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 $A(f \neq g) \in \mathcal{M}$.