9.2: More on L-Integrals and Absolute Continuity

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.$\quad \square$