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8.5: Integration of Extended-Real Functions

( \newcommand{\kernel}{\mathrm{null}\,}\)

We shall now define integrals for arbitrary functions f:SE in a measure
space (S,M,m). We start with the case f0.

Definition

Given f0 on AM, we define the upper and lower integrals,
¯ and _,
of f on A (with respect to m) by
¯Af=¯Afdm=infhAh
over all elementary maps hf on A, and
Af=Afdm=supgAg
over all elementary and nonnegative maps gf on A.
If f is not nonnegative, we use f+=f0 and f=(f)0 (§2), and set
\[
Af=¯Afdm=¯Af+A_f andA_f=A_fdm=A_f+¯Af.

\]
By our conventions, these expressions are always defined. The integral ¯Af( or Af) is called orthodox iff it does not have the form in (1), e.g., ¯ if f0 (i.e., f=0), or if Af<. An unorthodox integral equals +.
We often write for ¯ and call it simply the integral (of f), even if
¯AfAf.
"Classical" notation is Af(x)dm(x).

Definition

The function f is called integrable (or m-integrable, or Lebesgue integrable, with respect to m) on A, iff
¯Afdm=Afdm±

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 (1) and (1) to be simple step functions only (see §9). It is inferior to Lebesgue theory.
The values of
¯Afdm and A_fdm
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 LS-integrals, and so on.
Note 1. If f is elementary and (extended) real, our present definition of
¯Af
agrees with that of §4. For if f0,f itself is the least of all elementary and nonnegative functions
hf
and the greatest of all elementary and nonnegative functions
gf.
Thus by Problem 5 in §4,
Af=minhfAh=maxgfAg,
i.e.,
Af=¯Af=A_f.
If, however, f 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


This page titled 8.5: Integration of Extended-Real Functions is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform.

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