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8.3: Measurable Functions in (S,M,m)

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

I. Henceforth we shall presuppose not just a measurable space (§1) but a measure space (S,M,m), where m:ME is a measure on a σ -ring M2S.

We saw in Chapter 7 that one could often neglect sets of Lebesgue measure zero on En 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 xA (with respect to the measure m) or almost everywhere (a.e. (m)) on A iff it holds on AQ for some QM with mQ=0.

Thus we write

fnf(a.e.) or f=limfn(a.e.(m)) on A

iff fnf( pointwise ) on AQ,mQ=0. Of course, "pointwise" implies "a.e."( take Q=), but the converse fails.

Definition

We say that f:S(T,ρ) is almost measurable on A iff AM and f is M -measurable on AQ,mQ=0.

We then also say that f is m -measurable (m being the measure involved ) as opposed to M-measurable.

Observe that we may assume QA here (replace Q by AQ).

*Note 1. If m is a generalized measure (Chapter 7, §11), replace mQ=0 by vmQ=0(vm= total variation of m) in Definitions 1 and 2 and in the following proofs.

Corollary 8.3.1

If the functions

fn:S(T,ρ),n=1,2,

are m-measurable on A, and if

fnf(a.e.(m))

on A, then f is m-measurable on A.

Proof

By assumption, fnf( pointwise ) on AQ0,mQ0=0. Also, fn is M-measurable on

AQn,mQn=0,n=1,2,

(The Qn need not be the same.)

Let

Q=n=0Qn;

so

mQn=0mQn=0.

By Corollary 2 in §1, all fn are M-measurable on AQ (why?), and fnf
(pointwise) on AQ, as AQAQ0.

Thus by Theorem 4 in §1, f is M -measurable on AQ. As mQ=0, this
is the desired result.

Corollary 8.3.2

If f=g(a.e.(m)) on A and f is m-measurable on A, so is g.

Proof

By assumption, f=g on AQ1 and f is M-measurable on AQ2, with mQ1=mQ2=0.

Let Q=Q1Q2. Then mQ=0 and g=f on AQ. (Why? )

By Corollary 2 of §1, f is M-measurable on AQ. Hence so is g, as claimed.

Corollary 8.3.3

If f:S(T,ρ) is m -measurable on A, then

f=limnfn(uniformly) on AQ(mQ=0),

for some maps fn, all elementary on AQ.

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 "M -measurable." This implies m-measurability (take Q=), but the converse fails. (See Note 2, however.)

We still obtain the following result.

Corollary 8.3.4

If the functions

fn:SE(n=1,2,)

are m-measurable on a set A, so also are

supfn,inffn,¯limfn, and lim_fn

(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:SE(En,Cn),m and 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 8.3.1

Amapf:SEn(Cn) is M-measurable on A iff
Af1[B]M
for each Borel set (equivalently, open set) B in En(Cn).

Proof

See Problems 810 in §2 for a sketch of the proof.

Lemma 8.3.2

Amapf:(S,ρ)(T,ρ) is relatively continuous on AS iff for any open set B(T,ρ), the set Af1[B] is open in (A,ρ) as a subspace of (S,ρ).
(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:AT) is continuous in the ordinary sense.
Now, by Problem 15(iv)(v) in Chapter 4, §2, with S replaced by A, this means that g1[B] is open (closed) in(A,ρ) when B is so in (T,ρ). But
g1[B]={xA|f(x)B}=Af1[B].
(Why?) Hence the result follows.

Theorem 8.3.1

Let m:ME be a topological measure in (S,ρ). If f:S En(Cn) is relatively continuous on a set AM, it is M -measurable on A.

Proof

Let B be open in En(Cn). By Lemma 2,
Af1[B]
is open in(A,ρ). Hence by Theorem 4 of Chapter 3, §12,
Af1[B]=AU
for some open set U in (S,ρ).
Now, by assumption, A is in M. So is U, as M is topological (MG).
Hence
Af1[B]=AUM
for any open BEn(Cn). The result follows by Lemma 1.

Note 3. The converse fails. For example, the Dirichlet function (Example (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(T,ρ), too, provided f[A] is separable, i.e.,
f[A]¯D
for a countable set DT (cf. Problem 11 in §2).
*III. For strongly regular measures (Definition 5 in Chapter 7, §7), we obtain the following theorem.

*Theorem 8.3.2

(Luzin). Let m:ME be a strongly regular measure in (S,ρ). Let f:S(T,ρ) be m-measurable on A.
Then given ε>0, there is a closed set FA(FM) such that
m(AF)<ε
and f is relatively continuous on F.
(Note that if T=E,ρ is as in Problem 5 of Chapter 3, §11.)

Proof

By assumption, f is M-measurable on a set
H=AQ,mQ=0;
so by Problem 7 in §1, f[H] is separable in T. We may safely assume that f is 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 M-sets inside H; see also Problems 7 and 8 below.)
Then by Problem 12 of §2, we can fix globes G1,G2, in T such that
 each open set B in T is the union of a subsequence of {Gk}.
Now let ε>0, and set
Sk=Sf1[Gk]=f1[Gk],k=1,2,
By Corollary 2 in §2, SkM. As m is strongly regular, we find for each Sk an open set
UkSk,
with UkM and
m(UkSk)<ε2k+1.
Let Bk=UkSk,D=kBk; so DM and
mDkmBkkε2k+112ε
and
UkBk=Sk=f1[Gk].
As D=Bk, we have
(k)BkD=Bk(D)=.
Hence by (2),
(k)f1[Gk](D)=(UkBk)(D)=(Uk(D))(Bk(D))=Uk(D).
Combining this with (1), we have, for each open set B=iGkiinT,
f1[B](D)=if1[Gki](D)=iUki(D).
since the Uki are open in S (by construction), the set (3) is open in SD as a subspace of S. By Lemma 2, then, f is relatively continuous on SD, or rather on
HD=AQD
(since we actually substituted S for H in the course of the proof). As mQ=0 and mD<12ε by (2),
m(HD)<mA12ε.
Finally, as m is strongly regular and HDM, there is a closed M-set
FHDA
such that
m(HDF)<12ε.
since f is relatively continuous on HD, it is surely so on F. Moreover,
AF=(A(HD))(HDF);
so
m(AF)m(A(HD))+m(HDF)<12ε+12ε=ε.
This completes the proof.

*Lemma 8.3.3

Given [a,b]E1 and disjoint closed sets A,B(S,ρ), there always is a continuous map g:S[a,b] such that g=a on A and g=b.

Proof

If A= or B=, set g=b or g=a on all of S.
If, however, A and B are both nonempty, set
g(x)=a+(ba)ρ(x,A)ρ(x,A)+ρ(x,B).
As A is closed, ρ(x,A)=0 iff xA (Problem 15 in Chapter 3, §14); similarly for B. Thus ρ(x,A)+ρ(x,B)0.
Also, g=a on A,g=b on B, and agb on S.
For continuity, see Chapter 4, §8, Example (e).

*Lemma 8.3.4

(Tietze). If f:(S,ρ)E is relatively continuous on a closed set FS, there is a function g:SE such that g=f on F,
infg[S]=inff[F],supg[S]=supf[F],
and g is continuous on all of S.
(We assume E metrized as in Problem 5 of Chapter 3, §11. If |f|<, the standard metric in E1 may be used.)

Proof Outline

First, assume inf f[F]=0 and supf[F]=1. Set
A=F(f13)=Ff1[[0,13]]
and
B=F(f23)=Ff1[[23,1]].
As F is closed inS, so are A and B by Lemma 2. (Why? )
As BA=, Lemma 3 yields a continuous map g1:S[0,13], with g1=0 on A, and g1=13 on B. Set f1=fg1 on F; so |f1|23, and f1 is continuous on F.
Applying the same steps to f1 (with suitable sets A1,B1F), find a continuous map g2, with 0g22313 on S. Then f2=f1g2 is continuous, and 0f2(23)2 on F.
Continuing, obtain two sequences {gn} and {fn} of real functions such that each gn is continuous on S,
0gn13(23)n1,
and fn=fn1gn is defined and continuous on F, with
0fn(23)n
there (f0=f).
We claim that
g=n=1gn
is the desired map.
Indeed, the series converges uniformly on S (Theorem 3 of Chapter 4, §12).
As all gn are continuous, so is g (Theorem 2 in Chapter 4, §12). Also,
|fnk=1gk|(23)n0
on F( why? ); so f=g on F. Moreover,
0g1gn=113(23)n=1 on S.
Hence inf g[S]=0 and supg[S]=1, as required.
Now assume
inff[F]=a<supf[F]=b(a,bE1)
Set
h(x)=f(x)aba
so that inf h[F]=0 and suph[F]=1. (Why?)
As shown above, there is a continuous map g0 on S, with
g0=h=faba
on F, inf g0[S]=0, and supg0[S]=1. Set
a+(ba)g0=g.
Then g is the required function. (Verify!)
Finally, if a,bE(a<b), all reduces to the bounded case by considering H(x)=arctanf(x).

*Theorem 8.3.3

(Fréchet). Let m:ME be a strongly regular measure in (S,ρ). If f:SE(En,Cn) is m -measurable on A, then
f=limifi(ae.(m)) on A
for some sequence of maps fi continuous on S. (We assume E to be metrized as in Lemma 4.)

Proof

We consider f:SE (the other cases reduce to E1 via components).
Taking ε=1i(i=1,2,) in Theorem 2, we obtain for each i a closed M-set FiA such that
m(AFi)<1i
and f is relatively continuous on each Fi. We may assume that FiFi+1 (if not, replace Fi by ik=1Fk).
Now, Lemma 4 yields for each i a continuous map fi:SE such that fi=f on Fi. We complete the proof by showing that fif (pointwise) on the set
B=i=1Fi
and that m(AB)=0.
Indeed, fix any xB. Then xFi for some i=i0, hence also for i>i0 (since {Fi}). As fi=f on Fi, we have
(i>i0)fi(x)=f(x),
and so fi(x)f(x) for xB. As FiB, we get
m(AB)m(AFi)<1i
for all i. Hence m(AB)=0, and all is proved.


8.3: Measurable Functions in (S,M,m) is shared under a CC BY license and was authored, remixed, and/or curated by LibreTexts.

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