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8.2: Measurability of Extended-Real Functions

( \newcommand{\kernel}{\mathrm{null}\,}\)

Henceforth we presuppose a measurable space (S,M), where M is a σ-ring in S. Our aim is to prove the following basic theorem, which is often used as a definition, for extended-real functions f:SE.

Theorem 8.2.1

A function f:SE is measurable on a set AM iff it satisfies one of the following equivalent conditions (hence all of them):

(i)(aE)A(f>a)M;(ii)(aE)A(fa)M;(iii)(aE)A(f<a)M;(iv)(aE)A(fa)M.

We first prove the equivalence of these conditions by showing that (i) (ii)(iii)(iv)(i), closing the "circle."

(i)(ii). Assume (i). If a=,

A(fa)=AM

by assumption. If a=+,

A(fa)=A(f=)=n=1A(f>n)M

by (i). And if aE1,

A(fa)=n=1A(f>a1n).

(Verify!) By (i),

A(f>a1n)M;

so A(fa)M( a σ-ring! ).

(ii)(iii). For (i) and AM imply

A(f<a)=AA(fa)M.

(iii)(iv). If aE1,

A(fa)=n=1A(f<a+1n)M.

What if a=±?

(iv)(i). Indeed, (iv) and AM imply

A(f>a)=AA(fa)M.

Thus, indeed, each of (i) to (iv) implies the others. To finish, we need two lemmas that are of interest in their own right.

Lemma 8.2.1

If the maps fm:SE(m=1,2,) satisfy conditions (i)(iv), so also do the functions

supfm,inffm,¯limfm, and lim_fm,

defined pointwise, i.e.,

(supfm)(x)=supfm(x),

and similarly for the others.

Proof

Let f=supfm. Then

A(fa)=m=1A(fma) (Why?)

But by assumption,

A(fma)M

(fm satisfies (iv)). Hence A(fa)M( for M is a σ-ring ).

Thus sup fm satisfies (i)(iv).

So does inf fm; for

A(inffma)=m=1A(fma)M.

(Explain!)

So also do lim_fm and ¯limfm; for by definition,

lim_tfm=supkgk,

where

gk=infmkfm

satisfies (i)(iv), as was shown above; hence so does sup gk=lim_fm.

Similarly for ¯limfm.

Lemma 8.2.2

If f satisfies (i)(iv), then

f=limmfm (uniformly) on A

for some sequence of finite functions fm, all M -elementary on A.

Moreover, if f0 on A, the fm can be made nonnegative, with {fm} on A.

Proof

Let H=A(f=+),K=A(f=), and

Amk=A(k12mf<k2m)

for m=1,2, and k=0,±1,±2,,±n,

By(i)(iv),

H=A(f=+)=A(f+)M,

KM, and

Amk=A(fk12m)A(f<k2m)M.

Now define

(m)fm=k12m on Amk,

fm=m on H, and fm=m on K. Then each fm is finite and elementary on A since

(m)A=HKk=Amk(disjoint)

and fm is constant on H,K, and Amk.

We now show that fmf (uniformly) on H,K, and

J=k=Amk,

hence on A.

Indeed, on H we have

limfm=limm=+=f,

and the limit is uniform since the fm are constant on H.

Similarly,

fm=m=f on K.

Finally, on Amk we have

(k1)2mf<k2m

and fm=(k1)2m; hence

|fmf|<k2m(k1)2m=2m.

Thus

|fmf|<2m0

on each Amk, hence on

J=k=Amk.

By Theorem 1 of Chapter 4, §12, it follows that fmf( uniformly ) on J. Thus, indeed, fmf( uniformly ) on A.

If, further, f0 on A, then K= and Amk= for k0. Moreover, on passage from m to m+1, each Amk(k>0) splits into two sets. On one, fm+1=fm; on the other, fm+1>fm. (Why?)

Thus 0fmf( uniformly ) on A, and all is proved.

Theorem 8.2.1 (Restated)

A function f:SE is measurable on a set AM iff it satisfies one of the following equivalent conditions (hence all of them):

(i)(aE)A(f>a)M;(ii)(aE)A(fa)M;(iii)(aE)A(f<a)M;(iv)(aE)A(fa)M.

Proof

If f is measurable on A, then by definition, f=limfm (pointwise) for some elementary maps fm on A.

By Problem 4 (ii) in §1, all fm satisfy (i*)-(iv*). Thus so does fby Lemma 1, for here f=limfm=¯limfm.

The converse follows by Lemma 2. This completes the proof.

Note 1. Lemmas 1 and 2 prove Theorems 3 and 4 of $1, for f:SE. By using also Theorem 2 in §1, one easily extends this to f:SEn(Cn). Verify!

Corollary 8.2.1

If f:SE is measurable on A, then

(aE)A(f=a)M and A(fa)M.

Indeed,

A(f=a)=A(fa)A(fa)M

and

A(fa)=AA(f=a)M.

Corollary 8.2.2

If f:S(T,ρ) is measurable on A in (S,M), then

Af1[G]M

for every globe G=Gq(δ) in (T,ρ).

Proof

Define h:SE1 by

h(x)=ρ(f(x),q).
Then h is measurable on A by Problem 6 in §1. Thus by Theorem 1,
A(h<δ)M.

But as is easily seen,

A(h<δ)={xA|ρ(f(x),q)<δ}=Af1[Gq(δ)].

Hence the result.

Definition

Given f,g:SE, we define the maps fg and fg on S by

(fg)(x)=max{f(x),g(x)}

and

(fg)(x)=min{f(x),g(x)};

similarly for fgh,fgh, etc.

We also set

f+=f0 and f=f0.

Clearly, f+0 and f0 on S. Also, f=f+f and |f|=f++f.

(Why?) We now obtain the following theorem.

Theorem 8.2.2

If the functions f,g:SE are simple, elementary, or measurable on A, so also are f±g,fg,fg,fg,f+,f, and |f|a(a0).

Proof

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

fg=12(f+g+|fg|)

and

fg=12(f+g|fg|)

on S. (Check it!)

Otherwise, consider

A(f=+),A(f=),A(g=+), and A(g=).

By Theorem 1, these are M-sets; hence so is their union U.

On each of them fg and fg equal f or g; so by Corollary 3 in §1, fg and fg have the desired properties on U. So also have f+=f0 and f=f0( take g=0).

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

For example, on A(f=+),

f±g=+( constant )

by our conventions (2) in Chapter 4, §4. For fg, split A(f=+) into three sets A1,A2,A3M, with g>0 on A1,g<0 on A2, and g=0 on A3; so fg=+ on A1,fg= on A2, and fg=0 on A3. Hence fg is simple on A(f=+).

For |f|a, use U=A(|f|=). Again, the theorem holds on U, and also on AU, since f and g are finite on AUM. Thus it holds on A=(AU)U by Corollary 3 in §1.

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:SE is simple (elementary, measurable) iff f+ and f are. We also obtain the following result.

Theorem 8.2.3

If the functions f,g:SE are measurable on AM, then A(fg)M,A(f<g)M,A(f=g)M, and A(fg)M.


8.2: Measurability of Extended-Real Functions is shared under a CC BY license and was authored, remixed, and/or curated by LibreTexts.

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