8.10: Integration in Generalized Measure Spaces

Let $(S, \mathcal{M}, s)$ be a generalized measure space. By Note 1 in §3, a map $f$ is $s$-measurable iff it is $v_{s}$-measurable. This naturally leads us to the following definition.

Definition

A map $f : S \rightarrow E$ is $s$-integrable on a set $A$ iff it is $v_{s}$-integrable on $A.$ (Recall that $v_{s},$ the total variation of $s,$ is a measure.)

Note 1. Here the range spaces of $f$ and $s$ are assumed complete and such that $f(x) s A$ is defined for $x \in S$ and $A \in \mathcal{M}.$ Thus if $s$ is vector valued, $f$ must be scalar valued, and vice versa. Later, if a factor $p$ occurs, it must be such that $p f(x) s A$ is defined, i.e., at least two of $p, f(x),$ and $s A$ are scalars.

Note 2. If $s$ is a measure $(\geq 0),$ then $v_{s}=s^{+}=s$ (Corollary 3 in Chapter 7, §11); so our present definition agrees with the previous ones (as in Theorem 1 of §7).

Lemma $\PageIndex{1}$

If $m^{\prime}$ and $m^{\prime \prime}$ are measures, with $m^{\prime} \geq m^{\prime \prime}$ on $\mathcal{M},$ then

$$\int_{A}|f| d m^{\prime} \geq \int_{A}|f| d m^{\prime \prime}$$

for all $A \in \mathcal{M}$ and any $f : S \rightarrow E$.

Proof

First, take any elementary and nonnegative map $g \geq|f|$,

$$g=\sum_{i} C_{A_{i}} a_{i} \text { on } A.$$

Then (§4)

$$\int_{A} g d m^{\prime}=\sum a_{i} m^{\prime} A_{i} \geq \sum a_{i} m^{\prime \prime} A_{i}=\int_{A} g d m^{\prime \prime}.$$

Hence by Definition 1 in §5,

$$\int_{A}|f| d m^{\prime}=\inf _{g \geq|f|} \int_{A} g d m^{\prime} \geq \inf _{g \geq|f|} \int_{A} g d m^{\prime \prime}=\int_{A}|f| d m^{\prime \prime},$$

as claimed.$\quad \square$

Lemma $\PageIndex{2}$

(i) If $s : \mathcal{M} \rightarrow E^{n}\left(C^{n}\right)$ with $s=\left(s_{1}, \ldots, s_{n}\right),$ and if $f$ is s-integrable on $A \in \mathcal{M},$ then $f$ is $s_{k}$-integrable on $A$ for $k=1,2, \ldots, n.$

(ii) If $s$ is a signed measure and $f$ is s-integrable on $A,$ then $f$ is integrable on $A$ with respect to both $s^{+}$ and $s^{-}$ (with $s^{+}$ and $s^{-}$ as in formula (3) in Chapter 7, §11).

Note 3. The converse statements hold if $f$ is $\mathcal{M}$-measurable on $A$.

Proof

(i) If $s=\left(s_{1}, \ldots, s_{n}\right),$ then (Problem 4 of Chapter 7, §11)

$$v_{s} \geq v_{s_{k}}, \quad k=1, \ldots, n.$$

Hence by Definition 1 and Lemma 1, the $s$-integrability of $f$ implies

$$\infty>\int_{A}|f| d v_{s} \geq \int_{A}|f| d v_{s_{k}}.$$

Also, $f$ is $v_{s}$-measurable, i.e., $\mathcal{M}$-measurable on $A-Q,$ with

$$0=v_{s} Q \geq v_{s_{k}} Q \geq 0.$$

Thus $f$ is $s_{k}$ -integrable on $A, k=1, \ldots, n,$ as claimed.

(ii) If $s=s^{+}-s^{-},$ then by Theorem 4 in Chapter 7, §11, and Corollary 3 there, $s^{+}$ and $s^{-}$ are measures $(\geq 0)$ and $v_{s}=s^{+}+s^{-},$ so that both

$$v_{s} \geq s^{+}=v_{s^{+}} \text { and } v_{s} \geq s^{-}=v_{s^{-}}.$$

Thus the desired result follows exactly as in part (i) of the proof.$\quad \square$

We leave Note 3 as an exercise.

Definition

If $f$ is $s$-integrable on $A \in \mathcal{M},$ we set

(i) in the case $s : \mathcal{M} \rightarrow E^{*}$,

$$\int_{A} f d s=\int_{A} f d s^{+}-\int_{A} f d s^{-},$$

with $s^{+}$ and $s^{-}$ as in formula (3) of Chapter 7, §11;

(ii) in the case $s : \mathcal{M} \rightarrow E^{n}\left(C^{n}\right)$,

$$\int_{A} f d s=\sum_{k=1}^{n} \vec{e}_{k} \int_{A} f d s_{k},$$

with $\vec{e}_{k}$ as in Theorem 2 of Chapter 3, §§1-3;

(iii) if $s : \mathcal{M} \rightarrow C$,

$$\int_{A} f d s=\int_{A} f d s_{re}+i \cdot \int_{A} f d s_{im}.$$

(See also Problems 2 and 3.)

Note 4. If $s$ is a measure, then

$$s=s^{+}=s_{re}=s_{1}$$

and

$$0=s^{-}=s_{im}=s_{2};$$

so Definition 2 agrees with our previous definitions. Similarly for $s : \mathcal{M} \rightarrow$ $E^{n}\left(C^{n}\right).$

Below, $s, t,$ and $u$ are generalized measures on $\mathcal{M}$ as in Definition 2, while $f, g : S \rightarrow E$ are functions, with $E$ a complete normed space, as in Note 1.

Theorem $\PageIndex{1}$

The linearity, additivity, and $\sigma$-additivity properties (as in §7, Theorems 2 and 3) also apply to integrals

$$\int_{A} f ds,$$

with $s$ as in Definition 2.

Proof

(i) Linearity: Let $f, g : S \rightarrow E$ be s-integrable on $A \in \mathcal{M}.$ Let $p, q$ be suitable constants (see Note 1).

If $s$ is a signed measure, then by Lemma 2(ii) and Definitions 1 and 2, $f$ is integrable with respect to $v_{s}, s^{+},$ and $s^{-}.$ As these are measures, Theorem 2 in §7 shows that $pf+qg$ is integrable with respect to $v_{s}, s^{+},$ and $s^{-},$ and by Definition 2,

$$\begin{aligned} \int_{A}(p f+q g) d s &=\int_{A}(p f+q g) d s^{+}-\int_{A}(p f+q g) d s^{-} \\ &=p \int_{A} f d s^{+}+q \int_{A} g d s^{+}-p \int_{A} f d s^{-}-q \int_{A} g d s^{-} \\ &=p \int_{A} f d s+q \int_{A} g d s. \end{aligned}$$

Thus linearity holds for signed measures. Via components, it now follows for $s : \mathcal{M} \rightarrow E^{n}\left(C^{n}\right)$ as well. Verify!

(ii) Additivity and $\sigma$-additivity follow in a similar manner.$\quad \square$

Corollary $\PageIndex{1}$

Assume $f$ is s-integrable on $A,$ with $s$ as in Definition 2.

(i) If $f$ is constant $(f=c)$ on $A,$ we have

$$\int_{A} f d s=c \cdot s A.$$

(ii) If

$$f=\sum_{i} a_{i} C_{A_{i}}$$

for an $\mathcal{M}$-partition $\left\{A_{i}\right\}$ of $A,$ then

$$\int_{A} f d s=\sum_{i} a_{i} s A_{i} \text { and } \int_{A}|f| d s=\sum_{i}\left|a_{i}\right| s A_{i}$$

(both series absolutely convergent).

(iii) $|f|<\infty$ a.e. on $A$.

(iv) $\int_{A}|f| d v_{s}=0$ iff $f=0$ a.e. on $A$.

(v) The set $A$ $(f \neq 0)$ is $\left(v_{s}\right)$ $\sigma$-finite (Definition 4 in Chapter 7, §5).

(vi) $\int_{A} f d s=\int_{A-Q} f d s$ if $v_{s} Q=0$ or $f=0$ on $Q$ $(Q \in \mathcal{M})$.

(vii) $f$ is s-integrable on any $\mathcal{M}$-set $B \subseteq A$.

Proof

(i) If $s=s^{+}-s^{-}$ is a signed measure, we have by Definition 2 that

$$\int_{A} f d s=\int_{A} f d s^{+}-\int_{A} f d s^{-}=c\left(s^{+} A-s^{-} A\right)=c \cdot s A,$$

as required.

For $s : \mathcal{M} \rightarrow E^{n}\left(C^{n}\right),$ the result now follows via components. (Verify!)

(ii) As $f=a_{i}$ on $A_{i},$ clause (i) yields

$$\int_{A_{i}} f d s=a_{i} s A_{i}, \quad i=1,2, \ldots.$$

Hence by $\sigma$-additivity,

$$\int_{A} f d s=\sum_{i} \int_{A_{i}} f d s=\sum_{i} a_{i} s A_{i},$$

as claimed.

Clauses (iii), (iv), and (v) follow by Corollary 1 in §5 and Theorem 1(b)(h) there, as $v_{s}$ is a measure; (vi) is proved as §5, Corollary 2. We leave (vii) as an exercise.$\quad \square$

Theorem $\PageIndex{2}$ (dominated convergence)

If

$$f=\lim _{i \rightarrow \infty} f_{i} \text { (pointwise)}$$

on $A-Q\left(v_{s} Q=0\right)$ and if each $f_{i}$ is s-integrable on $A,$ so is $f,$ and

$$\int_{A} f d s=\lim _{i \rightarrow \infty} \int_{A} f_{i} d s,$$

all provided that

$$(\forall i) \quad\left|f_{i}\right| \leq g$$

for some map $g$ with $\int_{A} g d v_{s}<\infty$.

Proof

If $s$ is a measure, this follows by Theorem 5 in §6. Thus as $v_{s}$ is a measure, $f$ is $v_{s}$-integrable (hence $s$-integrable) on $A,$ as asserted.

Next, if $s=s^{+}-s^{-}$ is a signed measure, Lemma 2 shows that $f$ and the $f_{i}$ are $s^{+}$ and $s^{-}$-integrable as well, with

$$\int_{A}\left|f_{i}\right| d s^{+} \leq \int_{A}\left|f_{i}\right| d v_{s} \leq \int_{A} g d v_{s}<\infty;$$

similarly for

$$\int_{A}\left|f_{i}\right| d s^{-}.$$

As $s^{+}$ and $s^{-}$ are measures, Theorem 5 of §6 yields

$$\int_{A} f d s=\int_{A} f d s^{+}-\int_{A} f d s^{-}=\lim \left(\int_{A} f_{i} d s^{+}-\int_{A} f_{i} d s^{-}\right)=\lim \int_{A} f_{i} d s.$$

Thus all is proved for signed measures.

In the case $s : \mathcal{M} \rightarrow E^{n}\left(C^{n}\right),$ the result now easily follows by Definition 2(ii)(iii) via components.$quad \square$

Theorem $\PageIndex{3}$ (uniform convergence)

If $f_{i} \rightarrow f$ (uniformly) on $A-Q$ $(v_{s} A<\infty, v_{s} Q=0),$ and if each $f_{i}$ is s-integrable on $A,$ so is $f,$ and

$$\int_{A} f d s=\lim _{i \rightarrow \infty} \int_{A} f_{i} d s.$$

Proof

Argue as in Theorem 2, replacing §6, Theorem 5, by §7, Lemma 1.$\quad \square$

Our next theorem shows that integrals behave linearly with respect to measures.

Theorem $\PageIndex{4}$

Let $t, u : \mathcal{M} \rightarrow E^{*}\left(E^{n}, C^{n}\right),$ with $v_{t}<\infty$ on $\mathcal{M},$ and let

$$s=p t+q u$$

for finite constants $p$ and $q.$ Then the following statements are true.

(a) If $t$ and $u$ are generalized measures, so is $s.$

(b) If, further, $f$ is $\mathcal{M}$-measurable on a set $A$ and is both $t$-and $u$-integrable on $A,$ it is also s-integrable on $A,$ and

$$\int_{A} f d s=p \int_{A} f d t+q \int_{A} f d u.$$

Proof

We consider only assertion (b) for $s=t+u;$ the rest is easy.

First, let $f$ be $\mathcal{M}$-elementary on $A.$ By Corollary 1(ii), we set

$$\int_{A} f d t=\sum_{i} a_{i} t A_{i} \text { and } \int_{A} f d u=\sum_{i} a_{i} u A_{i}.$$

Also, by integrability,

$$\infty>\int_{A}|f| d v_{t}=\sum\left|a_{i}\right| v_{t} A_{i} \text { and } \infty>\int_{A}|f| d v_{u}=\sum_{i}\left|a_{i}\right| v_{u} A_{i}.$$

Now, by Problem 4 in Chapter 7, §11,

$$v_{s}=v_{t+u} \leq v_{t}+v_{u};$$

so

$$\begin{aligned} \int_{A}|f| d v_{s} &=\sum_{i}\left|a_{i}\right| v_{s} A_{i} \\ & \leq \sum_{i}\left|a_{i}\right|\left(v_{t} A_{i}+v_{u} A_{i}\right)=\int_{A}|f| d v_{t}+\int_{A}|f| d v_{u}<\infty. \end{aligned}$$

As $f$ is also $\mathcal{M}$-measurable (even elementary), it is $s$-integrable on $A$ (by Definition 1), and

$$\int_{A} f d s=\sum_{i} a_{i} s A_{i}=\sum_{i} a_{i}\left(t A_{i}+u A_{i}\right)=\int_{A} f d t+\int_{A} f d u,$$

as claimed.

Next, suppose $f$ is $\mathcal{M}$-measurable on $A$ and $v_{u} A<\infty.$ By assumption, $v_{t} A<\infty,$ too; so

$$v_{s} A \leq v_{t} A+v_{u} A<\infty.$$

Now, by Theorem 3 in §1,

$$f=\lim _{i \rightarrow \infty} f_{i} \text { (uniformly)}$$

for some $\mathcal{M}$-elementary maps $f_{i}$ on $A.$ By Lemma 2 in §7, for large $i,$ the $f_{i}$ are integrable with respect to both $v_{t}$ and $v_{u}$ on $A.$ By what was shown above, they are also $s$-integrable, with

$$\int_{A} f_{i} d s=\int_{A} f_{i} d t+\int_{A} f_{i} d u.$$

With $i \rightarrow \infty,$ Theorem 3 yields the result.

Finally, let $v_{u} A=\infty.$ By Corollary 1(v), we may assume (as in Lemma 3 of §7) that $A_{i} \nearrow A,$ with $v_{u} A_{i}<\infty,$ and $v_{t} A_{i}<\infty$ (since $v_{t}<\infty,$ by assumption). Set

$$f_{i}=f C_{A_{i}} \rightarrow f \text { (pointwise)}$$

on $A,$ with $\left|f_{i}\right| \leq|f|.$ (Why?)

As $f_{i}=f$ on $A_{i}$ and $f_{i}=0$ on $A-A_{i},$ all $f_{i}$ are both $t$- and $u$-integrable on $A$ (for $f$ is). Since $v_{t} A_{i}<\infty$ and $v_{u} A_{i}<\infty,$ the $f_{i}$ are also $s$-integrable (as shown above), with

$$\int_{A} f_{i} d s=\int_{A_{i}} f_{i} d s=\int_{A_{i}} f_{i} d t+\int_{A_{i}} f_{i} d u=\int_{A} f_{i} d t+\int_{A} f_{i} d u.$$

With $i \rightarrow \infty,$ Theorem 2 now yields the result.

To complete the proof of (b), it suffices to consider, along similar lines, the case $s=p t$ (or $s=q u$). We leave this to the reader.

For (a), see Chapter 7, §11.$\quad \square$

Theorem $\PageIndex{5}$

If $f$ is s-integrable on $A,$ so is $|f|,$ and

$$\left|\int_{A} f d s\right| \leq \int_{A}|f| d v_{s}.$$

Proof

By Definition 1, and Theorem 1 of §1, $f$ and $|f|$ are $\mathcal{M}$-measurable on $A-Q, v_{s} Q=0,$ and

$$\int_{A}|f| d v_{s}<\infty;$$

so $|f|$ is $s$-integrable on $A$.

The desired inequality is immediate by Corollary 1(ii) if $f$ is elementary.

Next, exactly as in Theorem 4, one obtains it for the case $v_{s} A<\infty,$ and then for $v_{s} A=\infty.$ We omit the details.$\quad \square$

Definition

We write

$$"ds=g dt \text { in } A"$$

or

$$"s=\int g dt \text{ in } A"$$

iff $g$ is $t$-integrable on $A,$ and

$$sX=\int_{X} g dt$$

for $A \supseteq X, X \in \mathcal{M}$.

We then call $s$ the indefinite integral of $g$ in $A.$ ($\int_{X} g dt$ may be interpreted as in Problems 2-4 below.)

Lemma $\PageIndex{3}$

If $A \in \mathcal{M}$ and

$$ds=g dt \text{ in } A,$$

then

$$dv_{s}=|g| dv_{t} \text { in } A.$$

Proof

By assumption, $g$ and $|g|$ are $v_{t}$-integrable on $X,$ and

$$sX=\int_{X} g dt$$

for $A \supseteq X, X \in \mathcal{M}.$ We must show that

$$v_{s} X=\int_{X}|g| dv_{t}$$

for such $X$.

This is easy if $g=c$ (constant) on $X.$ For by definition,

$$v_{s} X=\sup _{\mathcal{P}} \sum_{i}\left|s X_{i}\right|,$$

over all $\mathcal{M}$-partitions $\mathcal{P}=\left\{X_{i}\right\}$ of $X.$ As

$$sX_{i}=\int_{X_{i}} g dt=c \cdot t X_{i},$$

we have

$$v_{s} X=\sup _{\mathcal{P}} \sum_{i}|c|\left|t X_{i}\right|=|c| \sup _{\mathcal{P}} \sum_{i}\left|t X_{i}\right|=|c| v_{t} X;$$

so

$$v_{s} X=\int_{X}|g| dv_{t}.$$

Thus all is proved for constant $g$.

Hence by $\sigma$-additivity, the lemma holds for $\mathcal{M}$-elementary maps $g.$ (Why?)

In the general case, $g$ is $t$-integrable on $X,$ hence $\mathcal{M}$-measurable and finite on $X-Q, v_{t} Q=0.$ By Corollary 1(iii), we may assume $g$ finite and measurable on $X;$ so

$$g=\lim _{k \rightarrow \infty} g_{k} \text { (uniformly)}$$

on $X$ for some $\mathcal{M}$-elementary maps $g_{k},$ all integrable on $X,$ with respect to $v_{t}$ (and $t$).

Let

$$s_{k}=\int g_{k} dt$$

in $X.$ By what we just proved for elementary and integrable maps,

$$v_{s_{k}} X=\int_{X}\left|g_{k}\right| dv_{t}, \quad k=1,2, \ldots.$$

Now, if $v_{t} X<\infty,$ Theorem 3 yields

$$\int_{X}|g| d v_{t}=\lim _{k \rightarrow \infty} \int_{X}\left|g_{k}\right| d v_{t}=\lim _{k \rightarrow \infty} v_{s_{k}} X=v_{s} X$$

(see Problem 6). Thus all is proved if $v_{t} X<\infty$.

If, however, $v_{t} X=\infty,$ argue as in Theorem 4 (the last step), using the left continuity of $v_{s}$ and of

$$\int|g| dv_{t}.$$

Verify!$\quad \square$

Theorem $\PageIndex{6}$ (change of measure)

If $f$ is s-integrable on $A \in \mathcal{M},$ with

$$ds=g dt \text { in } A,$$

then (subject to Note 1) $fg$ is t-integrable on $A$ and

$$\int_{A} f ds=\int_{A} fg dt.$$

(Note the formal substitution of "$g dt$" for "$ds.$")

Proof

The proof is easy if $f$ is constant or elementary on $A$ (use Corollary 1(ii)). We leave this case to the reader, and next we assume $g$ is bounded and $v_{t} A<\infty.$

By s-integrability, $f$ is $\mathcal{M}$-measurable and finite on $A-Q,$ with

$$0=v_{s} Q=\int_{Q}|g| dv_{t}$$

by Lemma 3. Hence $0=g=f g$ on $Q-Z, v_{t} Z=0.$ Therefore,

$$\int_{Q} fg dt=0=\int_{Q} f ds$$

for $v_{s} Q=0.$ Thus we may neglect $Q$ and assume that $f$ is finite and $\mathcal{M}$-measurable on $A.$

As $ds=g dt,$ Definition 3 and Lemma 3 yield

$$v_{s} A=\int_{A}|g| dv_{t}<\infty.$$

Also (Theorem 3 in Chapter 8, §1),

$$f=\lim _{k \rightarrow \infty} f_{k} \quad \text {(uniformly)}$$

for elementary maps $f_{k},$ all $v_{s}$-integrable on $A$ (Lemma 2 in §7). As $g$ is bounded, we get on $A$

$$fg=\lim _{k \rightarrow \infty} f_{k} g \quad \text {(uniformly).}$$

Moreover, as the theorem holds for elementary and integrable maps, $f_{k} g$ is $t$-integrable on $A,$ and

$$\int_{A} f_{k} ds=\int_{A} f_{k} g dt, \quad k=1,2, \ldots.$$

Since $v_{s} A<\infty$ and $v_{t} A<\infty,$ Theorem 3 shows that $f g$ is $t$-integrable on $A,$ and

$$\int_{A} f ds=\lim _{k \rightarrow \infty} \int_{A} f_{k} ds=\lim _{k \rightarrow \infty} \int_{A} f_{k} g dt=\int_{A} fg dt.$$

Thus all is proved if $v_{t} A<\infty$ and $g$ is bounded on $A$.

In the general case, we again drop a null set to make $f$ and $g$ finite and $\mathcal{M}$-measurable on $A.$ By Corollary 1(v), we may again assume $A_{i} \nearrow A,$ with $v_{t} A_{i}<\infty$ $(\forall i).$

Now for $i=1,2, \ldots$ set

$$g_{i}=\left\{\begin{array}{ll}{g} & {\text { on } A_{i}(|g| \leq i),} \\ {0} & {\text { elsewhere.}}\end{array}\right.$$

Then each $g_{i}$ is bounded,

$$g_{i} \rightarrow g \text { (pointwise),}$$

and

$$\left|g_{i}\right| \leq|g|$$

on $A.$ We also set $f_{i}=f C_{A_{i}};$ so $f_{i} \rightarrow f$ (pointwise) and $\left|f_{i}\right| \leq|f|$ on $A.$ Then

$$\int_{A} f_{i} d s=\int_{A_{i}} f_{i} ds=\int_{A_{i}} f_{i} g_{i} dt=\int_{A} f_{i} g_{i} dt.$$

(Why?) Since $\left|f_{i} g_{i}\right| \leq|f g|$ and $f_{i} g_{i} \rightarrow f g,$ the result follows by Theorem 2.$\quad \square$