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

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