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.
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 \(E^{n}. \quad \square\)
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.\(\quad \square\)
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).
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.\(\quad \square\)
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!)\(\quad \square\)
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.\(\quad \square\)
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.
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.
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.\(\quad \square\)
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.