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.
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).
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.\(\quad \square\)
(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.
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.
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.\(\quad \square\)
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.\(\quad \square\)
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.\(quad \square\)
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.
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.\(\quad \square\)
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.\(\quad \square\)
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.)
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!\(\quad \square\)
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.\(\quad \square\)