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.

Definition

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.

Definition

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.

Corollary \(\PageIndex{1}\)

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. \(\square\)

Corollary \(\PageIndex{2}\)

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. \(\square\)

Corollary \(\PageIndex{3}\)

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.

Corollary \(\PageIndex{4}\)

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.

Lemma \(\PageIndex{1}\)

\(A \operatorname{map} f : S \rightarrow E^{n}\left(C^{n}\right)\) 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.

Lemma \(\PageIndex{2}\)

\(A \operatorname{map} f :(S, \rho) \rightarrow\left(T, \rho^{\prime}\right)\) 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. \(\square\)

Theorem \(\PageIndex{1}\)

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.

*Theorem \(\PageIndex{2}\)

(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. \(\square\)

*Lemma \(\PageIndex{3}\)

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 \((\mathrm{e}) .\) \(\square\)

*Lemma \(\PageIndex{4}\)

(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 \(H(x)=\arctan f(x)\). \(\square\)

*Theorem \(\PageIndex{3}\)

(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. \(\square\)