
# 9.1: L-Integrals and Antiderivatives


I. Lebesgue theory makes it possible to strengthen many calculus theorems. We shall start with functions on $$E^{1}, f : E^{1} \rightarrow E.$$ (A reader who has omitted the "starred" part of Chapter 8, §7, will have to set $$E=E^{*}\left(E^{n}, C^{n}\right)$$ throughout.)

By $$L$$-integrals of such functions, we mean integrals with respect to Lebesgue measure $$m$$ in $$E^{1}.$$ Notation:

$L \int_{a}^{b} f=L \int_{a}^{b} f(x) d x=L \int_{[a, b]} f$

and

$L \int_{b}^{a} f=-L \int_{a}^{b} f.$

For Riemann integrals, we replace "$$L$$" by "$$R.$$" We compare such integrals with antiderivatives (Chapter 5, §5), denoted

$\int_{a}^{b} f,$

without the "$$L$$" or "$$R.$$" Note that

$L \int_{[a, b]} f=L \int_{(a, b)} f,$

etc., since $$m\{a\}=m\{b\}=0$$ here.

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

Let $$f : E^{1} \rightarrow E$$ be $$L$$-integrable on $$A=[a, b].$$ Set

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

Then the following are true.

(i) The function $$f$$ is the derivative of $$H$$ at any $$p \in A$$ at which $$f$$ is finite and continuous. (At $$a$$ and $$b$$, continuity and derivatives may be one-sided from within.)

(ii) The function $$H$$ is absolutely continuous on $$A;$$ hence $$V_{H}[A]<\infty.$$

Proof

(i) Let $$p \in(a, b], q=f(p) \neq \pm \infty.$$ Let $$f$$ be left continuous at $$p;$$ so, given $$\varepsilon>0,$$ we can fix $$c \in (a, p)$$ such that

$|f(x)-q|<\varepsilon \text { for } x \in (c, p).$

Then

\begin{aligned}(\forall x \in(c, p)) &\left|L \int_{x}^{p}(f-q)\right| \leq L \int_{x}^{p}|f-q| \\ & \leq L \int_{x}^{p}(\varepsilon)=\varepsilon \cdot m[x, p]=\varepsilon(p-x). \end{aligned}

But

\begin{aligned} L \int_{x}^{p}(f-q) &=L \int_{x}^{p} f-L \int_{x}^{p} q \\ L \int_{x}^{p} q &=q(p-x), \quad \text { and } \\ L \int_{x}^{p} f &=L \int_{a}^{p} f-L \int_{a}^{x} f \\ &=H(p)-H(x). \end{aligned}

Thus

$|H(p)-H(x)-q(p-x)| \leq \varepsilon(p-x);$

i.e.,

$\left|\frac{H(p)-H(x)}{p-x}-q\right| \leq \varepsilon \quad(c<x<p).$

Hence

$f(p)=q=\lim _{x \rightarrow p^{-}} \frac{\Delta H}{\Delta x}=H_{-}^{\prime}(p).$

If $$f$$ is right continuous at $$p \in[a, b),$$ a similar formula results for $$H_{+}^{\prime}(p).$$ This proves clause (i).

(ii) Let $$\varepsilon>0$$ be given. Then Theorem 6 in Chapter 8, §6, yields a $$\delta>0$$ such that

$\left|L \int_{X} f\right| \leq L \int_{X}|f|<\varepsilon$

whenever

$m X<\delta \text { and } A \supseteq X, X \in \mathcal{M}.$

Here we may set

$X=\bigcup_{i=1}^{r} A_{i} \text { (disjoint)}$

for some intervals

$A_{i}=\left(a_{i}, b_{i}\right) \subseteq A$

so that

$m X=\sum_{i} m A_{i}=\sum_{i}\left(b_{i}-a_{i}\right)<\delta.$

Then (1) implies that

$\varepsilon>L \int_{X}|f|=\sum_{i} L \int_{A_{i}}|f| \geq \sum_{i}\left|L \int_{a_{i}}^{b_{i}} f\right|=\sum_{i}\left|H\left(b_{i}\right)-H\left(a_{i}\right)\right|.$

Thus

$\sum_{i}\left|H\left(b_{i}\right)-H\left(a_{i}\right)\right|<\varepsilon$

whenever

$\sum_{i}\left(b_{i}-a_{i}\right)<\delta$

and

$A \supseteq \bigcup_{i}\left(a_{i}, b_{i}\right) \text { (disjoint).}$

(This is what we call "absolute continuity in the stronger sense.") By Problem 2 in Chapter 5, §8, this implies "absolute continuity" in the sense of Chapter 5, §8, hence $$V_{H}[A]<\infty . \quad \square$$

Note 1. The converse to (i) fails: the differentiability of $$H$$ at $$p$$ does not imply the continuity of its derivative $$f$$ at $$p$$ (Problem 6 in Chapter 5, §2).

Note 2. If $$f$$ is continuous on $$A-Q$$ ($$Q$$ countable), Theorem 1 shows that $$H$$ is a primitive (antiderivative): $$H=\int f$$ on $$A.$$ Recall that "Q countable" implies $$m Q=0,$$ but not conversely. Observe that we may always assume $$a, b \in Q.$$

We can now prove a generalized version of the so-called fundamental theorem of calculus, widely used for computing integrals via antiderivatives.

## Theorem $$\PageIndex{2}$$

If $$f : E^{1} \rightarrow E$$ has a primitive $$F$$ on $$A=[a, b],$$ and if $$f$$ is bounded on $$A-P$$ for some $$P$$ with $$m P=0,$$ then $$f$$ is $$L$$-integrable on $$A,$$ and

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

Proof

By Definition 1 of Chapter 5, §5, $$F$$ is relatively continuous and finite on $$A=[a, b],$$ hence bounded on $$A$$ (Theorem 2 in Chapter 4, §8).

It is also differentiable, with $$F^{\prime}=f,$$ on $$A-Q$$ for a countable set $$Q \subseteq A,$$ with $$a, b \in Q.$$ We fix this $$Q$$ along with $$P.$$

As we deal with $$A$$ only, we surely may redefine $$F$$ and $$f$$ on $$-A:$$

$F(x)=\left\{\begin{array}{ll}{F(a)} & {\text { if } x<a,} \\ {F(b)} & {\text { if } x>b,}\end{array}\right.$

and $$f=0$$ on $$-A.$$ Then $$f$$ is bounded on $$-P,$$ while $$F$$ is bounded and
continuous on $$E^{1},$$ and $$F^{\prime}=f$$ on $$-Q;$$ so $$F=\int f$$ on $$E^{1}.$$

Also, for $$n=1,2, \ldots$$ and $$t \in E^{1},$$ set

$f_{n}(t)=n\left[F\left(t+\frac{1}{n}\right)-F(t)\right]=\frac{F(t+1 / n)-F(t)}{1 / n}.$

Then

$f_{n} \rightarrow F^{\prime}=f \quad \text {on } -Q;$

i.e., $$f_{n} \rightarrow f$$ (a.e.) on $$E^{1}$$ (as $$m Q=0$$).

By (3), each $$f_{n}$$ is bounded and continuous (as $$F$$ is). Thus by Theorem 1 of Chapter 8, §3, $$F$$ and all $$f_{n}$$ are $$m$$-measurable on $$A$$ (even on $$E^{1}$$). So is $$f$$ by Corollary 1 of Chapter 8, §3.

Moreover, by boundedness, $$F$$ and $$f_{n}$$ are $$L$$-integrable on finite intervals. So is $$f.$$ For example, let

$|f| \leq K<\infty \text { on } A-P;$

as $$m P=0$$,

$\int_{A}|f| \leq \int_{A}(K)=K \cdot m A<\infty,$

proving integrability. Now, as

$F=\int f \text { on any interval }\left[t, t+\frac{1}{n}\right],$

Corollary 1 in Chapter 5, §4 yields

$\left(\forall t \in E^{1}\right) \quad\left|F\left(t+\frac{1}{n}\right)-F(t)\right| \leq \sup _{t \in-Q}\left|F^{\prime}(t)\right| \frac{1}{n} \leq \frac{K}{n}.$

Hence

$\left|f_{n}(t)\right|=n\left|F\left(t+\frac{1}{n}\right)-F(t)\right| \leq K;$

i.e., $$\left|f_{n}\right| \leq K$$ for all $$n$$.

Thus $$f$$ and $$f_{n}$$ satisfy Theorem 5 of Chapter 8, §6, with $$g=K.$$ By Note 1 there,

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

In the next lemma, we show that also

$\lim _{n \rightarrow \infty} L \int_{a}^{x} f_{n}=F(x)-F(a),$

which will complete the proof.$$\quad \square$$

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

Given a finite continuous $$F : E^{1} \rightarrow E$$ and given $$f_{n}$$ as in (3), we have

$\lim _{n \rightarrow \infty} L \int_{a}^{x} f_{n}=F(x)-F(a) \quad \text {for all } x \in E^{1}.$

Proof

As before, $$F$$ and $$f_{n}$$ are bounded, continuous, and $$L$$-integrable on any $$[a, x]$$ or $$[x, a].$$ Fixing $$a,$$ let

$H(x)=L \int_{a}^{x} F, \quad x \in E^{1}.$

By Theorem 1 and Note 2, $$H=\int F$$ also in the sense of Chapter 5, §5, with $$F=H^{\prime}$$ (derivative of $$H$$) on $$E^{1}$$.

Hence by Definition 2 the same section,

$\int_{a}^{x} F=H(x)-H(a)=H(x)-0=L \int_{a}^{x} F;$

i.e.,

$L \int_{a}^{x} F=\int_{a}^{x} F,$

and so

\begin{aligned} L \int_{a}^{x} f_{n}(t) d t &=n \int_{a}^{x} F\left(t+\frac{1}{n}\right) d t-n \int_{a}^{x} F(t) d t \\ &=n \int_{a+1 / n}^{b+1 / n} F(t) d t-n \int_{a}^{x} F(t) dt. \end{aligned}

(We computed

$\int F(t+1 / n) d t$

by Theorem 2 in Chapter 5, §5, with $$g(t)=t+1 / n$$.) Thus by additivity,

$L \int_{a}^{x} f_{n}=n \int_{a+1 / n}^{x+1 / n} F-n \int_{a}^{x} F=n \int_{x}^{x+1 / n} F-n \int_{a}^{a+1 / n} F.$

But

$n \int_{x}^{x+1 / n} F=\frac{H\left(x+\frac{1}{n}\right)-H(x)}{\frac{1}{n}} \rightarrow H^{\prime}(x)=F(x).$

Similarly,

$\lim _{n \rightarrow \infty} n \int_{a}^{a+1 / n} F=F(a).$

This combined with (5) proves (4), and hence Theorem 2, too.$$\quad \square$$

We also have the following corollary.

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

If $$f : E^{1} \rightarrow E^{*}\left(E^{n}, C^{n}\right)$$ is $$R$$-integrable on $$A=[a, b],$$ then

$(\forall x \in A) \quad R \int_{a}^{x} f=L \int_{a}^{b} f=F(x)-F(a),$

provided $$F$$ is primitive to $$f$$ on $$A.$$

Proof

This follows from Theorem 2 by Definition (c) and Theorem 2 of Chapter 8, §9.

Caution. Formulas (2) and (6) may fail if $$f$$ is unbounded, or if $$F$$ is not a primitive in the sense of Definition 1 of Chapter 5, §5: We need $$F^{\prime}=f$$ on $$A-Q, Q$$ countable ($$m Q=0$$ is not enough!). Even $$R$$-integrability (which makes $$f$$ bounded and a.e. continuous) does not suffice if

$F \neq \int f.$

For examples, see Problems 2-5.

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

If $$f$$ is relatively continuous and finite on $$A=[a, b]$$ and has a bounded derivative on $$A-Q$$ ($$Q$$ countable), then $$f^{\prime}$$ is $$L$$-integrable on $$A$$ and

$L \int_{a}^{x} f^{\prime}=f(x)-f(a) \quad \text { for } x \in A.$

This is simply Theorem 2 with $$F, f, P$$ replaced by $$f, f^{\prime}, Q,$$ respectively

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

If in Theorem 2 the primitive

$F=\int f$

is exact on some $$B \subseteq A,$$ then

$f(x)=\frac{d}{d x} L \int_{a}^{x} f, \quad x \in B.$

(Recall that $$\frac{d}{d x} F(x)$$ is classical notation for $$F^{\prime}(x)$$.)

Proof

By (2), this holds on $$B \subseteq A$$ if $$F^{\prime}=f$$ there.$$\quad \square$$

II. Note that under the assumptions of Theorem 2,

$L \int_{a}^{x} f=F(x)-F(a)=\int_{a}^{x} f.$

Thus all laws governing the primitive $$\int f$$ apply to $$L \int f.$$ For example, Theorem 2 of Chapter 5, §5, yields the following corollary.

## Corollary $$\PageIndex{4}$$ (change of variable)

Let $$g : E^{1} \rightarrow E^{1}$$ be relatively continuous on $$A=[a, b]$$ and have a bounded derivative on $$A-Q$$ ($$Q$$ countable).

Suppose that $$f : E^{1} \rightarrow E$$ (real or not) has a primitive on $$g[A],$$ exact on $$g[A-Q],$$ and that $$f$$ is bounded on $$g[A-Q]$$.

Then $$f$$ is $$L$$-integrable on $$g[A],$$ the function

$(f \circ g) g^{\prime}$

is $$L$$-integrable on $$A,$$ and

$L \int_{a}^{b} f(g(x)) g^{\prime}(x) d x=L \int_{p}^{q} f(y) d y,$

where $$p=g(a)$$ and $$q=g(b)$$.

For this and other applications of primitives, see Problem 9. However, often a direct approach is stronger (though not simpler), as we illustrate next.

## Lemma $$\PageIndex{2}$$ (Bonnet)

Suppose $$f : E^{1} \rightarrow E^{1}$$ is $$\geq 0$$ and monotonically decreasing on $$A=[a, b].$$ Then, if $$g : E^{1} \rightarrow E^{1}$$ is $$L$$-integrable on $$A,$$ so also is $$f g,$$ and

$L \int_{a}^{b} f g=f(a) \cdot L \int_{a}^{c} g \quad \text { for some } c \in A.$

Proof

The $$L$$-integrability of $$f g$$ follows by Theorem 3 in Chapter 8, §6, as $$f$$ is monotone and bounded, hence even $$R$$-integrable (Corollary 3 in Chapter 8, §9).

Using this and Lemma 1 of the same section, fix for each $$n$$ a $$\mathcal{C}$$-partition

$\mathcal{P}_{n}=\left\{A_{n i}\right\} \quad\left(i=1,2, \ldots, q_{n}\right)$

of $$A$$ so that

$(\forall n) \quad \frac{1}{n}>\overline{S}\left(f, \mathcal{P}_{n}\right)-\underline{S}\left(f, \mathcal{P}_{n}\right)=\sum_{i=1}^{q_{n}} w_{n i} m A_{n i},$

where we have set

$w_{n i}=\sup f\left[A_{n i}\right]-\inf f\left[A_{n i}\right].$

Consider any such $$\mathcal{P}=\left\{A_{i}\right\}, i=1, \ldots, q$$ (we drop the "$$n$$" for brevity). If $$A_{i}=\left[a_{i-1}, a_{i}\right],$$ then since $$f \downarrow$$,

$w_{i}=f\left(a_{i-1}\right)-f\left(a_{i}\right) \geq\left|f(x)-f\left(a_{i-1}\right)\right|, \quad x \in A_{i}.$

Under Lebesgue measure (Problem 8 of Chapter 8, §9), we may set

$A_{i}=\left[a_{i-1}, a_{i}\right] \quad(\forall i)$

and still get

\begin{aligned} L \int_{A} f g &=\sum_{i=1}^{q} f\left(a_{i-1}\right) L \int_{A_{i}} g(x) d x \\ &+\sum_{i=1}^{q} L \int_{A_{i}}\left[f(x)-f\left(a_{i-1}\right)\right] g(x) d x. \end{aligned}

(Verify!) Here $$a_{0}=a$$ and $$a_{q}=b$$.

Now, set

$G(x)=L \int_{a}^{x} g$

and rewrite the first sum (call it $$r$$ or $$r_{n}$$) as

\begin{aligned} r &=\sum_{i=1}^{q} f\left(a_{i-1}\right)\left[G\left(a_{i}\right)-G\left(a_{i-1}\right)\right] \\ &=\sum_{i=1}^{q-1} G\left(a_{i}\right)\left[f\left(a_{i-1}\right)-f\left(a_{i}\right)\right]+G(b) f\left(a_{q-1}\right), \end{aligned}

or

$r=\sum_{i=1}^{q-1} G\left(a_{i}\right) w_{i}+G(b) f\left(a_{q-1}\right),$

because $$f\left(a_{i-1}\right)-f\left(a_{i}\right)=w_{i}$$ and $$G(a)=0$$.

Now, by Theorem 1 (with $$H, f$$ replaced by $$G, g$$), $$G$$ is continuous on $$A=$$ $$[a, b];$$ so $$G$$ attains a largest value $$K$$ and a least value $$k$$ on $$A.$$

As $$f \downarrow$$ and $$f \geq 0$$ on $$A,$$ we have

$w_{i} \geq 0 \text { and } f\left(a_{q-1}\right) \geq 0.$

Thus, replacing $$G(b)$$ and $$G\left(a_{i}\right)$$ by $$K(\text { or } k)$$ in $$(13)$$ and noting that

$\sum_{i=1}^{q-1} w_{i}=f(a)-f\left(a_{q-1}\right),$

we obtain

$k f(a) \leq r \leq K f(a);$

more fully, with $$k=\min G[A]$$ and $$K=\max G[A]$$,

$(\forall n) \quad k f(a) \leq r_{n} \leq K f(a).$

Next, let $$s$$ (or rather $$s_{n}$$ be the second sum in (12). Noting that

$w_{i} \geq\left|f(x)-f\left(a_{i-1}\right)\right|,$

suppose first that $$|g| \leq B$$ (bounded) on $$A$$.

Then for all $$n$$,

$\left|s_{n}\right| \leq \sum_{i=1}^{q_{n}} L \int_{A_{n i}}\left(w_{n i} B\right)=B \sum_{i=1}^{q_{n}} w_{n i} m A_{n i}<\frac{B}{n} \rightarrow 0 \quad \ \text{(by (11)).}$

But by (12),

$L \int_{A} f g=r_{n}+s_{n} \quad(\forall n).$

As $$s_{n} \rightarrow 0$$,

$L \int_{A} f g=\lim _{n \rightarrow \infty} r_{n},$

and so by (14),

$k f(a) \leq L \int_{A} f g \leq K f(a).$

By continuity, $$f(a) G(x)$$ takes on the intermediate value $$L \int_{A} f g$$ at some $$c \in A;$$ so

$L \int_{A} f g=f(a) G(c)=f(a) L \int_{a}^{c} g,$

since

$G(x)=L \int_{a}^{x} f.$

Thus all is proved for a bounded $$g.$$

The passage to an unbounded $$g$$ is achieved by the so-called truncation method described in Problems 12 and 13. (Verify!)$$\quad \square$$

## Corollary$$\PageIndex{5}$$ (second law of the mean)

Let $$f : E^{1} \rightarrow E^{1}$$ be monotone on $$A=[a, b].$$ Then if $$g : E^{1} \rightarrow E^{1}$$ is $$L$$-integrable on $$A,$$ so also is $$f g,$$ and

$L \int_{a}^{b} f g=f(a) L \int_{a}^{c} g+f(b) L \int_{c}^{b} g \quad \text {for some } c \in A.$

Proof

If, say, $$f \downarrow$$ on $$A,$$ set

$h(x)=f(x)-f(b).$

Then $$h \geq 0$$ and $$h \downarrow$$ on $$A;$$ so by Lemma 2,

$\int_{a}^{b} g h=h(a) L \int_{a}^{c} g \quad \text {for some } c \in A.$

As

$h(a)=f(a)-f(b),$

this easily implies (15).

If $$f \uparrow,$$ apply this result to $$-f$$ to obtain (15) again.$$\quad \square$$

Note 3. We may restate (15) as

$(\exists c \in A) \quad L \int_{a}^{b} f g=p L \int_{a}^{c} g+q L \int_{c}^{b} g,$

provided either

(i) $$f \uparrow$$ and $$p \leq f(a+) \leq f(b-) \leq q,$$ or

(ii) $$f \downarrow$$ and $$p \geq f(a+) \geq f(b-) \geq q$$.

This statement slightly strengthens (15).

To prove clause (i), redefine

$f(a)=p \text { and } f(b)=q.$

Then still $$f \uparrow;$$ so (15) applies and yields the desired result. Similarly for (ii). For a continuous $$g,$$ see also Problem 13(ii') in Chapter 8, §9, based on Stieltjes theory.

III. We now give a useful analogue to the notion of a primitive.

## Definition

A map $$F : E^{1} \rightarrow E$$ is called an $$L$$-primitive or an indefinite $$L$$-integral of $$f : E^{1} \rightarrow E,$$ on $$A=[a, b]$$ iff $$f$$ is $$L$$-integrable on $$A$$ and

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

for all $$x \in A$$ and some fixed finite $$c \in E$$.

Notation:

$F=L \int f \quad\left(\text {not } F=\int f\right)$

or

$F(x)=L \int f(x) d x \quad \text {on } A.$

By (16), all $$L$$-primitives of $$f$$ on $$A$$ differ by finite constants only.

If $$E=E^{*}\left(E^{n}, C^{n}\right),$$ one can use this concept to lift the boundedness restriction on $$f$$ in Theorem 2 and the corollaries of this section. The proof will be given in §2. However, for comparison, we state the main theorems already now.

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

Let

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

for some $$f : E^{1} \rightarrow E^{*}\left(E^{n}, C^{n}\right)$$.

Then $$F$$ is differentiable, with

$F^{\prime}=f \quad \text {a.e. on } A.$

In classical notation,

$f(x)=\frac{d}{d x} L \int_{a}^{x} f(t) d t \quad \text {for almost all } x \in A.$

A proof was sketched in Problem 6 of Chapter 8, §12. (It is brief but requires more "starred" material than used in §2.)

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

Let $$F : E^{1} \rightarrow E^{n}\left(C^{n}\right)$$ be differentiable on $$A=[a, b]$$ (at $$a$$ and $$b$$ differentiability may be one sided). Let $$F^{\prime}=f$$ be $$L$$-integrable on $$A$$.

Then

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