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 \(f \geq 0\).
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 \(\int_{A} f(x) d m(x)\).
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
\[
\int_{-A} f^{+} \leq \overline{\int}_{A} f^{+} \text { and } \int_{-A} f^{-} \leq \overline{J}_{A} f^{-}.
\]
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.
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 \(h\).
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 \((-f) \geq 0\). \(\square\)
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 \(f . \square\)
(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. \(\square\)
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).
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 \(\mathcal{M}\).
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). \(\square\)