
# 8.6: Integrable Functions. Convergence Theorems


I. Some important theorems apply to integrable functions.

## Theorem $$\PageIndex{1}$$ (linearity of the integral)

If $$f, g : S \rightarrow E^{*}$$ are integrable on a set $$A \in \mathcal{M}$$ in $$(S, \mathcal{M}, m),$$ so is

$p f+q g$

for any $$p, q \in E^{1},$$ and

$\int_{A}(p f+q g)=p \int_{A} f+q \int_{A} g;$

in particular,

$\int_{A}(f \pm g)=\int_{A} f \pm \int_{A} g.$

Proof

By Problem 5 in §5,

$\overline{\int}_{A} f+\overline{\int}_{A} g \geq \overline{\int}_{A}(f+g) \geq \underline{\int}_{A}(f+g) \geq \underline{\int}_{A} f + \underline{\int}_{A} g.$

(Here

$\overline{\int}_{A} f, \underline{\int}_{A} f, \overline{\int}_{A} g, \text { and } \underline{\int}_{A} g$

are finite by integrability; so all is orthodox.)

As

$\overline{\int}_{A} f=\underline{\int}_{A} f \text { and } \overline{\int}_{A} g=\underline{\int}_{A} g,$

the inequalities turn into equalities, so that

$\int_{A} f+\int_{A} g=\overline{\int}_{A}(f+g)=\underline{\int}_{A}(f+g).$

Using also Theorem 1(e)(e') from §5, we obtain the desired result for any $$p, q \in E^{1}. \quad \square$$

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

A function $$f : S \rightarrow E^{*}$$ is integrable on $$A$$ in $$(S, \mathcal{M}, m)$$ iff

(i) it is $$m$$-measurable on $$A,$$ and

(ii) $$\overline{\int}_{A} f$$ (equivalently $$\overline{\int}_{A}|f|$$) is finite.

Proof

If these conditions hold, $$f$$ is integrable on $$A$$ by Theorem 3 of §5.

Conversely, let

$\overline{\int}_{A} f=\underline{\int}_{A} f \neq \pm \infty.$

Using Lemma 2 in §5, fix measurable maps $$g$$ and $$h$$ ($$g \leq f \leq h$$) on $$A,$$ with

$\int_{A} g=\int_{A} f=\int_{A} h \neq \pm \infty.$

By Theorem 3 in §5, $$g$$ and $$h$$ are integrable on $$A;$$ so by Theorem 1,

$\int_{A}(h-g)=\int_{A} h-\int_{A} g=0.$

As

$h-g \geq h-f \geq 0,$

we get

$\int_{A}(h-f)=0,$

and so by Theorem 1(h) of §5, $$h-f=0$$ a.e. on $$A$$.

Hence $$f$$ is almost measurable on $$A,$$ and

$\int_{A} f \neq \pm \infty$

by assumption. From formula (1), we then get

$\int_{A} f^{+} \text { and } \int_{A} f^{-}<\infty,$

and hence

$\int_{A}|f|=\int_{A}\left(f^{+}+f^{-}\right)=\int_{A} f^{+}+\int_{A} f^{-}<\infty$

by Theorem 1 and by Theorem 2 of §2. Thus all is proved.$$\quad \square$$

Simultaneously, we also obtain the following corollary.

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

A function $$f : S \rightarrow E^{*}$$ is integrable on $$A$$ iff $$f^{+}$$ and $$f^{-}$$ are.

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

If $$f, g : S \rightarrow E^{*}$$ are integrable on $$A,$$ so also are

$f \vee g, f \wedge g,|f|, \text { and } k f \text { for } k \in E^{1},$

with

$\int_{A} k f=k \int_{A} f.$

Exercise!

For products $$f g,$$ this holds if $$f$$ or $$g$$ is bounded. In fact, we have the following theorem.

## Theorem $$\PageIndex{3}$$ (weighted law of the mean)

Let $$f$$ be $$m$$-measurable and bounded on A. Set

$p=\inf f[A] \text { and } q=\sup f[A].$

Then if $$g$$ is $$m$$-integrable on $$A,$$ so is $$f g,$$ and

$\int_{A} f|g|=c \int_{A}|g|$

for some $$c \in[p, q]$$.

If, further, $$f$$ also has the Darboux property on A (Chapter 4, §9), then $$c=f(x_{0})$$ for some $$x_{0} \in A.$$

Proof

By assumption,

$\left(\exists k \in E^{1}\right) \quad|f| \leq k$

on $$A.$$ Hence if $$\int_{A}|g|=0$$,

$\left|\int_{A} f|g|\right| \leq \int_{A}|f g| \leq k \int_{A}|g|=0;$

so any $$c \in[p, q]$$ yields

$\int_{A} f|g|=c \int_{A}|g|=0.$

If, however, $$\int_{A}|g| \neq 0,$$ the number

$c=\left(\int_{A} f|g|\right) / \int_{A}|g|$

is the required constant.

Moreover, as $$f$$ and $$g$$ are $$m$$-measurable on $$A,$$ so is $$f g;$$ and as

$\left|\int_{A} f g\right| \leq|c| \int_{A}|g|<\infty,$

$$f g$$ is integrable on $$A$$ by Theorem 2.

Finally, if $$f$$ has the Darboux property and if $$p<c<q$$ (with $$p, q$$ as above), then

$f(x)<c<f(y)$

for some $$x, y \in A$$ (why?); hence by the Darboux property, $$f\left(x_{0}\right)=c$$ for some $$x_{0} \in A.$$

If, however,

$c \leq \inf f[A]=p,$

then

$(f-c)|g| \geq 0$

and

$\int_{A}(f-c)|g|=\int_{A} f|g|-c \int_{A}|g|=0 \text { (why?);}$

so by Theorem 1(h) in §5, $$f-c=0$$ a.e. on $$A.$$ Then surely $$f\left(x_{0}\right)=c$$ for some $$x_{0} \in A$$ (except the trivial case $$m A=0$$). This also implies $$c \in f[A] \in[p, q]$$.

Proceed similarly in the case $$c \geq q. \quad \square$$

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

If $$f$$ is integrable on $$A \in \mathcal{M},$$ it is so on any $$B \subseteq A(B \in \mathcal{M})$$.

Proof

Apply Theorem 1(f) in §5, and Theorem 3 of §5, to $$f^{+}$$ and $$f^{-}. \quad \square$$

II. Convergence Theorems. If $$f_{n} \rightarrow f$$ on $$A$$ (pointwise, a.e., or uniformly), does it follow that

$\int_{A} f_{n} \rightarrow \int_{A} f ?$

To give some answers, we need a lemma.

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

If $$f \geq 0$$ on $$A \in \mathcal{M}$$ and if

$\underline{\int}_{A} f>p \in E^{*},$

there is an elementary and nonnegative map $$g$$ on $$A$$ such that

$\int_{A} g>p,$

and $$g<f$$ on $$A$$ except only at those $$x \in A$$ (if any) at which

$f(x)=g(x)=0.$

(We then briefly write $$g \subset f$$ on $$A.$$)

Proof

By Lemma 1 in §5, there is an elementary and nonnegative map $$G \leq f$$ on $$A,$$ with

$\underline{\int}_{A} f \geq \int_{A} G>p.$

For the rest, proceed as in Problem 7 of §4, replacing $$f$$ by $$G$$ there.$$\quad \square$$

## Theorem $$\PageIndex{4}$$ (monotone convergence)

If $$0 \leq f_{n} \nearrow f$$ (a.e.) on $$A \in \mathcal{M},$$ i.e.,

$0 \leq f_{n} \leq f_{n+1} \quad (\forall n),$

and $$f_{n} \rightarrow f$$ (a.e.) on $$A,$$ then

$\overline{\int}_{A} f_{n} \nearrow \overline{\int}_{A} f.$

Proof for $$\mathcal{M}$$-measurable $$f_{n}$$ and $$f$$ on $$A.$$

By Corollary 2 in §5, we may assume that $$f_{n} \nearrow f$$ (pointwise) on $$A$$ (otherwise, drop a null set).

By Theorem 1(c) of §5, $$0 \leq f_{n} \nearrow f$$ implies

$0 \leq \int_{A} f_{n} \leq \int_{A} f,$

and so

$\lim _{n \rightarrow \infty} \int_{A} f_{n} \leq \int_{A} f.$

The limit, call it $$p,$$ exists in $$E^{*},$$ as $$\left\{\int_{A} f_{n}\right\} \uparrow.$$ It remains to show that

$p \geq \overline{\int}_{A} f=\underline{\int}_{A} f.$

(We know that

$\overline{\int}_{A} f=\underline{\int}_{A} f,$

by the assumed measurability of $$f;$$ see Theorem 3 in §5.)

Suppose

$\underline{\int}_{A} f>p.$

Then Lemma 1 yields an elementary and nonnegative map $$g \subset f$$ on $$A,$$ with

$p<\int_{A} g.$

Let

$A_{n}=A\left(f_{n} \geq g\right), \quad n=1,2, \ldots.$

Then $$A_{n} \in \mathcal{M}$$ and

$A_{n} \nearrow A=\bigcup_{n=1}^{\infty} A_{n}.$

For if $$f(x)=0,$$ then $$x \in A_{1},$$ and if $$f(x)>0,$$ then $$f(x)>g(x),$$ so that $$f_{n}(x)>g(x)$$ for large $$n;$$ hence $$x \in A_{n}.$$

By Note 4 in §5, the set function $$s=\int g$$ is a measure, hence continuous by Theorem 2 in Chapter 7, §4. Thus

$\int_{A} g=s A=\lim _{n \rightarrow \infty} s A_{n}=\lim _{n \rightarrow \infty} \int_{A_{n}} g.$

But as $$g \leq f_{n}$$ on $$A_{n},$$ we have

$\int_{A_{n}} g \leq \int_{A_{n}} f_{n} \leq \int_{A} f_{n}.$

Hence

$\int_{A} g=\lim \int_{A_{n}} g \leq \lim \int_{A} f_{n}=p,$

contrary to $$p<\int_{A} g.$$ This contradiction completes the proof.$$\quad \square$$

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

If $$f_{n} \geq 0$$ on $$A \in \mathcal{M}(n=1,2, \ldots),$$ then

$\overline{\int}_{A} \underline{\lim} f_{n} \leq \underline{\lim } \overline{\int}_{A} f_{n}.$

Proof

Let

$g_{n}=\inf _{k \geq n} f_{k}, \quad n=1,2, \ldots;$

so $$f_{n} \geq g_{n} \geq 0$$ and $$\left\{g_{n}\right\} \uparrow$$ on $$A.$$ Thus by Theorem 4,

$\overline{\int}_{A} \lim g_{n} = \lim \overline{\int}_{A} g_{n} = \underline{\lim} \overline{\int}_{A} g_{n} \leq \underline{\lim} \overline{\int}_{A} f_{n}.$

But

$\lim _{n \rightarrow \infty} g_{n}=\sup _{n} g_{n}=\sup _{n} \inf _{k \geq n} f_{k}=\lim _{n} f_{n}.$

Hence

$\overline{\int}_{A} \underline{\lim} f_{n} = \overline{\int}_{A} \lim g_{n} \leq \underline{\lim} \overline{\int}_{A} f_{n},$
as claimed.$$\quad \square$$

## Theorem $$\PageIndex{5}$$ (dominated convergence)

Let $$f_{n} : S \rightarrow E$$ be $$m$$-measurable on $$A \in \mathcal{M} (n=1,2,\ldots).$$ Let

$f_{n} \rightarrow f \text{ (a.e.) on } A.$

Then

$\lim _{n \rightarrow \infty} \int_{A}\left|f_{n}-f\right|=0,$

provided that there is a map $$g : S \rightarrow E^{1}$$ such that

$\int_{A} g<\infty$

and

$(\forall n) \quad\left|f_{n}\right| \leq g \text { a.e. on } A.$

Proof

Neglecting null sets, we may assume that

$\left|f_{n}\right| \leq g<\infty$

on $$A$$ and $$f_{n} \rightarrow f$$ (pointwise) on $$A;$$ so $$|f| \leq g$$ and

$\left|f_{n}-f\right| \leq\left|f_{n}\right|+|f| \leq 2 g$

on $$A.$$ As $$|f|<\infty,$$ we have

$\left|f_{n}-f\right| \rightarrow 0$

on $$A.$$ Hence, setting

$h_{n}=2 g-\left|f_{n}-f\right| \geq 0,$

we get

$2 g=\lim _{n \rightarrow \infty} h_{n}=\underline{\lim} h_{n}.$

We may also assume that $$g$$ is measurable on $$A.$$ (If not, replace it by a measurable $$G \geq g,$$ with

$\int_{A} G=\int_{A} g<\infty,$

by Lemma 2 in §5.) Then all

$h_{n}=2 g-\left|f_{n}-f\right|$

are measurable (even integrable) on $$A$$.

Thus by Lemma 2,

\begin{aligned} \int_{A} 2 g = \int_{A} \underline{\lim} h_{n} & \leq \underline{\lim} \int_{A}\left(2 g-\left|f_{n}-f\right|\right) \\ &=\underline{\lim}\left(\int_{A} 2 g+\int_{A}\left(-\left|f_{n}-f\right|\right)\right) \\ &= \int_{A} 2 g+\underline{\lim}\left(-\int_{A}\left|f_{n}-f\right|\right) \\ &= \int_{A} 2 g-\overline{\lim} \int_{A}\left|f_{n}-f\right|. \end{aligned}
(See Problems 5 and 8 in Chapter 2, §13.)

Canceling $$\int_{A} 2 g$$ (finite!), we have

$0 \leq-\overline{\lim } \int_{A}\left|f_{n}-f\right|.$

Hence

$0 \geq \overline{\lim } \int_{A}\left|f_{n}-f\right| \geq \underline{\lim }\int_{A}\left|f_{n}-f\right| \geq 0,$

as $$\left|f_{n}-f\right| \geq 0.$$ This yields

$0=\overline{\lim } \int_{A}\left|f_{n}-f\right|=\underline{\lim} \int_{A}\left|f_{n}-f\right|=\lim \int_{A}\left|f_{n}-f\right|,$

as required.$$\quad \square$$

Note 1. Theorem 5 holds also for complex and vector-valued functions (for $$\left|f_{n}-f\right|$$ is real).

In the extended-real case, Theorems 1(g) in §5 and Theorems 1 and 2 in §6 yield

$\left|\int_{A} f_{n}-\int_{A} f\right|=\left|\int_{A}\left(f_{n}-f\right)\right| \leq \int_{A}\left|f_{n}-f\right| \rightarrow 0,$

i.e.,

$\int_{A} f_{n} \rightarrow \int_{A} f.$

Moreover, $$f$$ is integrable on $$A,$$ being measurable (why?), with

$\int_{A}|f| \leq \int_{A} g<\infty.$

For complex and vector-valued functions, this will follow from §7. Observe that Theorem 5, unlike Theorem 4, requires the $$m$$-measurability of the $$f_{n}$$.

Note 2. Theorem 5 fails if there is no "dominating"

$g \geq\left|f_{n}\right| \text { with } \int_{A} g<\infty,$

even if $$f$$ and the $$f_{n}$$ are integrable.

## Example

Let $$m$$ be Lebesgue measure in $$A=E^{1}, f=0,$$ and

$f_{n}=\left\{\begin{array}{ll}{1} & {\text { on }[n, n+1],} \\ {0} & {\text { elsewhere. }}\end{array}\right.$

Then $$f_{n} \rightarrow f$$ and $$\int_{A} f_{n}=1;$$ so

$\lim _{n \rightarrow \infty} \int_{A} f_{n}=1 \neq 0=\int_{A} f.$

The trouble is that any

$g \geq f_{n} \quad(n=1,2, \ldots)$

would have to be $$\geq 1$$ on $$B=[1, \infty);$$ so

$\int_{A} g \geq \int_{B} g=1 \cdot m B=\infty,$

instead of $$\int_{A} g<\infty$$.

This example also shows that $$f_{n} \rightarrow f$$ alone does not imply

$\int_{A} f_{n} \rightarrow \int_{A} f.$

## Theorem $$\PageIndex{6}$$ (absolute continuity of the integral)

Given $$f : S \rightarrow E$$ with

$\overline{\int}_{A}|f|<\infty$

and $$\varepsilon>0,$$ there is $$\delta>0$$ such that

$\overline{\int}_{X}|f|<\varepsilon$

whenever

$m X<\delta \quad(A \supseteq X, X \in \mathcal{M}).$

Proof

By Lemma 2 in §5, fix $$h \geq|f|,$$ measurable on $$A,$$ with

$\int_{A} h=\overline{\int}_{A}|f|<\infty.$

Neglecting a null set, we assume that $$|h|<\infty$$ on $$A$$ (Corollary 1 of §5). Now, $$(\forall n)$$ set

$g_{n}(x)=\left\{\begin{array}{ll}{h(x),} & {x \in A_{n}=A(h<n),} \\ {0,} & {x \in-A_{n}.}\end{array}\right.$

Then $$g_{n} \leq n$$ and $$g_{n}$$ is measurable on $$A$$. (Why?)

Also, $$g_{n} \geq 0$$ and $$g_{n} \rightarrow h$$ (pointwise) on $$A$$.

For let $$\varepsilon>0,$$ fix $$x \in A,$$ and find $$k>h(x).$$ Then

$(\forall n \geq k) \quad h(x) \leq n \text { and } g_{n}(x)=h(x).$

So

$(\forall n \geq k) \quad\left|g_{n}(x)-h(x)\right|=0<\varepsilon.$

Clearly, $$g_{n} \leq h.$$ Hence by Theorem 5

$\lim _{n \rightarrow \infty} \int_{A}\left|h-g_{n}\right|=0.$

Thus we can fix $$n$$ so large that

$\int_{A}\left(h-g_{n}\right)<\frac{1}{2} \varepsilon.$

For that $$n,$$ let

$\delta=\frac{\varepsilon}{2 n}$

and take any $$X \subseteq A(X \in \mathcal{M}),$$ with $$m X<\delta$$.

As $$g_{n} \leq n$$ (see above), Theorem 1(c) in §5 yields

$\int_{X} g_{n} \leq \int_{X}(n)=n \cdot m X<n \delta=\frac{1}{2} \varepsilon.$

Hence as $$|f| \leq h$$ and

$\int_{X}\left(h-g_{n}\right) \leq \int_{A}\left(h-g_{n}\right)<\frac{1}{2} \varepsilon$

(Theorem 1(f) of §5), we obtain

$\overline{\int}_{X}|f| \leq \int_{X} h=\int_{X}\left(h-g_{n}\right)+\int_{X} g_{n}<\frac{1}{2} \varepsilon+\frac{1}{2} \varepsilon=\varepsilon,$

as required.$$\quad \square$$