_{k} \rightarrow \overline{p} \in J$

terminates inside $$J$$ anyway.

When so restricted,

$s=\int f$

is a generalized measure in $$J;$$ for $$\mathcal{M}_{0}$$ is a $$\sigma$$-ring (verify!), and $$f$$ is integrable on $$J.$$ Also, $$m$$ is strongly regular, and $$s$$ is $$m$$-continuous.

First, suppose $$f$$ is $$\mathcal{M}_{0}$$-simple on $$J,$$ say,

$f=\sum_{i=1}^{q} a_{i} C_{A_{i}},$

say, with $$0<a_{i}<\infty, A_{i} \in \mathcal{M}^{*},$$ and

$J=\bigcup_{i=1}^{q} A_{i} \text { (disjoint).}$

Then

$s=\int f=\sum_{i=1}^{q} a_{i} \int C_{A_{i}}.$

Hence by Lemma 6 above and by Theorem 1 in Chapter 7, §12, $$s$$ is differentiable a.e. (as each $$\int C_{A_{i}}$$ is), and

$s^{\prime}=\sum_{i=1}^{q} a_{i}\left(\int C_{A_{i}}\right)^{\prime}=\sum_{i=1}^{q} a_{i} C_{A_{i}}=f \text { (a.e.),}$

as required.

The general case reduces (via components and the formula $$f=f^{+}-f^{-}$$) to the case $$f \geq 0,$$ with $$f$$ measurable (even integrable) on $$J.$$

By Problem 6 in §2, then, we have $$f_{k} \nearrow f$$ for some simple maps $$f_{k} \geq 0.$$ Let

$s_{k}=\int f_{k} \text { on } M_{0}, k=1,2, \ldots.$

Then all $$s_{k}$$ and $$s=\int f$$ are finite measures and $$s_{k} \nearrow s,$$ by Theorem 4 in §6. Also, by what was shown above, each $$s_{k}$$ is differentiable a.e. on $$J,$$ with $$s_{k}^{\prime}=f_{k}$$ (a.e.). Thus as in Lemma 5,

$\overline{D} s=\underline{D} s=s^{\prime}=\lim _{k \rightarrow \infty} s_{k}^{\prime}=\lim f_{k}=f \text { (a.e.) on } J,$

with $$s^{\prime}=f \neq \pm \infty$$ (a.e.), as $$f$$ is integrable on $$J.$$ Thus all is proved.$$\quad \square$$

II. So far we have considered Lebesgue $$(\overline{\mathcal{K}})$$ differentiation. However, our results easily extend to $$\Omega$$-differentiation (Definition 2 in Chapter 7, §12).

The proof is even simpler. Thus in Lemma 1, the union in formula (1) is countable (as $$\overline{\mathcal{K}}$$ is replaced by the countable set family $$\Omega$$); hence it is $$\mu$$-measurable. In Lemma 2, the use of the Vitali theorem is replaced by Theorem 3 in Chapter 7, §12. Otherwise, one only has to replace Lebesgue measure $$m$$ by $$\mu$$ on $$\mathcal{M}.$$ Once the lemmas are established (reread the proofs!), we obtain the following.

## Theorem $$\PageIndex{2}$$

Let $$S, \rho, \Omega,$$ and $$\mu : \mathcal{M} \rightarrow E^{*}$$ be as in Definition 2 of Chapter 7, §12. Let $$f : S \rightarrow E^{*}\left(E^{r}, C^{r}\right)$$ be $$mu$$-integrable on each $$A \in \mathcal{M}$$ with $$\mu A<\infty.$$

Then the set function

$s=\int f d \mu$

is $$\Omega$$-differentiable, with $$s^{\prime}=f,$$ (a.e.) on $$S$$.

Proof

Recall that $$S$$ is a countable union of sets $$U_{n}^{i} \in \Omega$$ with $$0<\mu U_{n}^{i}<\infty.$$ As $$\mu^{*}$$ is $$\mathcal{G}$$-regular, each $$U_{n}^{i}$$ lies in an open set $$J_{n}^{i} \in \mathcal{M}$$ with

$\mu J_{n}^{i}<\mu U_{n}^{i}+\varepsilon_{n}^{i}<\infty.$

Also, $$f$$ is $$\mu$$-measurable (even integrable) on $$J_{n}^{i}.$$ Dropping a null set, assume that $$f$$ is $$\mathcal{M}$$-measurable on $$J=J_{n}^{i}$$.

From here, proceed exactly as in Theorem 1, replacing $$m$$ by $$\mu.\quad \square$$

Both theorems combined yield the following result.

## Corollary $$\PageIndex{1}$$

If $$s : \mathcal{M}^{\prime} \rightarrow E^{*}\left(E^{r}, C^{r}\right)$$ is an $$m$$-continuous and $$m$$-finite generalized measure in $$E^{n},$$ then $$s$$ is $$\overline{\mathcal{K}}$$-differentiable a.e. on $$E^{n},$$ and $$d s=s^{\prime} d m$$ (see Definition 3 in §10) in any $$A \in \mathcal{M}^{*}(m A<\infty).$$

Similarly for $$\Omega$$-differentiation.

Proof

Given $$A \in \mathcal{M}^{*}(m A<\infty),$$ there is an open set $$J \supseteq A$$ such that

$mJ<mA+\varepsilon<\infty.$

As before, restrict $$s$$ and $$m$$ to

$\mathcal{M}_{0}=\left\{X \in \mathcal{M}^{*} | X \subseteq J\right\}.$

Then by assumption, $$s$$ is finite and $$m$$-continuous on $$\mathcal{M}_{0}$$ (a $$\sigma$$-ring); so by Theorem 1 in §11,

$s=\int f dm$

on $$\mathcal{M}_{0}$$ for some $$m$$-integrable map $$f$$ on $$J$$.

Hence by our present Theorem 1, $$s$$ is differentiable, with $$s^{\prime}=f$$ a.e. on $$J$$ and so

$s=\int f=\int s^{\prime} \text { on } \mathcal{M}_{0}.$

This implies $$d s=s^{\prime} d m$$ in $$A$$.

For $$\Omega$$-differentiation, use Theorem 2.$$\quad \square$$

## Corollary $$\PageIndex{2}$$ (change of measure)

Let $$s$$ be as in Corollary 1. Subject to Note 1 in §10, if $$f$$ is $$s$$-integrable on $$A \in \mathcal{M}^{*}(m A<\infty),$$ then $$f s^{\prime}$$ is $$m$$-integrable on $$A$$ and

$\int_{A} f d s=\int_{A} f s^{\prime} dm.$

Similarly for $$\Omega$$-derivatives, with $$m$$ replaced by $$\mu$$.

Proof

By Corollary 1, $$d s=s^{\prime} d m$$ in $$A.$$ Thus Theorem 6 of §10 yields the result.$$\quad \square$$

Note 1. In particular, Corollary 2 applies to $$m$$-continuous signed LS measures $$s=s_{\alpha}$$ in $$E^{1}$$ (see end of §11). If $$A=[a, b],$$ then $$s_{\alpha}$$ is surely finite on $$s_{\alpha}$$-measurable subsets of $$A;$$ so Corollaries 1 and 2 show that

$\int_{A} f ds_{\alpha}=\int_{A} f s_{\alpha}^{\prime} dm=\int_{A} f \alpha^{\prime} dm,$

since $$s_{\alpha}^{\prime}=\alpha^{\prime}.$$ (See Problem 9 in Chapter 7, §12.)

Note 2. Moreover, $$s=s_{\alpha}$$ (see Note 1) is absolutely $$m$$-continuous iff $$\alpha$$ is absolutely continuous in the stronger sense (Problem 2 in Chapter 4, §8).

Indeed, assuming the latter, fix $$\varepsilon>0$$ and choose $$\delta$$ as in Definition 3 of Chapter 7, §11. Then if $$m X<\delta,$$ we have

$X \subseteq \bigcup I_{k} \text { (disjoint)}$

for some intervals $$I_{k}=\left(a_{k}, b_{k}\right],$$ with

$\delta>\sum m I_{k}=\sum\left(b_{k}-a_{k}\right).$

Hence

$|s X| \leq \sum\left|s I_{k}\right|<\varepsilon.$

(Why?) Similarly for the converse.

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