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. $\square$