
# 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 $$s : \mathcal{M} \rightarrow C$$,

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

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.$$\quad \square$$