8.12: Integration and Differentiation

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.

We start with several lemmas that go back to Lebesgue.

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.