8.3: Measurable Functions in \((S, \mathcal{M}, m)\)
I. Henceforth we shall presuppose not just a measurable space (§1) but a measure space \((S, \mathcal{M}, m),\) where \(m : \mathcal{M} \rightarrow E^{*}\) is a measure on a \(\sigma\) -ring \(\mathcal{M} \subseteq 2^{S}\).
We saw in Chapter 7 that one could often neglect sets of Lebesgue measure zero on \(E^{n}-\) if a property held everywhere except on a set of Lebesgue measure zero, we said it held "almost everywhere." The following definition generalizes this usage.
We say that a property \(P(x)\) holds for almost all \(x \in A\) (with respect to the measure \(m )\) or almost everywhere (a.e. \((m) )\) on \(A\) iff it holds on \(A-Q\) for some \(Q \in \mathcal{M}\) with \(m Q=0\).
Thus we write
\[
f_{n} \rightarrow f(a . e .) \text { or } f=\lim f_{n}(a . e .(m)) \text { on } A
\]
iff \(f_{n} \rightarrow f(\text { pointwise })\) on \(A-Q, m Q=0 .\) Of course, "pointwise" implies \(" a . e . "(\text { take } Q=\emptyset),\) but the converse fails.
We say that \(f : S \rightarrow\left(T, \rho^{\prime}\right)\) is almost measurable on \(A\) iff \(A \in \mathcal{M}\) and \(f\) is \(\mathcal{M}\) -measurable on \(A-Q, m Q=0\).
We then also say that \(f\) is \(m\) -measurable (\(m\) being the measure involved ) as opposed to \(\mathcal{M}\)-measurable.
Observe that we may assume \(Q \subseteq A\) here (replace \(Q\) by \(A \cap Q )\).
*Note 1. If \(m\) is a generalized measure (Chapter 7, §11), replace \(m Q=0\) by \(v_{m} Q=0\left(v_{m}=\text { total variation of } m\right)\) in Definitions 1 and 2 and in the following proofs.
If the functions
\[
f_{n} : S \rightarrow\left(T, \rho^{\prime}\right), \quad n=1,2, \ldots
\]
are \(m\)-measurable on \(A,\) and if
\[
f_{n} \rightarrow f(a . e .(m))
\]
on \(A,\) then \(f\) is \(m\)-measurable on \(A\).
- Proof
-
By assumption, \(f_{n} \rightarrow f (\text { pointwise })\) on \(A-Q_{0}, m Q_{0}=0 .\) Also, \(f_{n}\) is \(\mathcal{M}\)-measurable on
\[
A-Q_{n}, m Q_{n}=0, \quad n=1,2, \dots
\](The \(Q_{n}\) need not be the same.)
Let
\[
Q=\bigcup_{n=0}^{\infty} Q_{n};
\]so
\[
m Q \leq \sum_{n=0}^{\infty} m Q_{n}=0.
\]By Corollary 2 in §1, all \(f_{n}\) are \(\mathcal{M}\)-measurable on \(A-Q\) (why?), and \(f_{n} \rightarrow f\)
(pointwise) on \(A-Q,\) as \(A-Q \subseteq A-Q_{0} .\)Thus by Theorem 4 in §1, \(f\) is \(\mathcal{M}\) -measurable on \(A-Q .\) As \(m Q=0\), this
is the desired result. \(\square\)
If \(f=g (a . e .(m))\) on \(A\) and \(f\) is \(m\)-measurable on \(A,\) so is \(g\).
- Proof
-
By assumption, \(f=g\) on \(A-Q_{1}\) and \(f\) is \(\mathcal{M}\)-measurable on \(A-Q_{2}\), with \(m Q_{1}=m Q_{2}=0\).
Let \(Q=Q_{1} \cup Q_{2} .\) Then \(m Q=0\) and \(g=f\) on \(A-Q .\) (Why? \()\)
By Corollary 2 of §1, \(f\) is \(\mathcal{M}\)-measurable on \(A-Q\). Hence so is \(g\), as claimed. \(\square\)
If \(f : S \rightarrow\left(T, \rho^{\prime}\right)\) is \(m\) -measurable on \(A,\) then
\[
f=\lim _{n \rightarrow \infty} f_{n} (\text {uniformly}) \text { on } A-Q(m Q=0),
\]
for some maps \(f_{n},\) all elementary on \(A-Q\).
- Proof
-
Add proof here and it will automatically be hidden
(Compare Corollary 3 with Theorem 3 in §1).
Quite similarly all other propositions of §1 carry over to almost measurable (i.e., \(m\) -measurable) functions. Note, however, that the term "measurable" in §§1 and 2 always meant \(" \mathcal{M}\) -measurable." This implies \(m\)-measurability (take \(Q=\emptyset ),\) but the converse fails. (See Note \(2,\) however.)
We still obtain the following result.
If the functions
\[
f_{n} : S \rightarrow E^{*} \quad(n=1,2, \ldots)
\]
are \(m\)-measurable on a set \(A,\) so also are
\[
\sup f_{n}, \inf f_{n}, \overline{\lim } f_{n}, \text { and } \underline{\lim} f_{n}
\]
(Use Lemma 1 of §2).
Similarly, Theorem 2 in §2 carries over to \(m\)-measurable functions.
Note 2. If \(m\) is complete (such as Lebesgue measure and LS measures) then, for \(f : S \rightarrow E^{*}\left(E^{n}, C^{n}\right), m-\) and \(\mathcal{M}\)-measurability coincide (see Problem 3 below).
II. Measurability and Continuity. To study the connection between these notions, we first state two lemmas, often treated as definitions.
\(A \operatorname{map} f : S \rightarrow E^{n}\left(C^{n}\right)\) is \(\mathcal{M}\)-measurable on \(A\) iff
\[
A \cap f^{-1}[B] \in \mathcal{M}
\]
for each Borel set (equivalently, open set) \(B\) in \(E^{n}\left(C^{n}\right)\).
- Proof
-
See Problems \(8-10\) in §2 for a sketch of the proof.
\(A \operatorname{map} f :(S, \rho) \rightarrow\left(T, \rho^{\prime}\right)\) is relatively continuous on \(A \subseteq S\) iff for any open set \(B \subseteq\left(T, \rho^{\prime}\right),\) the set \(A \cap f^{-1}[B]\) is open in \((A, \rho)\) as a subspace of \((S, \rho)\).
(This holds also with "open" replaced by "closed.")
- Proof
-
By Chapter 4, §1, footnote \(4, f\) is relatively continuous on \(A\) iff its restriction to \(A\) (call it \(g : A \rightarrow T )\) is continuous in the ordinary sense.
Now, by Problem 15\((\mathrm{iv})(\mathrm{v})\) in Chapter 4, §2, with \(S\) replaced by \(A,\) this means that \(g^{-1}[B]\) is open (closed) \(i n(A, \rho)\) when \(B\) is so in \(\left(T, \rho^{\prime}\right) .\) But
\[
g^{-1}[B]=\{x \in A | f(x) \in B\}=A \cap f^{-1}[B].
\]
(Why?) Hence the result follows. \(\square\)
Let \(m : \mathcal{M} \rightarrow E^{*}\) be a topological measure in \((S, \rho) .\) If \(f : S \rightarrow\) \(E^{n}\left(C^{n}\right)\) is relatively continuous on a set \(A \in \mathcal{M},\) it is \(\mathcal{M}\) -measurable on \(A\).
- Proof
-
Let \(B\) be open in \(E^{n}\left(C^{n}\right) .\) By Lemma 2,
\[
A \cap f^{-1}[B]
\]
is open \(i n(A, \rho) .\) Hence by Theorem 4 of Chapter 3, §12,
\[
A \cap f^{-1}[B]=A \cap U
\]
for some open set \(U\) in \((S, \rho)\).
Now, by assumption, \(A\) is in \(\mathcal{M} .\) So is \(U,\) as \(\mathcal{M}\) is topological \((\mathcal{M} \supseteq \mathcal{G})\).
Hence
\[
A \cap f^{-1}[B]=A \cap U \in \mathcal{M}
\]
for any open \(B \subseteq E^{n}\left(C^{n}\right) .\) The result follows by Lemma 1. \(\square\)
Note 3.
The converse fails. For example, the Dirichlet function (Example \((\mathrm{c})\) in Chapter 4, §1) is L-measurable (even simple) but discontinuous everywhere.
Note 4.
Lemma 1 and Theorem 1 hold for a map \(f : S \rightarrow\left(T, \rho^{\prime}\right),\) too, provided \(f[A]\) is separable, i.e.,
\[
f[A] \subseteq \overline{D}
\]
for a countable set \(D \subseteq T\) (cf. Problem 11 in §2).
*III.
For strongly regular measures (Definition 5 in Chapter 7, §7), we obtain the following theorem.
(Luzin). Let \(m : \mathcal{M} \rightarrow E^{*}\) be a strongly regular measure in \((S, \rho)\). Let \(f : S \rightarrow\left(T, \rho^{\prime}\right)\) be \(m\)-measurable on \(A\).
Then given \(\varepsilon>0,\) there is a closed set \(F \subseteq A(F \in \mathcal{M})\) such that
\[
m(A-F)<\varepsilon
\]
and \(f\) is relatively continuous on \(F\).
(Note that if \(T=E^{*}, \rho^{\prime}\) is as in Problem 5 of Chapter 3, §11.)
- Proof
-
By assumption, \(f\) is \(\mathcal{M}\)-measurable on a set
\[
H=A-Q, m Q=0;
\]
so by Problem 7 in §1, \(f[H]\) is separable in \(T\). We may safely assume that \(f\) is \(\mathcal{M}\)-measurable on \(S\) and that all of \(T\) is separable. (If not, replace \(S\) and \(T\) by \(H\) and \(f[H],\) restricting \(f\) to \(H,\) and \(m\) to \(\mathcal{M}\)-sets inside \(H ;\) see also Problems 7 and 8 below.)
Then by Problem 12 of §2, we can fix globes \(G_{1}, G_{2}, \ldots\) in \(T\) such that
\[
\text { each open set } B \neq \emptyset \text { in } T \text { is the union of a subsequence of }\left\{G_{k}\right\}.
\]
Now let \(\varepsilon>0,\) and set
\[
S_{k}=S \cap f^{-1}\left[G_{k}\right]=f^{-1}\left[G_{k}\right], \quad k=1,2, \ldots
\]
By Corollary 2 in §2, \(S_{k} \in \mathcal{M} .\) As \(m\) is strongly regular, we find for each \(S_{k}\) an open set
\[
U_{k} \supseteq S_{k},
\]
with \(U_{k} \in \mathcal{M}\) and
\[
m\left(U_{k}-S_{k}\right)<\frac{\varepsilon}{2^{k+1}}.
\]
Let \(B_{k}=U_{k}-S_{k}, D=\bigcup_{k} B_{k} ;\) so \(D \in \mathcal{M}\) and
\[
m D \leq \sum_{k} m B_{k} \leq \sum_{k} \frac{\varepsilon}{2^{k+1}} \leq \frac{1}{2} \varepsilon
\]
and
\[
U_{k}-B_{k}=S_{k}=f^{-1}\left[G_{k}\right].
\]
As \(D=\bigcup B_{k},\) we have
\[
(\forall k) \quad B_{k}-D=B_{k} \cap(-D)=\emptyset.
\]
Hence by \(\left(2^{\prime}\right)\),
\[
\begin{aligned}(\forall k) \quad f^{-1}\left[G_{k}\right] \cap(-D) &=\left(U_{k}-B_{k}\right) \cap(-D) \\ &=\left(U_{k} \cap(-D)\right)-\left(B_{k} \cap(-D)\right)=U_{k} \cap(-D) \end{aligned}.
\]
Combining this with \((1),\) we have, for each open set \(B=\bigcup_{i} G_{k_{i}} \operatorname{in} T\),
\[
f^{-1}[B] \cap(-D)=\bigcup_{i} f^{-1}\left[G_{k_{i}}\right] \cap(-D)=\bigcup_{i} U_{k_{i}} \cap(-D).
\]
since the \(U_{k_{i}}\) are open in \(S\) (by construction), the set \((3)\) is open in \(S-D\) as a subspace of \(S .\) By Lemma \(2,\) then, \(f\) is relatively continuous on \(S-D,\) or rather on
\[
H-D=A-Q-D
\]
(since we actually substituted \(S\) for \(H\) in the course of the proof). As \(m Q=0\) and \(m D<\frac{1}{2} \varepsilon\) by \((2)\),
\[
m(H-D)<m A-\frac{1}{2} \varepsilon.
\]
Finally, as \(m\) is strongly regular and \(H-D \in \mathcal{M},\) there is a closed \(\mathcal{M}\)-set
\[
F \subseteq H-D \subseteq A
\]
such that
\[
m(H-D-F)<\frac{1}{2} \varepsilon.
\]
since \(f\) is relatively continuous on \(H-D,\) it is surely so on \(F .\) Moreover,
\[
A-F=(A-(H-D)) \cup(H-D-F);
\]
so
\[
m(A-F) \leq m(A-(H-D))+m(H-D-F)<\frac{1}{2} \varepsilon+\frac{1}{2} \varepsilon=\varepsilon.
\]
This completes the proof. \(\square\)
Given \([a, b] \subset E^{1}\) and disjoint closed sets \(A, B \subseteq(S, \rho),\) there always is a continuous map \(g : S \rightarrow[a, b]\) such that \(g=a\) on \(A\) and \(g=b\).
- Proof
-
If \(A=\emptyset\) or \(B=\emptyset,\) set \(g=b\) or \(g=a\) on all of \(S\).
If, however, \(A\) and \(B\) are both nonempty, set
\[
g(x)=a+\frac{(b-a) \rho(x, A)}{\rho(x, A)+\rho(x, B)}.
\]
As \(A\) is closed, \(\rho(x, A)=0\) iff \(x \in A\) (Problem 15 in Chapter 3, §14); similarly for \(B .\) Thus \(\rho(x, A)+\rho(x, B) \neq 0\).
Also, \(g=a\) on \(A, g=b\) on \(B,\) and \(a \leq g \leq b\) on \(S\).
For continuity, see Chapter 4, §8, Example \((\mathrm{e}) .\) \(\square\)
(Tietze). If \(f :(S, \rho) \rightarrow E^{*}\) is relatively continuous on a closed set \(F \subseteq S,\) there is a function \(g : S \rightarrow E^{*}\) such that \(g=f\) on \(F\),
\[
\inf g[S]=\inf f[F], \sup g[S]=\sup f[F],
\]
and \(g\) is continuous on all of \(S\).
(We assume \(E^{*}\) metrized as in Problem 5 of Chapter 3, §11. If \(|f|<\infty,\) the standard metric in \(E^{1}\) may be used.)
- Proof Outline
-
First, assume inf \(f[F]=0\) and \(\sup f[F]=1 .\) Set
\[
A=F\left(f \leq \frac{1}{3}\right)=F \cap f^{-1}\left[\left[0, \frac{1}{3}\right]\right]
\]
and
\[
B=F\left(f \geq \frac{2}{3}\right)=F \cap f^{-1}\left[\left[\frac{2}{3}, 1\right]\right].
\]
As \(F\) is closed \(i n S,\) so are \(A\) and \(B\) by Lemma \(2 .\) (Why? \()\)
As \(B \cap A=\emptyset,\) Lemma 3 yields a continuous map \(g_{1} : S \rightarrow\left[0, \frac{1}{3}\right],\) with \(g_{1}=0\) on \(A,\) and \(g_{1}=\frac{1}{3}\) on \(B .\) Set \(f_{1}=f-g_{1}\) on \(F ;\) so \(\left|f_{1}\right| \leq \frac{2}{3},\) and \(f_{1}\) is continuous on \(F .\)
Applying the same steps to \(f_{1}\) (with suitable sets \(A_{1}, B_{1} \subseteq F ),\) find a continuous map \(g_{2},\) with \(0 \leq g_{2} \leq \frac{2}{3} \cdot \frac{1}{3}\) on \(S .\) Then \(f_{2}=f_{1}-g_{2}\) is continuous, and \(0 \leq f_{2} \leq\left(\frac{2}{3}\right)^{2}\) on \(F\).
Continuing, obtain two sequences \(\left\{g_{n}\right\}\) and \(\left\{f_{n}\right\}\) of real functions such that each \(g_{n}\) is continuous on \(S\),
\[
0 \leq g_{n} \leq \frac{1}{3}\left(\frac{2}{3}\right)^{n-1},
\]
and \(f_{n}=f_{n-1}-g_{n}\) is defined and continuous on \(F,\) with
\[
0 \leq f_{n} \leq\left(\frac{2}{3}\right)^{n}
\]
there \(\left(f_{0}=f\right)\).
We claim that
\[
g=\sum_{n=1}^{\infty} g_{n}
\]
is the desired map.
Indeed, the series converges uniformly on \(S\) (Theorem 3 of Chapter 4, §12).
As all \(g_{n}\) are continuous, so is \(g\) (Theorem 2 in Chapter 4, §12). Also,
\[
\left|f-\sum_{k=1}^{n} g_{k}\right| \leq\left(\frac{2}{3}\right)^{n} \rightarrow 0
\]
on \(F(\text { why? }) ;\) so \(f=g\) on \(F .\) Moreover,
\[
0 \leq g_{1} \leq g \leq \sum_{n=1}^{\infty} \frac{1}{3}\left(\frac{2}{3}\right)^{n}=1 \text { on } S.
\]
Hence inf \(g[S]=0\) and \(\sup g[S]=1,\) as required.
Now assume
\[
\inf f[F]=a<\sup f[F]=b \quad\left(a, b \in E^{1}\right)
\]
Set
\[
h(x)=\frac{f(x)-a}{b-a}
\]
so that inf \(h[F]=0\) and \(\sup h[F]=1 .\) (Why?)
As shown above, there is a continuous map \(g_{0}\) on \(S,\) with
\[
g_{0}=h=\frac{f-a}{b-a}
\]
on \(F,\) inf \(g_{0}[S]=0,\) and \(\sup g_{0}[S]=1 .\) Set
\[
a+(b-a) g_{0}=g.
\]
Then \(g\) is the required function. (Verify!)
Finally, if \(a, b \in E^{*}(a<b),\) all reduces to the bounded case by considering \(H(x)=\arctan f(x)\). \(\square\)
(Fréchet). Let \(m : \mathcal{M} \rightarrow E^{*}\) be a strongly regular measure in \((S, \rho) .\) If \(f : S \rightarrow E^{*}\left(E^{n}, C^{n}\right)\) is \(m\) -measurable on \(A,\) then
\[
f=\lim _{i \rightarrow \infty} f_{i}(a \cdot e .(m)) \text { on } A
\]
for some sequence of maps \(f_{i}\) continuous on \(S .\) (We assume \(E^{*}\) to be metrized as in Lemma 4.)
- Proof
-
We consider \(f : S \rightarrow E^{*}\) (the other cases reduce to \(E^{1}\) via components).
Taking \(\varepsilon=\frac{1}{i}(i=1,2, \ldots)\) in Theorem \(2,\) we obtain for each \(i\) a closed \(\mathcal{M}\)-set \(F_{i} \subseteq A\) such that
\[
m\left(A-F_{i}\right)<\frac{1}{i}
\]
and \(f\) is relatively continuous on each \(F_{i} .\) We may assume that \(F_{i} \subseteq F_{i+1}\) (if not, replace \(F_{i}\) by \(\bigcup_{k=1}^{i} F_{k} )\).
Now, Lemma 4 yields for each \(i\) a continuous map \(f_{i} : S \rightarrow E^{*}\) such that \(f_{i}=f\) on \(F_{i} .\) We complete the proof by showing that \(f_{i} \rightarrow f\) (pointwise) on the set
\[
B=\bigcup_{i=1}^{\infty} F_{i}
\]
and that \(m(A-B)=0\).
Indeed, fix any \(x \in B .\) Then \(x \in F_{i}\) for some \(i=i_{0},\) hence also for \(i>i_{0}\) (since \(\left\{F_{i}\right\} \uparrow ) .\) As \(f_{i}=f\) on \(F_{i},\) we have
\[
\left(\forall i>i_{0}\right) \quad f_{i}(x)=f(x),
\]
and so \(f_{i}(x) \rightarrow f(x)\) for \(x \in B .\) As \(F_{i} \subseteq B,\) we get
\[
m(A-B) \leq m\left(A-F_{i}\right)<\frac{1}{i}
\]
for all \(i .\) Hence \(m(A-B)=0,\) and all is proved. \(\square\)