# 8.3: Measurable Functions in $$(S, \mathcal{M}, m)$$

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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. $$\square$$

This page titled 8.3: Measurable Functions in $$(S, \mathcal{M}, m)$$ is shared under a CC BY license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) .