# 8.2: Measurability of Extended-Real Functions


Henceforth we presuppose a measurable space $$(S, \mathcal{M}),$$ where $$\mathcal{M}$$ is a $$\sigma$$-ring in $$S .$$ Our aim is to prove the following basic theorem, which is often used as a definition, for extended-real functions $$f : S \rightarrow E^{*} .$$

## Theorem $$\PageIndex{1}$$

$$A$$ function $$f : S \rightarrow E^{*}$$ is measurable on a set $$A \in \mathcal{M}$$ iff $$i t$$ satisfies one of the following equivalent conditions (hence all of them):

$\begin{array}{ll}{\left(\mathfrak{i}^{*}\right)\left(\forall a \in E^{*}\right) A(f>a) \in \mathcal{M} ;} & {\left(\mathrm{ii}^{*}\right)\left(\forall a \in E^{*}\right) A(f \geq a) \in \mathcal{M}}; \\ {\left(\mathrm{iii}^{*}\right)\left(\forall a \in E^{*}\right) A(f<a) \in \mathcal{M} ;} & {\left(\mathrm{iv}^{*}\right)\left(\forall a \in E^{*}\right) A(f \leq a) \in \mathcal{M}}.\end{array}$

We first prove the equivalence of these conditions by showing that $$\left(\mathrm{i}^{*}\right) \Rightarrow$$ $$\left(\mathrm{ii}^{*}\right) \Rightarrow\left(\mathrm{iii}^{*}\right) \Rightarrow\left(\mathrm{i} v^{*}\right) \Rightarrow\left(\mathrm{i}^{*}\right),$$ closing the "circle."

$$\left(\mathrm{i}^{*}\right) \Rightarrow\left(\mathrm{ii}^{*}\right) .$$ Assume $$\left(\mathrm{i}^{*}\right) .$$ If $$a=-\infty$$,

$A(f \geq a)=A \in \mathcal{M}$

by assumption. If $$a=+\infty$$,

$A(f \geq a)=A(f=\infty)=\bigcap_{n=1}^{\infty} A(f>n) \in \mathcal{M}$

by $$\left(\mathrm{i}^{*}\right) .$$ And if $$a \in E^{1}$$,

$A(f \geq a)=\bigcap_{n=1}^{\infty} A\left(f>a-\frac{1}{n}\right).$

(Verify!) By $$\left(\mathrm{i}^{*}\right)$$,

$A\left(f>a-\frac{1}{n}\right) \in \mathcal{M};$

so $$A(f \geq a) \in \mathcal{M}(\text { a } \sigma \text {-ring! })$$.

$$\left(\mathrm{ii}^{*}\right) \Rightarrow\left(\mathrm{iii}^{*}\right) .$$ For $$\left(\mathrm{i}^{*}\right)$$ and $$A \in \mathcal{M}$$ imply

$A(f<a)=A-A(f \geq a) \in \mathcal{M}.$

$$\left(\mathrm{iii}^{*}\right) \Rightarrow\left(\mathrm{iv}^{*}\right) .$$ If $$a \in E^{1}$$,

$A(f \leq a)=\bigcap_{n=1}^{\infty} A\left(f<a+\frac{1}{n}\right) \in \mathcal{M}.$

What if $$a=\pm \infty ?$$

$$\left(\mathrm{i} v^{*}\right) \Rightarrow\left(\mathrm{i}^{*}\right) .$$ Indeed, $$\left(\mathrm{iv}^{*}\right)$$ and $$A \in \mathcal{M}$$ imply

$A(f>a)=A-A(f \leq a) \in \mathcal{M}.$

Thus, indeed, each of $$\left(\mathrm{i}^{*}\right)$$ to $$\left(\mathrm{iv}^{*}\right)$$ implies the others. To finish, we need two lemmas that are of interest in their own right.

## Lemma $$\PageIndex{1}$$

If the maps $$f_{m} : S \rightarrow E^{*}(m=1,2, \ldots)$$ satisfy conditions $$\left(\mathrm{i}^{*}\right)-$$$$\left(\mathrm{iv}^{*}\right),$$ so also do the functions

$\sup f_{m}, \inf f_{m}, \overline{\lim } f_{m}, \text { and } \underline{\lim} f_{m},$

defined pointwise, i.e.,

$\left(\sup f_{m}\right)(x)=\sup f_{m}(x),$

and similarly for the others.

Proof

Let $$f=\sup f_{m} .$$ Then

$A(f \leq a)=\bigcap_{m=1}^{\infty} A\left(f_{m} \leq a\right) \quad \text{ (Why?)}$

But by assumption,

$A\left(f_{m} \leq a\right) \in \mathcal{M}$

$$\left(f_{m} \text { satisfies }\left(\mathrm{i} \mathrm{v}^{*}\right)\right) .$$ Hence $$A(f \leq a) \in \mathcal{M}(\text { for } \mathcal{M} \text { is a } \sigma \text {-ring })$$.

Thus sup $$f_{m}$$ satisfies $$\left(i^{*}\right)-\left(i v^{*}\right) .$$

So does inf $$f_{m} ;$$ for

$A\left(\inf f_{m} \geq a\right)=\bigcap_{m=1}^{\infty} A\left(f_{m} \geq a\right) \in \mathcal{M}.$

(Explain!)

So also do $$\underline{\lim} f_{m}$$ and $$\overline{\lim } f_{m} ;$$ for by definition,

$\underline{\lim} {t} f_{m}=\sup _{k} g_{k},$

where

$g_{k}=\inf _{m \geq k} f_{m}$

satisfies $$\left(\mathrm{i}^{*}\right)-\left(\mathrm{i} \mathrm{v}^{*}\right),$$ as was shown above; hence so does sup $$g_{k}= \underline{\lim} f_{m}$$.

Similarly for $$\overline{\lim } f_{m} . \square$$

## Lemma $$\PageIndex{2}$$

If $$f$$ satisfies $$\left(\mathrm{i}^{*}\right)-\left(\mathrm{iv}^{*}\right),$$ then

$f=\lim _{m \rightarrow \infty} f_{m} \text { (uniformly) on } A$

for some sequence of finite functions $$f_{m},$$ all $$\mathcal{M}$$ -elementary on $$A$$.

Moreover, if $$f \geq 0$$ on $$A,$$ the $$f_{m}$$ can be made nonnegative, with $$\left\{f_{m}\right\} \uparrow$$ on $$A$$.

Proof

Let $$H=A(f=+\infty), K=A(f=-\infty),$$ and

$A_{m k}=A\left(\frac{k-1}{2^{m}} \leq f<\frac{k}{2^{m}}\right)$

for $$m=1,2, \ldots$$ and $$k=0, \pm 1, \pm 2, \ldots, \pm n, \ldots$$

$$\mathrm{By}\left(\mathrm{i}^{*}\right)-\left(\mathrm{iv}^{*}\right)$$,

$H=A(f=+\infty)=A(f \geq+\infty) \in \mathcal{M},$

$$K \in \mathcal{M},$$ and

$A_{m k}=A\left(f \leq \frac{k-1}{2^{m}}\right) \cap A\left(f<\frac{k}{2^{m}}\right) \in \mathcal{M}.$

Now define

$(\forall m) \quad f_{m}=\frac{k-1}{2^{m}} \text { on } A_{m k},$

$$f_{m}=m$$ on $$H,$$ and $$f_{m}=-m$$ on $$K .$$ Then each $$f_{m}$$ is finite and elementary on $$A$$ since

$(\forall m) \quad A=H \cup K \cup \bigcup_{k=-\infty}^{\infty} A_{m k}(d i s j o i n t)$

and $$f_{m}$$ is constant on $$H, K,$$ and $$A_{m k}$$.

We now show that $$f_{m} \rightarrow f$$ (uniformly) on $$H, K,$$ and

$J=\bigcup_{k=-\infty}^{\infty} A_{m k},$

hence on $$A$$.

Indeed, on $$H$$ we have

$\lim f_{m}=\lim m=+\infty=f,$

and the limit is uniform since the $$f_{m}$$ are constant on $$H$$.

Similarly,

$f_{m}=-m \rightarrow-\infty=f \text { on } K.$

Finally, on $$A_{m k}$$ we have

$(k-1) 2^{-m} \leq f<k 2^{-m}$

and $$f_{m}=(k-1) 2^{-m} ;$$ hence

$\left|f_{m}-f\right|<k 2^{-m}-(k-1) 2^{-m}=2^{-m}.$

Thus

$\left|f_{m}-f\right|<2^{-m} \rightarrow 0$

on each $$A_{m k},$$ hence on

$J=\bigcup_{k=-\infty}^{\infty} A_{m k}.$

By Theorem 1 of Chapter 4, §12, it follows that $$f_{m} \rightarrow f (\text { uniformly })$$ on $$J$$. Thus, indeed, $$f_{m} \rightarrow f (\text { uniformly })$$ on $$A$$.

If, further, $$f \geq 0$$ on $$A,$$ then $$K=\emptyset$$ and $$A_{m k}=\emptyset$$ for $$k \leq 0 .$$ Moreover, on passage from $$m$$ to $$m+1,$$ each $$A_{m k}(k>0)$$ splits into two sets. On one, $$f_{m+1}=f_{m} ;$$ on the other, $$f_{m+1}>f_{m} .$$ (Why?)

Thus $$0 \leq f_{m} \nearrow f (\text { uniformly })$$ on $$A,$$ and all is proved. $$\square$$

## Theorem $$\PageIndex{1}$$ (Restated)

$$A$$ function $$f : S \rightarrow E^{*}$$ is measurable on a set $$A \in \mathcal{M}$$ iff $$i t$$ satisfies one of the following equivalent conditions (hence all of them):

$\begin{array}{ll}{\left(\mathfrak{i}^{*}\right)\left(\forall a \in E^{*}\right) A(f>a) \in \mathcal{M} ;} & {\left(\mathrm{ii}^{*}\right)\left(\forall a \in E^{*}\right) A(f \geq a) \in \mathcal{M}}; \\ {\left(\mathrm{iii}^{*}\right)\left(\forall a \in E^{*}\right) A(f<a) \in \mathcal{M} ;} & {\left(\mathrm{iv}^{*}\right)\left(\forall a \in E^{*}\right) A(f \leq a) \in \mathcal{M}}.\end{array}$

Proof

If $$f$$ is measurable on $$A,$$ then by definition, $$f=\lim f_{m}$$ (pointwise) for some elementary maps $$f_{m}$$ on $$A$$.

By Problem 4 (ii) in §1, all $$f_{m}$$ satisfy (i*)-(iv*). Thus so does $$f$$by Lemma 1, for here $$f=\lim f_{m}=\overline{\lim } f_{m}$$.

The converse follows by Lemma 2. This completes the proof. $$\square$$

Note 1. Lemmas 1 and 2 prove Theorems 3 and 4 of $$\ 1,$$ for $$f : S \rightarrow E^{*}$$. By using also Theorem 2 in §1, one easily extends this to $$f : S \rightarrow E^{n} (C^{n})$$. Verify!

## Corollary $$\PageIndex{1}$$

If $$f : S \rightarrow E^{*}$$ is measurable on $$A,$$ then

$\left(\forall a \in E^{*}\right) \quad A(f=a) \in \mathcal{M} \text { and } A(f \neq a) \in \mathcal{M}.$

Indeed,

$A(f=a)=A(f \geq a) \cap A(f \leq a) \in \mathcal{M}$

and

$A(f \neq a)=A-A(f=a) \in \mathcal{M}.$

## Corollary $$\PageIndex{2}$$

If $$f : S \rightarrow\left(T, \rho^{\prime}\right)$$ is measurable on $$A$$ in $$(S, \mathcal{M}),$$ then

$A \cap f^{-1}[G] \in \mathcal{M}$

for every globe $$G=G_{q}(\delta)$$ in $$\left(T, \rho^{\prime}\right)$$.

Proof

Define $$h : S \rightarrow E^{1}$$ by

$h(x)=\rho^{\prime}(f(x), q).$
Then $$h$$ is measurable on $$A$$ by Problem 6 in §1. Thus by Theorem 1,
$A(h<\delta) \in \mathcal{M}.$

But as is easily seen,

$A(h<\delta)=\left\{x \in A | \rho^{\prime}(f(x), q)<\delta\right\}=A \cap f^{-1}\left[G_{q}(\delta)\right].$

Hence the result. $$\square$$

## Definition

Given $$f, g : S \rightarrow E^{*},$$ we define the maps $$f \vee g$$ and $$f \wedge g$$ on $$S$$ by

$(f \vee g)(x)=\max \{f(x), g(x)\}$

and

$(f \wedge g)(x)=\min \{f(x), g(x)\};$

similarly for $$f \vee g \vee h, f \wedge g \wedge h,$$ etc.

We also set

$f^{+}=f \vee 0 \text { and } f^{-}=-f \vee 0.$

Clearly, $$f^{+} \geq 0$$ and $$f^{-} \geq 0$$ on $$S .$$ Also, $$f=f^{+}-f^{-}$$ and $$|f|=f^{+}+f^{-} .$$

(Why?) We now obtain the following theorem.

## Theorem $$\PageIndex{2}$$

If the functions $$f, g : S \rightarrow E^{*}$$ are simple, elementary, or measurable on $$A,$$ so also are $$f \pm g, f g, f \vee g, f \wedge g, f^{+}, f^{-},$$ and $$|f|^{a}(a \neq 0)$$.

Proof

If $$f$$ and $$g$$ are finite, this follows by Theorem 1 of §1 on verifying that

$f \vee g=\frac{1}{2}(f+g+|f-g|)$

and

$f \wedge g=\frac{1}{2}(f+g-|f-g|)$

on $$S .$$ (Check it!)

Otherwise, consider

$A(f=+\infty), A(f=-\infty), A(g=+\infty), \text { and } A(g=-\infty).$

By Theorem $$1,$$ these are $$\mathcal{M}$$-sets; hence so is their union $$U$$.

On each of them $$f \vee g$$ and $$f \wedge g$$ equal $$f$$ or $$g ;$$ so by Corollary 3 in §1, $$f \vee g$$ and $$f \wedge g$$ have the desired properties on $$U .$$ So also have $$f^{+}=f \vee 0$$ and $$f^{-}=-f \vee 0 (\text { take } g=0)$$.

We claim that the maps $$f \pm g$$ and $$f g$$ are simple (hence elementary and measurable) on each of the four sets mentioned above, hence on $$U .$$

For example, on $$A(f=+\infty)$$,

$f \pm g=+\infty(\text { constant })$

by our conventions $$\left(2^{*}\right)$$ in Chapter 4, §4. For $$f g,$$ split $$A(f=+\infty)$$ into three sets $$A_{1}, A_{2}, A_{3} \in \mathcal{M},$$ with $$g>0$$ on $$A_{1}, g<0$$ on $$A_{2},$$ and $$g=0$$ on $$A_{3} ;$$ so $$f g=+\infty$$ on $$A_{1}, f g=-\infty$$ on $$A_{2},$$ and $$f g=0$$ on $$A_{3} .$$ Hence $$f g$$ is simple on $$A(f=+\infty)$$.

For $$|f|^{a},$$ use $$U=A(|f|=\infty) .$$ Again, the theorem holds on $$U,$$ and also on $$A-U,$$ since $$f$$ and $$g$$ are finite on $$A-U \in \mathcal{M} .$$ Thus it holds on $$A=(A-U) \cup U$$ by Corollary 3 in §1. $$\square$$

Note 2. Induction extends Theorem 2 to any finite number of functions.

Note 3. Combining Theorem 2 with $$f=f^{+}-f^{-},$$ we see that $$f : S \rightarrow E^{*}$$ is simple (elementary, measurable) iff $$f^{+}$$ and $$f^{-}$$ are. We also obtain the following result.

## Theorem $$\PageIndex{3}$$

If the functions $$f, g : S \rightarrow E *$$ are measurable on $$A \in \mathcal{M},$$ then $$A(f \geq g) \in \mathcal{M}, A(f<g) \in \mathcal{M}, A(f=g) \in \mathcal{M},$$ and $$A(f \neq g) \in \mathcal{M}$$.

This page titled 8.2: Measurability of Extended-Real Functions is shared under a CC BY license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) .