# 8.5: Integration of Extended-Real Functions

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We shall now define integrals for arbitrary functions $$f : S \rightarrow E^{*}$$ in a measure
space $$(S, \mathcal{M}, m) .$$ We start with the case $$f \geq 0$$.

## Definition

Given $$f \geq 0$$ on $$A \in \mathcal{M},$$ we define the upper and lower integrals,
$\overline{\int} \text{ and } \underline{\int},$
of $$f$$ on $$A$$ (with respect to $$m )$$ by
$\overline{\int_{A}} f=\overline{\int_{A}} f d m=\inf _{h} \int_{A} h$
over all elementary maps $$h \geq f$$ on $$A,$$ and
$\int_{-A} f=\int_{-A} f d m=\sup _{g} \int_{A} g$
over all elementary and nonnegative maps $$g \leq f$$ on $$A$$.
If $$f$$ is not nonnegative, we use $$f^{+}=f \vee 0$$ and $$f^{-}=(-f) \vee 0$$ (§2), and set
$\begin{array}[c] \overline{\int_{A}} f &= \overline{\int_{A}} f dm = \overline{\int_{A}} f^{+} - \underline{\int_{A}} f^{-} \text{ and} \\ \underline{\int_{A}} f &= \underline{\int_{A}} f dm = \underline{\int_{A}} f^{+} - \overline{\int_{A}} f^{-} . \end{array}$
By our conventions, these expressions are always defined. The integral $$\overline{\int}_{A} f\left(\text { or } \int_{-A} f\right)$$ is called orthodox iff it does not have the form $$\infty-\infty$$ in (1), e.g., $$\overline{\text { if }} f \geq 0$$ (i.e., $$f^{-}=0 ),$$ or if $$\int_{A} f<\infty .$$ An unorthodox integral equals $$+\infty .$$
We often write $$\int$$ for $$\overline{\int}$$ and call it simply the integral (of $$f ),$$ even if
$\overline{\int_{A}} f \neq \int_{A} f.$
"Classical" notation is $$\int_{A} f(x) d m(x)$$.

## Definition

The function $$f$$ is called integrable (or $$m$$-integrable, or Lebesgue integrable, with respect to $$m )$$ on $$A,$$ iff
$\overline{\int_{A}} f d m=\int_{A} f d m \neq \pm \infty$

The process described above is called (abstract) Lebesgue integration as opposed to Riemann integration (B. Riemann, 1826-1866). The latter deals with bounded functions only and allows $$h$$ and $$g$$ in $$\left(1^{\prime}\right)$$ and $$\left(1^{\prime \prime}\right)$$ to be simple step functions only (see §9). It is inferior to Lebesgue theory.
The values of
$\overline{\int_{A}} f d m \text { and } \underline{\int_{A}} f d m$
depend on $$m .$$ If $$m$$ is Lebesgue measure, we speak of Lebesgue integrals, in the stricter sense. If $$m$$ is Lebesgue-Stieltjes measure, we speak of $$L S$$-integrals, and so on.
Note 1. If $$f$$ is elementary and (extended) real, our present definition of
$\overline{\int_{A}} f$
agrees with that of §4. For if $$f \geq 0, f$$ itself is the least of all elementary and nonnegative functions
$h \geq f$
and the greatest of all elementary and nonnegative functions
$g \leq f.$
Thus by Problem 5 in §4,
$\int_{A} f=\min _{h \geq f} \int_{A} h=\max _{g \leq f} \int_{A} g,$
i.e.,
$\int_{A} f=\overline{\int}_{A} f= \underline{\int_{A}} f.$
If, however, $$f \ngeq 0,$$ this follows by Definition 2 in §4. This also shows that for elementary and (extended) real maps,
$\overline{\int_{A}} f= \underline{\int_{A}} f \text { always.}$
Note 2. By Definition 1,
$\underline{\int_{A}} f \leq \overline{\int_{A}} f \text { always.}$
For if $$f \geq 0,$$ then for any elementary and nonnegative maps $$g, h$$ with
$g \leq f \leq h,$
we have
$\int_{A} g \leq \int_{A} h$
by Problem 5 in §4. Thus
$\underline{\int_{A}} f=\sup _{g} \int_{A} g$
is a lower bound of all such $$\int_{A} h,$$ and so
$\underline{\int_{A}} f \leq \operatorname{glb} \int_{A} h=\overline{\int}_{A} f.$
In the general formula $$(1),$$ too
$\underline{\int_{A}} f \leq \overline{\int_{A}} f,$
since
$\int_{-A} f^{+} \leq \overline{\int}_{A} f^{+} \text { and } \int_{-A} f^{-} \leq \overline{J}_{A} f^{-}.$

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

For any functions $$f, g : S \rightarrow E^{*}$$ and any set $$A \in \mathcal{M},$$ we have the following results.
(a) If $$f=a(\text {constant})$$ on $$A,$$ then
$$\overline{\int_{A}} f= \underline{\int_{A}} f=a \cdot m A$$.
(b) If $$f=0$$ on $$A$$ or $$m A=0,$$ then
$\overline{\int_{A}} f= \underline{\int_{A}} f=0.$
(c) If $$f \geq g$$ on $$A,$$ then
$\overline{\int_{A}} f \geq \overline{\int}_{A} g \text { and } \underline{\int_{A}} f \geq \underline{\int_{A}} g.$
(d) If $$f \geq 0$$ on $$A,$$ then
$\overline{\int_{A}} f \geq 0 \text { and } \underline{\int_{A}} f \geq 0.$
Similarly if $$f \leq 0$$ on $$A$$.
(e) If $$0 \leq p < \infty,$$ then
$\overline{\int_{A}} p f=p \overline{\int_{A}} f \text { and } \underline{\int_{A}} p f=p \underline{\int_{A}} f.$
(e') We have
$\overline{\int_{A}}(-f)=-\underline{\int_{A}} f \text { and } \underline{\int_{A}}(-f)=-\overline{\int_{A}} f$
if one of the integrals involved in each case is orthodox. Otherwise,
$\overline{\int_{A}}(-f)=\infty=\underline{\int_{A}} f \text { and } \underline{\int_{A}}(-f)=\infty=\overline{\int_{A}} f.$
(f) If $$f \geq 0$$ on $$A$$ and
$A \supseteq B, B \in \mathcal{M},$
then
$\overline{\int_{A}} f \geq \overline{\int_{B}} f \text { and } \underline{\int_{A}} f \geq \underline{\int_{B}} f.$
(g) We have
$\left|\overline{\int}_{A} f\right| \leq \overline{\int}_{A}|f| \text { and }\left| \underline{\int_{A}} f\right| \leq \overline{\int_{A}}|f|$
(but not
$\left|\underline{\int_{A}} f\right| \leq \underline{\int_{A}}|f|$
in general).
(h) If $$f \geq 0$$ on $$A$$ and $$\overline{\int_{A}} f=0$$ (or $$f \leq 0$$ and $$\underline{\int_{A}} f=0 ),$$ then $$f=0$$ a.e. on $$A$$.

Proof

We prove only some of the above, leaving the rest to the reader.
(a) This following by Corollary 1 (iv) in §4.
(b) Use (a) and Corollary 1 (v) in §4.
(c) First, let
$f \geq g \geq 0 \text { on } A.$
Take any elementary and nonnegative map $$H \geq f$$ on $$A .$$ Then $$H \geq g$$ as well; so by definition,
$\overline{\int_{A}} g=\inf _{h \geq g} \int_{A} h \leq \int_{A} H.$
Thus
$\overline{\int_{A}} f \leq \int_{A} H$
for any such $$H .$$ Hence also
$\overline{\int}_{A} g \leq \inf _{H \geq f} \int_{A} H=\overline{\int}_{A} f.$
Similarly,
$\underline{\int_{A}} f \geq \underline{\int_{A}} g$
if $$f \geq g \geq 0$$.
In the general case, $$f \geq g$$ implies
$f^{+} \geq g^{+} \text { and } f^{-} \leq g^{-} . \text { (Why?) }$
Thus by what was proved above,
$\overline{\int_{A}} f^{+} \geq \overline{\int_{A}} g^{+} \text { and } \underline{\int_{A}} f^{-} \leq \underline{\int_{A}} g^{-}.$
Hence
$\overline{\int_{A}} f^{+}- \underline{\int_{A}} f^{-} \geq \overline{\int_{A}} g^{+}- \underline{\int_{-A}} g^{-};$
i.e.,
$\overline{\int_{A}} \geq \overline{\int_{A}} g.$
Similarly, one obtains
$]underline{\int{A}} f \geq \underline{\int_{A}} g.$
(d) It is clear that (c) implies (d).
(e) Let $$0 \leq p<\infty$$ and suppose $$f \geq 0$$ on $$A .$$ Take any elementary and nonnegative map
$h \geq f \text { on } A.$
By Corollary 1 (vii) and Note 3 of §4,
$\int_{A} p h=p \int_{A} h$
for any such $$h .$$ Hence
$\overline{\int_{A}} p f =\inf _{h} \int_{A} p h=\inf _{h} p \int_{A} h=p \overline{\int_{A}} f.$
Similarly,
$\underline{\int_{A}} p f=p \underline{\int_{A}} f.$
The general case reduces to the case $$f \geq 0$$ by formula $$(1)$$.
(e') Assertion $$\left(\mathrm{e}^{\prime}\right)$$ follows from $$(1)$$ since
$(-f)^{+}=f^{-}, \quad(-f)^{-}=f^{+},$
and $$-(x-y)=y-x$$ if $$x-y$$ is orthodox. (Why?)
(f) Take any elementary and nonnegative map
$h \geq f \geq 0 \text { on } A.$
By Corollary 1 (ii) and Note 3 of §4,
$\int_{B} h \geq \int_{A} h$
for any such $$h .$$ Hence
$\overline{\int_{B}} f=\inf _{h} \int_{B} h \leq \inf _{h} \int_{A} h=\overline{\int}_{A} f.$
Similarly for $$\underline{\int}$$.
(g) This follows from $$(\mathrm{c})$$ and $$\left(\mathrm{e}^{\prime}\right)$$ since $$\pm f \leq|f|$$ implies
$\overline{\int}_{A}|f| \geq \overline{\int}_{A} f \geq \underline{\int_{A}} f$
and
$\overline{\int_{A}}|f| \geq \overline{\int_{A}}(-f) \geq - \underline{\int_{A}} f \geq - \overline{\int_{A}} f. \square$
For $$(\mathrm{h})$$ and later work, we need the following lemmas.

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

Let $$f : S \rightarrow E^{*}$$ and $$A \in \mathcal{M} .$$ Then the following are true.
(i) If
$\int_{A} f<q \in E^{*},$
there is an elementary and (extended) real map
$h \geq f \text { on } A,$
with
$\int_{A} h<q.$
(ii) If
$\int_{A} f>p \in E^{*},$
there is an elementary and (extended) real map
$g \leq f \text { on } A,$
with
$\int_{A} g>p;$
moreover, $$g$$ can be made elementary and nonnegative if $$f \geq 0$$ on $$A$$.

Proof

If $$f \geq 0,$$ this is immediate by Definition 1 and the properties of glb and lub.
If, however, $$f \ngeq 0,$$ and if
$q>\int_{A} f=\overline{\int_{A}} f^{+}- \underline{\int_{A}} f^{-},$
our conventions yield
$\infty>\int_{A} f^{+} .(\mathrm{Why} ?)$
Thus there are $$u, v \in E^{*}$$ such that $$q=u+v$$ and
$0 \leq \int_{A} f^{+}<u<\infty$
and
$-\int_{A} f^{-}<v.$
To see why this is so, choose $$u$$ so close to $$\overline{\int}_{A} f^{+}$$ that
$q-u>-\underline{\int_{A}} f^{-}$
and set $$v=q-u$$.
As the lemma holds for positive functions, we find elementary and nonnegative maps $$h^{\prime}$$ and $$h^{\prime \prime},$$ with
$h^{\prime} \geq f^{+}, h^{\prime \prime} \leq f^{-},$
$\int_{A} h^{\prime}<u<\infty \text { and } \int_{A} h^{\prime \prime}>-v.$
Let $$h=h^{\prime}-h^{\prime \prime} .$$ Then
$h \geq f^{+}-f^{-}=f,$
and by Problem 6 in §4,
$\int_{A} h=\int_{A} h^{\prime}-\int_{A} h^{\prime \prime} \quad\left(\text { for } \int_{A} h^{\prime} \text { is finite! }\right).$
Hence
$\int_{A} h>u+v=q,$
and clause (i) is proved in full.
Clause (ii) follows from (i) by Theorem 1$$\left(\mathrm{e}^{\prime}\right)$$ if
$\underline{\int_{A}} f<\infty.$
(Verify!) For the case $$\underline{\int_{A}} f=\infty,$$ see Problem $$3 . \square$$

Note 3. The preceding lemma shows that formulas $$\left(1^{\prime}\right)$$ and $$\left(1^{\prime \prime}\right)$$ hold (and might be used as definitions) even for sign-changing $$f, g,$$ and $$h$$.

## Lemma $$\PageIndex{2}$$

If $$f : S \rightarrow E^{*}$$ and $$A \in \mathcal{M},$$ there are $$\mathcal{M}$$ -measurable maps $$g$$ and $$h,$$ with
$g \leq f \leq h \text { on } A,$
such that
$\overline{\int_{A}} f=\overline{int_{A}} h \text { and } \underline{\int_{A}} f= \underline{\int_{A}} g.$
We can take $$g, h \geq 0$$ if $$f \geq 0$$ on $$A$$.

Proof

If
$\overline{\int_{A}} f=\infty,$
the constant map $$h=\infty$$ satisfies the statement of the theorem.
If
$-\infty<\overline{\int_{A}} f<\infty,$
let
$q_{n}=\overline{\int_{A}} f+\frac{1}{n}, \quad n=1,2, \ldots;$
so
$q_{n} \rightarrow \overline{\int_{A}} f<q_{n}.$
By Lemma $$1,$$ for each $$n$$ there is an elementary and (extended) real (hence measurable) map $$h_{n} \geq f$$ on $$A,$$ with
$q_{n} \geq \int_{A} h_{n} \geq \overline{\int_{A}} f.$
Let
$h=\inf _{n} h_{n} \geq f.$
By Lemma 1 in §2, $$h$$ is $$\mathcal{M}$$-measurable on $$A .$$ Also,
$(\forall n) \quad q_{n}>\int_{A} h_{n} \geq \overline{\int_{A}} h \geq \overline{\int_{A}} f$
by Theorem 1$$(\mathrm{c}) .$$ Hence
$\overline{\int_{A}} f=\lim _{n \rightarrow \infty} q_{n} \geq \overline{\int_{A}} h \geq \overline{\int_{A}} f,$
so
$\overline{\int_{A}} f=\overline{\int_{A}} h,$
as required.
Finally, if
$\overline{\int_{A}} f=-\infty,$
the same proof works with $$q_{n}=-n .$$ (Verify! $$)$$
Similarly, one finds a measurable map $$g \leq f,$$ with
$\underline{\int_{A}} f=\underline{\int_{A}} g. \square$

Proof of Theorem 1(h). If $$f \geq 0,$$ choose $$h \geq f$$ as in Lemma $$2 .$$ Let
$D=A(h>0) \text { and } A_{n}=A\left(h>\frac{1}{n}\right);$
so
$D=\bigcup_{n=1}^{\infty} A_{n}(\text { why } ?)$
and $$D, A_{n} \in \mathcal{M}$$ by Theorem 1 of §2. Also,
$0=\overline{\int_{A}} f=\overline{\int_{A}} h \geq \int_{A_{n}}\left(\frac{1}{n}\right)=\frac{1}{n} m A_{n} \geq 0.$
Thus $$(\forall n) m A_{n}=0 .$$ Hence
$m D=m \bigcup_{n=1}^{\infty} A_{n}=m A(h>0)=0;$
so $$0 \leq f \leq h \leq 0 (\text { i.e., } f=0)$$ a.e. on $$A$$.
The case $$f \leq 0$$ reduces to $$(-f) \geq 0$$. $$\square$$

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

If
$\overline{\int_{A}}|f|<\infty,$
then $$|f|<\infty$$ a.e. on $$A,$$ and $$A(f \neq 0)$$ is $$\sigma$$-finite.

Proof

By Lemma $$1,$$ fix an elementary and nonnegative $$h \geq|f|$$ with
$\int_{A} h<\infty$
(so $$h$$ is elementary and integrable).
Now, by Corollary $$1(\mathrm{i})-(\text { iii })$$ in §4, our assertions apply to $$h,$$ hence certainly to $$f . \square$$

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

(additivity). Given $$f : S \rightarrow E^{*}$$ and an $$\mathcal{M}$$-partition $$\mathcal{P}=\left\{B_{n}\right\}$$ of $$A \in \mathcal{M},$$ we have
$\text { (a) } \overline{\int}_{A} f=\sum_{n} \overline{\int}_{B_{n}} f \quad \text { and } \quad \text { (b) } \underline{\int_{A}} f=\sum_{n} \underline{\int_{B_{n}}} f,$
provided
$\overline{\int_{A}} f\left(\underline{\int_{A}} f, \text { respectively }\right)$
is orthodox, or $$\mathcal{P}$$ is finite.
Hence if $$f$$ is integrable on each of finitely many disjoint M-sets $$B_{n},$$ it is so on
$A=\bigcup_{n} B_{n},$
and formulas $$(2)(\mathrm{a})(\mathrm{b})$$ apply.

Proof

Assume first $$f \geq 0$$ on $$A .$$ Then by Theorem $$1(\mathrm{f}),$$ if one of
$\overline{\int_{B_{n}}} f=\infty,$
so is $$\overline{\int}_{A} f,$$ and all is trivial. Thus assume all $$\int_{B_{n}} f$$ are finite.
Then for any $$\varepsilon>0$$ and $$n \in N,$$ there is an elementary and nonnegative map $$h_{n} \geq f$$ on $$B_{n},$$ with
$\int_{B_{n}} h_{n}<\overline{\int}_{B_{n}} f+\frac{\varepsilon}{2^{n}}.$
(Why?) Now define $$h : A \rightarrow E^{*}$$ by $$h=h_{n}$$ on $$B_{n}, n=1,2, \ldots$$
Clearly, $$h$$ is elementary and nonnegative on each $$B_{n},$$ hence on $$A$$ (Corollary 3 in §1), and $$h \geq f$$ on $$A .$$ Thus by Theorem 1 of §4,
$\overline{\int}_{A} f \leq \int_{A} h=\sum_{n} \int_{B_{n}} h_{n} \leq \sum_{n}\left(\overline{\int_{B_{n}}} f+\frac{\varepsilon}{2^{n}}\right) \leq \sum_{n} \overline{\int}_{B_{n}} f+\varepsilon.$
Making $$\varepsilon \rightarrow 0,$$ we get
$\overline{\int_{A}} f \leq \sum_{n} \overline{\int}_{B_{n}} f.$
To prove also
$\overline{\int_{A}} f \geq \sum_{n} \overline{\int}_{B_{n}} f,$
take any elementary and nonnegative map $$H \geq f$$ on $$A .$$ Then again,
$\int_{A} H=\sum_{n} \int_{B_{n}} H \geq \sum_{n} \overline{\int}_{B_{n}} f.$
As this holds for any such $$H,$$ we also have
$\overline{\int_{A}} f=\inf _{H} \int_{A} H \geq \sum_{n} \overline{\int}_{B_{n}} f.$
This proves formula (a) for $$f \geq 0 .$$ The proof of $$(\mathrm{b})$$ is quite similar.
If $$f \ngeq 0,$$ we have
$\overline{\int_{A}} f=\overline{\int_{A}} f^{+} - \underline{\int_{A}} f^{-},$
where by the first part of the proof,
$\overline{\int_{A}} f^{+}=\sum_{n} \overline{\int_{B_{n}}} f^{+} \text { and } \underline{\int_{A}} f^{-}=\sum_{n} \underline{\int_{B_{n}}} f^{-}.$
If
$\overline{\int_{A}} f$
is orthodox, one of these sums must be finite, and so their difference may be rearranged to yield
$\overline{\int_{A}} f=\sum_{n}\left(\overline{\int_{B_{n}}} f^{+}-\underline{\int_{B_{n}}} f^{-}\right)=\sum_{n} \overline{\int_{B_{n}}} f,$
proving (a). Similarly for (b).
This rearrangement works also if $$\mathcal{P}$$ is finite (i.e., the sums have a finite number of terms $$) .$$ For, then, all reduces to commutativity and associativity of addition, and our conventions $$\left(2^{*}\right)$$ of Chapter 4, §4. Thus all is proved. $$\square$$

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

If $$m Q=0 (Q \in \mathcal{M}),$$ then for $$A \in \mathcal{M}$$
$\overline{\int}_{A-Q} f=\overline{\int_{A}} f \text { and } \underline{\int_{A-Q}} f= \underline{\int_{A}} f.$
For by Theorem 2,
$\overline{\int_{A}} f=\overline{\int_{A-Q}} f+\overline{\int_{A \cap Q}} f,$
where
$\overline{\int_{A \cap Q}} f=0$
by Theorem 1(b).

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

If
$\overline{\int_{A}} f \left(o r \underline{\int_{A}} f\right)$
is orthodox, so is
$\overline{\int_{X}} f \left( \underline{\int_{X}} f \right)$
whenever $$A \supseteq X, X \in \mathcal{M}$$.
For if
$\overline{\int_{A}} f^{+}, \overline{\int_{A}} f^{-}, \underline{\int_{A}} f^{+}, \text { or } \underline{\int_{A}} f^{-} \text { is finite, }$
it remains so also when $$A$$ is reduced to $$X$$ (see Theorem 1$$(\mathrm{f})$$. Hence orthodoxy follows by formula $$(1)$$.
Note 4. Given $$f : S \rightarrow E^{*},$$ we can define two additive (by Theorem 2) set functions $$\overline{s}$$ and $$\underline{s}$$ by setting for $$X \in \mathcal{M}$$
$\overline{s} X=\overline{\int_{X}} f \text { and } \underline{s} X=\underline{\int_{X}} f.$
They are called, respectively, the upper and lower indefinite integrals of $$f,$$ also denoted by
$\overline{\int} f \text { and } \underline{\int} f$
$$\left(\text { or } \overline{s}_{f} \text { and } \underline{s}_{f}\right)$$.
By Theorem 2 and Corollary $$3,$$ if
$\overline{\int_{A}} f$
is orthodox, then $$\overline{s}$$ is $$\sigma$$ -additive (and semifinite) when restricted to $$\mathcal{M}$$-sets $$X \subseteq A .$$ Also
$\overline{s} \emptyset=\underline{s} \emptyset=0$
by Theorem 1(b).
Such set functions are called signed measures (see Chapter 7, §11). In particular, if $$f \geq 0$$ on $$S, \overline{s}$$ and $$\underline{s}$$ are $$\sigma$$-additive and nonnegative on all of $$\mathcal{M},$$ hence measures on $$\mathcal{M}$$.

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

If $$f : S \rightarrow E^{*}$$ is m-measurable (Definition 2 in §3) on $$A,$$ then
$\overline{\int_{A}} f = \underline{\int_{A}} f.$

Proof

First, let $$f \geq 0$$ on $$A .$$ By Corollary $$2,$$ we may assume that $$f$$ is $$\mathcal{M}$$-measurable on $$A$$ (drop a set of measure zero). Now fix $$\varepsilon>0$$.
Let $$A_{0}=A(f=0), A_{\infty}=A(f=\infty),$$ and
$A_{n}=A\left((1+\varepsilon)^{n} \leq f<(1+\varepsilon)^{n+1}\right), \quad n=0, \pm 1, \pm 2, \ldots$
Clearly, these are disjoint $$\mathcal{M}$$ -sets (Theorem 1 of §2), and
$A=A_{0} \cup A_{\infty} \cup \bigcup_{n=-\infty}^{\infty} A_{n}.$
Thus, setting
$g=\left\{\begin{array}{ll}{0} & {\text { on } A_{0}} \\ {\infty} & {\text { on } A_{\infty}, \text { and }} \\ {(1+\varepsilon)^{n}} & {\text { on } A_{n}(n=0, \pm 1, \pm 2, \ldots)}\end{array}\right.$
and
$$h=(1+\varepsilon) g$$ on $$A$$,
we obtain two elementary and nonnegative maps, with
$g \leq f \leq h \text { on } A .(\mathrm{Why} ?)$
By Note 1,
$\underline{\int_{A}} g = \overline{\int_{A}} g.$
Now, if $$\int_{A} g=\infty,$$ then
$\overline{\int_{A}} f \geq \underline{\int_{A}} f \geq \int_{A} g$
yields
$\overline{\int_{A}} f \geq \underline{\int_{A}} f=\infty.$
If, however, $$\int_{A} g<\infty,$$ then
$\int_{A} h=\int_{A}(1+\varepsilon) g=(1+\varepsilon) \int_{A} g<\infty;$
so $$g$$ and $$h$$ are elementary and integrable on $$A .$$ Thus by Theorem 2(ii) in §4,
$\int_{A} h-\int_{A} g=\int_{A}(h-g)=\int_{A}((1+\varepsilon) g-g)=\varepsilon \int_{A} g.$
Moreover, $$g \leq f \leq h$$ implies
$\int_{A} g \leq \underline{\int_{A}} f \leq \overline{\int_{A}} f \leq \int_{A} h;$
so
$\left|\overline{\int}_{A} f-\underline{\int_{A}} f\right| \leq \int_{A} h-\int_{A} g \leq \varepsilon \int_{A} g.$
As $$\varepsilon$$ is arbitrary, all is proved for $$f \geq 0$$.
The case $$f \ngeq 0$$ now follows by formula (1), since $$f^{+}$$ and $$f^{-}$$ are $$\mathcal{M}-$$measurable (Theorem 2 in §2). $$\square$$

This page titled 8.5: Integration of Extended-Real Functions is shared under a CC BY license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) .