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