# 9.2: More on L-Integrals and Absolute Continuity

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- Elias Zakon
- University of Windsor via The Trilla Group (support by Saylor Foundation)

**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.\(\quad \square\)