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# 9.2: More on L-Integrals and Absolute Continuity

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I. In this section, we presuppose the "starred" §10 in Chapter 7. First, however, we add some new ideas that do not require any starred material. The notation is as in §1.

## Definition

Given $$F : E^{1} \rightarrow E, p \in E^{1},$$ and $$q \in E,$$ we write

$q \sim D F(p)$

and call $$q$$ an $$F$$-derivate at $$p$$ iff

$q=\lim _{k \rightarrow \infty} \frac{F\left(x_{k}\right)-F(p)}{x_{k}-p}$

for at least one sequence $$x_{k} \rightarrow p\left(x_{k} \neq p\right)$$.

If $$F$$ has a derivative at $$p,$$ it is the only $$F$$-derivate at $$p;$$ otherwise, there may be many derivates at $$p$$ (finite or not).

Such derivates must exist if $$E=E^{1}\left(E^{*}\right).$$ Indeed, given any $$p \in E^{1},$$ let

$x_{k}=p+\frac{1}{k} \rightarrow p;$

let

$y_{k}=\frac{F\left(x_{k}\right)-F(p)}{x_{k}-p}, \quad k=1,2, \ldots$

By the compactness of $$E^{*}$$ (Chapter 4, §6, example (d)), $$\left\{y_{k}\right\}$$ must have a subsequence $$\left\{y_{k_{i}}\right\}$$ with a limit $$q \in E^{*}$$ (e.g., take $$q=\underline{\lim} y_{k}$$), and so $$q \sim D F(p)$$.

We also obtain the following lemma.

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

If $$F : E^{1} \rightarrow E^{*}$$ has no negative derivates on $$A-Q,$$ where $$A=$$ $$[a, b]$$ and $$m Q=0,$$ and if no derivate of $$F$$ on $$A$$ equals $$-\infty,$$ then $$F \uparrow$$ on $$A$$.

Proof

First, suppose $$F$$ has no negative derivates on $$A$$ at all. Fix $$\varepsilon>0$$ and set

$G(x)=F(x)+\varepsilon x.$

Seeking a contradiction, suppose $$a \leq p<q \leq b,$$ yet $$G(q)<G(p).$$ Then if

$r=\frac{1}{2}(p+q),$

one of the intervals $$[p, r]$$ and $$[r, q]$$ (call it $$[p_{1}, q_{1}]$$) satisfies $$G\left(q_{1}\right)<G\left(p_{1}\right)$$.

Let

$r_{1}=\frac{1}{2}\left(p_{1}+q_{1}\right).$

Again, one of $$[p_{1}, r_{1}]$$ and $$[r_{1}, q_{1}]$$ (call it $$[p_{2}, q_{2}]$$) satisfies $$G\left(q_{2}\right)<G\left(p_{2}\right).$$ Let

$r_{2}=\frac{1}{2}\left(p_{2}+q_{2}\right),$

and so on.

Thus obtain contracting intervals $$\left[p_{n}, q_{n}\right],$$ with

$G\left(q_{n}\right)<G\left(p_{n}\right), \quad n=1,2, \ldots.$

Now, by Theorem 5 of Chapter 4, §6, let

$p_{o} \in \bigcap_{n=1}^{\infty}\left[p_{n}, q_{n}\right].$

Then set $$x_{n}=q_{n}$$ if $$G\left(q_{n}\right)<G\left(p_{o}\right),$$ and $$x_{n}=p_{n}$$ otherwise. Then

$\frac{G\left(x_{n}\right)-G\left(p_{o}\right)}{x_{n}-p_{o}}<0$

and $$x_{n} \rightarrow p_{o}.$$ By the compactness of $$E^{*},$$ fix a subsequence

$\frac{G\left(x_{n_{k}}\right)-G\left(p_{o}\right)}{x_{n_{k}}-p_{o}} \rightarrow c \in E^{*},$

say. Then $$c \leq 0$$ is a $$G$$-derivate at $$p_{o} \in A$$.

But this is impossible; for by our choice of $$G$$ and our assumption, all derivates of $$G$$ are $$>0.$$ (Why?)

This contradiction shows that $$a \leq p<q \leq b$$ implies $$G(p) \leq G(q),$$ i.e.,

$F(p)+\varepsilon p \leq F(q)+\varepsilon q.$

Making $$\varepsilon \rightarrow 0,$$ we obtain $$F(p) \leq F(q)$$ when $$a \leq p<q \leq b,$$ i.e., $$F \uparrow$$ on $$A$$.

Now, for the general case, let $$Q$$ be the set of all $$p \in A$$ that have at least one $$D F(p)<0;$$ so $$m Q=0$$.

Let $$g$$ be as in Problem 8 of §1; so $$g^{\prime}=\infty$$ on $$Q.$$ Given $$\varepsilon>0,$$ set

$G=F+\varepsilon g.$

As $$g \uparrow,$$ we have

$(\forall x, p \in A) \quad \frac{G(x)-G(p)}{x-p} \geq \frac{F(x)-F(p)}{x-p}.$

Hence $$D G(p) \geq 0$$ if $$p \notin Q$$.

If, however, $$p \in Q,$$ then $$g^{\prime}(p)=\infty$$ implies $$D G(p) \geq 0.$$ (Why?) Thus all $$D G(p)$$ are $$\geq 0;$$ so by what was proved above, $$G \uparrow$$ on $$A.$$ It follows, as before, that $$F \uparrow$$ on $$A,$$ also. The lemma is proved.$$\quad \square$$

We now proceed to prove Theorems 3 and 4 of §1. To do this, we shall need only one "starred" theorem (Theorem 3 of Chapter 7, §10).

Proof of Theorem 3 of §1. (1) First, let $$f$$ be bounded:

$|f| \leq K \quad \text {on } A.$

Via components and by Corollary 1 of Chapter 8, §6, all reduces to the real positive case $$f \geq 0$$ on $$A.$$ (Explain!)

Then (Theorem 1(f) of Chapter 8, §5) $$a \leq x<y \leq b$$ implies

$L \int_{a}^{x} f \leq L \int_{a}^{y} f,$

i.e., $$F(x) \leq F(y);$$ so $$F \uparrow$$ and $$F^{\prime} \geq 0$$ on $$A$$.

Now, by Theorem 3 of Chapter 7, §10, $$F$$ is a.e. differentiable on $$A.$$ Thus exactly as in Theorem 2 in §1, we set

$f_{n}(t)=\frac{F\left(t+\frac{1}{n}\right)-F(t)}{\frac{1}{n}} \rightarrow F^{\prime}(t) \text { a.e.}$

Since all $$f_{n}$$ are $$m$$-measurable on $$A$$ (why?), so is $$F^{\prime}$$. Moreover, as $$|f| \leq K$$, we obtain (as in Lemma 1 of §1)

$\left|f_{n}(x)\right|=n\left(L \int_{x}^{x+1 / n} f\right) \leq n \cdot \frac{K}{n}=K.$

Thus by Theorem 5 from Chapter 8, §6 (with $$g=K$$),

$L \int_{a}^{x} F^{\prime}=\lim _{n \rightarrow \infty} L \int_{a}^{x} f_{n}=L \int_{a}^{x} f$

(Lemma 1 of §1). Hence

$L \int_{a}^{x}\left(F^{\prime}-f\right)=0, \quad x \in A,$

and so (Problem 10 in §1) $$F^{\prime}=f$$ (a.e.) as claimed.

(2) If $$f$$ is not bounded, we still can reduce all to the case $$f \geq 0, f : E^{1} \rightarrow E^{*}$$ so that $$F \uparrow$$ and $$F^{\prime} \geq 0$$ on $$A.$$

If so, we use "truncation": For $$n=1,2, \ldots,$$ set

$g_{n}=\left\{\begin{array}{ll}{f} & {\text { on } A(f \leq n), \text { and }} \\ {0} & {\text { elsewhere.}}\end{array}\right.$

Then (see Problem 12 in §1) the $$g_{n}$$ are $$L$$-measurable and bounded, hence $$L$$-integrable on $$A,$$ with $$g_{n} \rightarrow f$$ and

$0 \leq g_{n} \leq f$

on $$A.$$ By the first part of the proof, then,

$\frac{d}{d x} L \int_{a}^{x} g_{n}=g_{n} \quad \text { a.e. on } A, n=1,2, \ldots.$

Also, set $$(\forall n)$$

$F_{n}(x)=L \int_{a}^{x}\left(f-g_{n}\right) \geq 0;$

so $$F_{n}$$ is monotone ($$\uparrow$$) on $$A.$$ (Why?)

Thus by Theorem 3 in Chapter 7, §10, each $$F_{n}$$ has a derivative at almost every $$x \in A,$$

$F_{n}^{\prime}(x)=\frac{d}{d x}\left(L \int_{a}^{x} f-L \int_{a}^{x} g_{n}\right)=F^{\prime}(x)-g_{n}(x) \geq 0 \quad \text {a.e. on } A.$

Making $$n \rightarrow \infty$$ and recalling that $$g_{n} \rightarrow f$$ on $$A,$$ we obtain

$F^{\prime}(x)-f(x) \geq 0 \quad \text {a.e. on } A.$

Thus

$L \int_{a}^{x}\left(F^{\prime}-f\right) \geq 0.$

But as $$F \uparrow$$ (see above), Problem 11 of §1 yields

$L \int_{a}^{x} F^{\prime} \leq F(x)-F(a)=L \int_{a}^{x} f;$

so

$L \int_{a}^{x}\left(F^{\prime}-f\right)=L \int_{a}^{x} F^{\prime}-L \int_{a}^{x} f \leq 0.$

Combining, we get

$(\forall x \in A) \quad L \int_{a}^{x}\left(F^{\prime}-f\right)=0;$

so by Problem 10 of §1, $$F^{\prime}=f$$ a.e. on $$A,$$ as required.$$\quad \square$$

Proof of Theorem 4 of §1. Via components, all again reduces to a real $$f$$.

Let $$(\forall n)$$

$g_{n}=\left\{\begin{array}{ll}{f} & {\text { on } A(f \leq n),} \\ {0} & {\text { on } A(f>n);}\end{array}\right.$

so $$g_{n} \rightarrow f$$ (pointwise), $$g_{n} \leq f, g_{n} \leq n,$$ and $$\left|g_{n}\right| \leq|f|$$.

This makes each $$g_{n}$$ $$L$$-integrable on $$A.$$ Thus as before, by Theorem 5 of Chapter 8, §6,

$\lim _{n \rightarrow \infty} L \int_{a}^{x} g_{n}=L \int_{a}^{x} f, \quad x \in A.$

Now, set

$F_{n}(x)=F(x)-L \int_{a}^{x} g_{n}.$

Then by Theorem 3 of §1 (already proved),

$F_{n}^{\prime}(x)=F^{\prime}(x)-\frac{d}{d x} L \int_{a}^{x} g_{n}=f(x)-g_{n}(x) \geq 0 \quad \text {a.e. on } A$

(since $$g_{n} \leq f$$).

Thus $$F_{n}$$ has solely nonnegative derivates on $$A-Q(m Q=0).$$ Also, as $$g_{n} \leq n,$$ we get

$\frac{1}{x-p} L \int_{a}^{x} g_{n} \leq n,$

even if $$x<p.$$ (Why?) Hence

$\frac{\Delta F_{n}}{\Delta x} \geq \frac{\Delta F}{\Delta x}-n,$

as

$F_{n}(x)=F(x)-L \int_{a}^{x} g_{n}.$

Thus none of the $$F_{n}$$-derivates on $$A$$ can be $$-\infty$$.

By Lemma 1, then, $$F_{n}$$ is monotone ($$\uparrow$$) on $$A;$$ so $$F_{n}(x) \geq F_{n}(a),$$ i.e.,

$F(x)-L \int_{a}^{x} g_{n} \geq F(a)-L \int_{a}^{a} g_{n}=F(a),$

or

$F(x)-F(a) \geq L \int_{a}^{x} g_{n}, \quad x \in A, n=1,2, \ldots.$

Hence by (1),

$F(x)-F(a) \geq L \int_{a}^{x} f, \quad x \in A.$

For the reverse inequality, apply the same formula to $$-f.$$ Thus we obtain the desired result:

$F(x)=F(a)+L \int_{a}^{x} f \quad \text { for } x \in A. \quad \square$

Note 1. Formula (2) is equivalent to $$F=L \int f$$ on $$A$$ (see the last part of §1). For if (2) holds, then

$F(x)=c+L \int_{a}^{x} f,$

with $$c=F(a);$$ so $$F=L \int f$$ by definition.

Conversely, if

$F(x)=c+L \int_{a}^{x} f,$

set $$x=a$$ to find $$c=F(a)$$.

II. A bsolute continuity redefined.

## Definition

A map $$f : E^{1} \rightarrow E$$ is absolutely continuous on an interval $$I \subseteq E^{1}$$ iff for every $$\varepsilon>0,$$ there is $$\delta>0$$ such that

$\sum_{i=1}^{r}\left(b_{i}-a_{i}\right)<\delta \text { implies } \sum_{i=1}^{r}\left|f\left(b_{i}\right)-f\left(a_{i}\right)\right|<\varepsilon$

for any disjoint intervals $$\left(a_{i}, b_{i}\right),$$ with $$a_{i}, b_{i} \in I$$.

From now on, this replaces the "weaker" definition given in Chapter 5, §8. The reader will easily verify the next three "routine" propositions.

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

If $$f, g, h : E^{1} \rightarrow E^{*}(C)$$ are absolutely continuous on $$A=[a, b]$$ so are

$f \pm g, h f, \text { and }|f|.$

So also is $$f / h$$ if

$(\exists \varepsilon>0) \quad|h| \geq \varepsilon \text { on } A.$

All this also holds if $$f, g : E^{1} \rightarrow E$$ are vector valued and $$h$$ is scalar valued. Finally, if $$E \subseteq E^{*},$$ then

$f \vee g, f \wedge g, f^{+}, \text {and } f^{-}$

are absolutely continuous along with $$f$$ and $$g.$$

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

A function $$F : E^{1} \rightarrow E^{n}\left(C^{n}\right)$$ is absolutely continuous on $$A=$$ $$[a, b]$$ iff all its components $$F_{1}, \ldots, F_{n}$$ are.

Hence a complex function $$F : E^{1} \rightarrow C$$ is absolutely continuous iff its real and imaginary parts, $$F_{re}$$ and $$F_{im},$$ are.

## Corollary $$\PageIndex{2}$$

If $$f : E^{1} \rightarrow E$$ is absolutely continuous on $$A=[a, b],$$ it is bounded, is uniformly continuous, and has bounded variation, $$V_{f}[a, b]<\infty$$ all on $$A.$$

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

If $$F : E^{1} \rightarrow E^{n}\left(C^{n}\right)$$ is of bounded variation on $$A=[a, b],$$ then

(i) $$F$$ is a.e. differentiable on $$A,$$ and

(ii) $$F^{\prime}$$ is $$L$$-integrable on $$A$$.

Proof

Via components (Theorem 4 of Chapter 5, §7), all reduces to the real case, $$F : E^{1} \rightarrow E^{1}$$.

Then since $$V_{F}[A]<\infty,$$ we have

$F=g-h$

for some nondecreasing $$g$$ and $$h$$ (Theorem 3 in Chapter 5, §7).

Now, by Theorem 3 from Chapter 7, §10, $$g$$ and $$h$$ are a.e. differentiable on $$A.$$ Hence so is

$g-h=F.$

Moreover, $$g^{\prime} \geq 0$$ and $$h^{\prime} \geq 0$$ since $$g \uparrow$$ and $$h \uparrow$$.

Thus for the $$L$$-integrability of $$F^{\prime},$$ proceed as in Problem 11 in §1, i.e., show that $$F^{\prime}$$ is measurable on $$A$$ and that

$L \int_{a}^{b} F^{\prime}=L \int_{a}^{b} g^{\prime}-L \int_{a}^{b} h^{\prime}$

is finite. This yields the result.$$\quad \square$$

## Theorem $$\PageIndex{2}$$ (Lebesgue)

If $$F : E^{1} \rightarrow E^{n}\left(C^{n}\right)$$ is absolutely continuous on $$A=[a, b],$$ then the following are true:

(i*) $$F$$ is a.e. differentiable, and $$F^{\prime}$$ is $$L$$-integrable, on $$A$$.

(ii*) If, in addition, $$F^{\prime}=0$$ a.e. on $$A,$$ then $$F$$ is constant on $$A$$.

Proof

Assertion (i*) is immediate from Lemma 2, since any absolutely continuous function is of bounded variation by Corollary 2.

(ii*) Now let $$F^{\prime}=0$$ a.e. on $$A.$$ Fix any

$B=[a, c] \subseteq A$

and let $$Z$$ consist of all $$p \in B$$ at which the derivative $$F^{\prime}=0$$.

Given $$\varepsilon>0,$$ let $$\mathcal{K}$$ be the set of all closed intervals $$[p, x], p<x,$$ such that

$\left|\frac{\Delta F}{\Delta x}\right|=\left|\frac{F(x)-F(p)}{x-p}\right|<\varepsilon.$

By assumption,

$\lim _{x \rightarrow p} \frac{\Delta F}{\Delta x}=0 \quad(p \in Z),$

and $$m(B-Z)=0; B=[a, c] \in \mathcal{M}^{*}.$$ If $$p \in Z,$$ and $$x-p$$ is small enough, then

$\left|\frac{\Delta F}{\Delta x}\right|<\varepsilon,$

i.e., $$[p, x] \in \mathcal{K}$$.

It easily follows that $$\mathcal{K}$$ covers $$Z$$ in the Vitali sense (verify!); so for any
$$\delta>0,$$ Theorem 2 of Chapter 7, §10 yields disjoint intervals

$I_{k}=\left[p_{k}, x_{k}\right] \in \mathcal{K}, I_{k} \subseteq B,$

with

$m^{*}\left(Z-\bigcup_{k=1}^{q} I_{k}\right)<\delta,$

hence also

$m\left(B-\bigcup_{k=1}^{q} I_{k}\right)<\delta$

(for $$m(B-Z)=0$$). But

\begin{aligned} B-\bigcup_{k=1}^{q} I_{k} &=[a, c]-\bigcup_{k=1}^{q-1}\left[p_{k}, x_{k}\right] \\ &=\left[a, p_{1}\right) \cup \bigcup_{k=1}^{q-1}\left[x_{k}, p_{k+1}\right) \cup\left[x_{q}, c\right] \quad\left(\text { if } x_{k}<p_{k}<x_{k+1}\right); \end{aligned}

so

$m\left(B-\bigcup_{k=1}^{q} I_{k}\right)=\left(p_{1}-a\right)+\sum_{k=1}^{q-1}\left(p_{k+1}-x_{k}\right)+\left(c-x_{q}\right)<\delta.$

Now, as $$F$$ is absolutely continuous, we can choose $$\delta>0$$ so that (3) implies

$\left|F\left(p_{1}\right)-F(a)\right|+\sum_{k=1}^{q-1}\left|F\left(p_{k+1}\right)-F\left(x_{k}\right)\right|+\left|F(c)-F\left(x_{q}\right)\right|<\varepsilon.$

But $$I_{k} \in \mathcal{K}$$ also implies

$\left|F\left(x_{k}\right)-F\left(p_{k}\right)\right|<\varepsilon\left(x_{k}-p_{k}\right)=\varepsilon \cdot m I_{k}.$

Hence

$\left|\sum_{k=1}^{q}\left[F\left(x_{k}\right)-F\left(p_{k}\right)\right]\right|<\varepsilon \sum_{k=1}^{q} m I_{k} \leq \varepsilon \cdot m B=\varepsilon(c-p).$

Combining with (4), we get

$|F(c)-F(a)| \leq \varepsilon(1+c-a) \rightarrow 0 \text { as } \varepsilon \rightarrow 0;$

so $$F(c)=F(a).$$ As $$c \in A$$ was arbitrary, $$F$$ is constant on $$A,$$ as claimed.$$\quad \square$$

Note 2. This shows that Cantor's function (Problem 6 of Chapter 4, §5) is not absolutely continuous, even though it is continuous and monotone, hence of bounded variation on $$[0,1].$$ Indeed (see Problem 2 in §1), it has a zero derivative a.e. on $$[0,1]$$ but is not constant there. Thus absolute continuity, as now defined, differs from its "weak" counterpart (Chapter 5, §8).

## Theorem $$\PageIndex{3}$$

A map $$F : E^{1} \rightarrow E^{1}\left(C^{n}\right)$$ is absolutely continuous on $$A=$$ $$[a, b]$$ iff

$F=L \int f \quad \text { on } A$

for some function $$f;$$ and then

$F(x)=F(a)+L \int_{a}^{x} f, \quad x \in A.$

Briefly: Absolutely continuous maps are exactly all $$L$$-primitives.

Proof

If $$F=L \int f,$$ then by Theorem 1 of §1, $$F$$ is absolutely continuous on $$A,$$ and by Note 1,

$F(x)=F(a)+L \int_{a}^{x} f, \quad x \in A.$

Conversely, if $$F$$ is absolutely continuous, then by Theorem 2, it is a.e. differentiable and $$F^{\prime}=f$$ is $$L$$-integrable (all on $$A$$). Let

$H(x)=L \int_{a}^{x} f, \quad x \in A.$

Then $$H,$$ too, is absolutely continuous and so is $$F-H.$$ Also, by Theorem 3 of §1,

$H^{\prime}=f=F^{\prime},$

and so
$(F-H)^{\prime}=0 \quad \text {a.e. on } A.$

By Theorem 2, $$F-H=c;$$ i.e.,

$F(x)=c+H(x)=c+L \int_{a}^{x} f,$

and so $$F=L \int f$$ on $$A,$$ as claimed.$$\quad \square$$

## Corollary $$\PageIndex{3}$$

If $$f, F : E^{1} \rightarrow E^{*}\left(E^{n}, C^{n}\right),$$ we have

$F=L \int f$

on an interval $$I \subseteq E^{1}$$ iff $$F$$ is absolutely continuous on $$I$$ and $$F^{\prime}=f$$ a.e. on $$I$$.

(Use Problem 3 in §1 and Theorem 3.)

Note 3. This (or Theorem 3) could serve as a definition. Comparing ordinary primitives

$F=\int f$

with $$L$$-primitives

$F=L \int f,$

we see that the former require $$F$$ to be just relatively continuous but allow only a countable "exceptional" set $$Q,$$ while the latter require absolute continuity but allow $$Q$$ to even be uncountable, provided $$m Q=0$$.

The simplest and "strongest" kind of absolutely continuous functions are so-called Lipschitz maps (see Problem 6). See also Problems 7 and 10.

III. We conclude with another important idea, due to Lebesgue.

## Definition

We call $$p \in E^{1}$$ a Lebesgue point ("$$L$$-point") of $$f : E^{1} \rightarrow E$$ iff

(i) $$f$$ is $$L$$-integrable on some $$G_{p}(\delta)$$;

(ii) $$q=f(p)$$ is finite; and

(iii) $$\lim _{x \rightarrow p} \frac{1}{x-p} L \int_{p}^{x}|f-q|=0$$.

The Lebesgue set of $$f$$ consists of all such $$p$$.

## Corollary $$\PageIndex{4}$$

Let

$F=L \int f \quad \text { on } A=[a, b].$

If $$p \in A$$ is an $$L$$-point of $$f,$$ then $$f(p)$$ is the derivative of $$F$$ at $$p$$ (but the converse fails).

Proof

By assumption,

$F(x)=c+L \int_{p}^{x} f, \quad x \in G_{p}(\delta),$

and

$\frac{1}{|\Delta x|}\left|L \int_{p}^{x}(f-q)\right| \leq \frac{1}{|\Delta x|} L \int_{p}^{x}|f-q| \rightarrow 0$

as $$x \rightarrow p.$$ (Here $$q=f(p)$$ and $$\Delta x=x-p$$.)

Thus with $$x \rightarrow p,$$ we get

\begin{aligned}\left|\frac{F(x)-F(p)}{x-p}-q\right| &=\frac{1}{|x-p|}\left|L \int_{p}^{x} f-(x-p) q\right| \\ &=\frac{1}{|x-p|}\left|L \int_{p}^{x} f-L \int_{p}^{x}(q)\right| \rightarrow 0, \end{aligned}

as required.$$\quad \square$$

## Corollary $$\PageIndex{5}$$

Let $$f : E^{1} \rightarrow E^{n}\left(C^{n}\right).$$ Then $$p$$ is an $$L$$-point of $$f$$ iff it is an $$L$$-point for each of the $$n$$ components, $$f_{1}, \ldots, f_{n},$$ of $$f$$.

Proof

(Exercise!)

## Theorem $$\PageIndex{4}$$

If $$f : E^{1} \rightarrow E^{*}\left(E^{n}, C^{n}\right)$$ is $$L$$-integrable on $$A=[a, b],$$ then almost all $$p \in A$$ are Lebesgue points of $$f.$$

Note that this strengthens Theorem 3 of §1.

Proof

By Corollary 5, we need only consider the case $$f : E^{1} \rightarrow E^{*}$$.

For any $$r \in E^{1},|f-r|$$ is $$L$$-integrable on $$A;$$ so by Theorem 3 of §1, setting

$F_{r}(x)=L \int_{a}^{x}|f-r|,$

we get

$F_{r}^{\prime}(p)=\lim _{x \rightarrow p} \frac{1}{|x-p|} L \int_{p}^{x}|f-r|=|f(p)-r|$

for almost all $$p \in A$$.

Now, for each $$r,$$ let $$A_{r}$$ be the set of those $$p \in A$$ for which (5) fails; so $$m A_{r}=0.$$ Let $$\left\{r_{k}\right\}$$ be the sequence of all rationals in $$E^{1}.$$ Let

$Q=\bigcup_{k=1}^{\infty} A_{r_{k}} \cup\{a, b\} \cup A_{\infty},$

where

$A_{\infty}=A(|f|=\infty);$

so $$m Q=0.$$ (Why?)

To finish, we show that all $$p \in A-Q$$ are $$L$$-points of $$f.$$ Indeed, fix any $$p \in A-Q$$ and any $$\varepsilon>0.$$ Let $$q=f(p).$$ Fix a rational $$r$$ such that

$|q-r|<\frac{\varepsilon}{3}.$

Then

$| | f-r|-| f-q| | \leq|(f-r)-(f-q)|=|q-r|<\frac{\varepsilon}{3} \text { on } A-A_{\infty}.$

Hence as $$m A_{\infty}=0,$$ we have

$\left|L \int_{p}^{x}\right| f-r\left|-L \int_{p}^{x}\right| f-q| | \leq L \int_{p}^{x}\left(\frac{\varepsilon}{3}\right)=\frac{\varepsilon}{3}|x-p|.$

Since

$p \notin Q \supseteq \bigcup_{k} A_{r_{k}},$

formula (5) applies. So there is $$\delta>0$$ such that $$|x-p|<\delta$$ implies

$\left|\left(\frac{1}{|x-p|} L \int_{p}^{x}|f-r|\right)-\right| f(p)-r| |<\frac{\varepsilon}{3}.$

As

$|f(p)-r|=|q-r|<\frac{\varepsilon}{3},$

we get

\begin{aligned} \frac{1}{|x-p|} L \int_{p}^{x}|f-r| & \leq\left|\left(\frac{1}{|x-p|} L \int_{p}^{x}|f-r|\right)-\right| q-r| |+|q-r| \\ &<\frac{\varepsilon}{3}+\frac{\varepsilon}{3}=\frac{2 \varepsilon}{3}. \end{aligned}

Hence

$L \int_{p}^{x}|f-r|<\frac{2 \varepsilon}{3}|x-p|.$

Combining with (6), we have

$\frac{1}{|x-p|} L \int_{p}^{x}|f-q|<\frac{\varepsilon}{3}+\frac{2 \varepsilon}{3}=\varepsilon$

whenever $$|x-p|<\delta.$$ Thus

$\lim _{x \rightarrow p} \frac{1}{|x-p|} L \int_{p}^{x}|f-q|=0,$

as required.$$\quad \square$$

9.2: More on L-Integrals and Absolute Continuity is shared under a CC BY license and was authored, remixed, and/or curated by LibreTexts.

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