8.5: Integration of Extended-Real Functions

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,
$$\overline{\int_{A}} f= \underline{\int_{A}} f \text { always.}$$
(See also Theorem 3.)
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). $\square$