8.5.E: Problems on Integration of Extended-Real Functions

Exercise \(\PageIndex{1}\)

Using the formulas in ( 1) and our conventions, verify that
(i) \(\overline{\int}_{A} f=+\infty\) iff \(\overline{\int}_{A} f^{+}=\infty\);
(ii) \(\underline{\int}_{A} f=\infty\) iff \(\underline{\int}_{A} f^{+}=\infty ;\) and
(iii) \(\overline{f}_{A} f=-\infty\) iff \(\underline{\int}_{A} f^{-}=\infty\) and \(\overline{\int}_{A} f^{+}<\infty\).
(iv) Derive a condition similar to (iii) for \(\underline{\int}_{A} f=-\infty\).
(v) Review Problem 6 of Chapter 4, §4.

Exercise \(\PageIndex{2}\)

Fill in the missing proof details in Theorems 1 to 3 and Lemmas 1 and 2.

Exercise \(\PageIndex{3}\)

Prove that if \(\underline{\int_{A}} f=\infty,\) there is an elementary and (extended) real map \(g \leq f\) on \(A,\) with \(\int_{A} g=\infty\).
[Outline: By Problem \(1,\) we have
\[
\underline{\int_{A}} f^{+}=\infty .
\]
As Lemmas 1 and 2 surely hold for nonnegative functions, fix a measurable \(F \leq f^{+}\) \((F \geq 0),\) with
\[
\int_{A} F=\underline{\int_{A}} f^{+}=\infty .
\]
Arguing as in Theorem \(3,\) find an elementary and nonnegative map \(g \leq F,\) with
\[
(1+\varepsilon) \int_{A} g=\int_{A} F=\infty ;
\]
so \(\int_{A} g=\infty\) and \(0 \leq g \leq F \leq f^{+}\) on \(A\).
Let
\[
A_{+}=A(F>0) \in \mathcal{M}
\]
and
\[
A_{0}=A(F=0) \in \mathcal{M}
\]
(Theorem 1 in §2). On \(A_{+},\)
\[
g \leq F \leq f^{+}=f(\text { why? }) ,
\]
while on \(A_{0}, g=F=0 ;\) so
\[
\int_{A_{+}} g=\int_{A} g=\infty(\mathrm{why} ?) .
\]
Now redefine \(g=-\infty\) on \(A_{0}\) (only). Show that \(g\) is then the required function.]

Exercise \(\PageIndex{4}\)

For any \(f: S \rightarrow E^{*},\) prove the following.
(a) If \(\overline{\int}_{A} f<\infty,\) then \(f<\infty\) a.e. on \(A\).
(b) If \(\underline{\int_{A}} f\) is orthodox and \(>-\infty,\) then \(f>-\infty\) a.e. on \(A\).
[Hint: Use Problem 1 and apply Corollary 1 to \(f^{+} ;\) thus prove (a). Then for (b), use Theorem 1(e').]

Exercise \(\PageIndex{5}\)

\(\Rightarrow 5\). For any \(f, g: S \rightarrow E^{*},\) prove that
(i) \(\overline{\int}_{A} f+\overline{\int}_{A} g \geq \overline{\int}_{A}(f+g),\) and
(ii) \(\underline{\int}_{A}(f+g) \geq \underline{\int}_{A} f+\underline{\int}_{A} g \quad\) if \(\left|\underline{\int}_{A} g\right|<\infty\).
[Hint: Suppose that
\[
\overline{\int}_{A} f+\overline{\int}_{A} g<\overline{\int}_{A}(f+g) .
\]
Then there are numbers
\[
u>\overline{\int}_{A} f \text { and } v>\overline{\int}_{A} g ,
\]
with
\[
u+v \leq
overline{\int}_{A}(f+g) .
\]
(Why?) Thus Lemma 1 yields elementary and (extended) real maps \(F \geq f\) and \(G \geq g\) such that
\[
u>\overline{\int}_{A} F \text { and } v>\overline{\int}_{A} G
\]
As \(f+g \leq F+G\) on \(A,\) Theorem \(1(\mathrm{c})\) of §5 and Problem 6 of §4 show that
\[
\overline{\int}_{A}(f+g) \leq \int_{A}(F+G)=\int_{A} F+\int_{A} G<u+v ,
\]
contrary to
\[
u+v \leq \overline{\int}_{A}(f+g) .
\]
Similarly prove clause (ii).]

Exercise \(\PageIndex{6}\)

Continuing Problem \(5,\) prove that
\[
\overline{\int}_{A}(f+g) \geq \overline{\int}_{A} f+\underline{\int}_{A} g \geq \underline{\int}_{A}(f+g) \geq \underline{\int}_{A} f+\underline{\int}_{A} g
\]
provided \(\left|\underline{\int}_{A} g\right|<\infty\).
[Hint for the second inequality: We may assume that
\[
\overline{\int}_{A}(f+g)<\infty \text { and } \overline{\int}_{A} f>-\infty .
\]
(Why?) Apply Problems 5 and \(4(\mathrm{a})\) to
\[
\overline{\int}_{A}((f+g)+(-g)) .
\]
Use Theorem \(\left.1\left(\mathrm{e}^{\prime}\right) .\right]\)

Exercise \(\PageIndex{7}\)

Prove the following.
(i) \[
\overline{\int}_{A}|f|<\infty \text { iff }-\infty<\underline{\int}_{A} f \leq \overline{\int}_{A} f<\infty .
\]
(ii) If \(\overline{f}_{A}|f|<\infty\) and \(\overline{\int}_{A}|g|<\infty,\) then
\[
\left|\overline{\int}_{A} f-\overline{\int}_{A} g\right| \leq \overline{\int}_{A}|f-g|
\]
and
\[
\left|\underline{\int}_{A} f-\underline{\int}_{A} g\right| \leq \overline{\int}_{A}|f-g| .
\]
[Hint: Use Problems \(5 \text { and } 6 .]\)

Exercise \(\PageIndex{8}\)

Show that any signed measure \(\left.\bar{s}_{f} \text { (Note } 4\right)\) is the difference of two measures: \(\bar{s}_{f}=\bar{s}_{f+}-\bar{s}_{f-}\).