8.3: Measurable Functions in (S,M,m)
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( \newcommand{\kernel}{\mathrm{null}\,}\)
I. Henceforth we shall presuppose not just a measurable space (§1) but a measure space (S,M,m), where m:M→E∗ is a measure on a σ -ring M⊆2S.
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 x∈A (with respect to the measure m) or almost everywhere (a.e. (m)) on A iff it holds on A−Q for some Q∈M with mQ=0.
Thus we write
fn→f(a.e.) or f=limfn(a.e.(m)) on A
iff fn→f( pointwise ) on A−Q,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 A∈M and f is M -measurable on A−Q,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 Q⊆A here (replace Q by A∩Q).
*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
fn→f(a.e.(m))
on A, then f is m-measurable on A.
- Proof
-
By assumption, fn→f( pointwise ) on A−Q0,mQ0=0. Also, fn is M-measurable on
A−Qn,mQn=0,n=1,2,…
(The Qn need not be the same.)
Let
Q=∞⋃n=0Qn;
so
mQ≤∞∑n=0mQn=0.
By Corollary 2 in §1, all fn are M-measurable on A−Q (why?), and fn→f
(pointwise) on A−Q, as A−Q⊆A−Q0.Thus by Theorem 4 in §1, f is M -measurable on A−Q. 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 A−Q1 and f is M-measurable on A−Q2, with mQ1=mQ2=0.
Let Q=Q1∪Q2. Then mQ=0 and g=f on A−Q. (Why? )
By Corollary 2 of §1, f is M-measurable on A−Q. Hence so is g, as claimed. ◻
Corollary 8.3.3
If f:S→(T,ρ′) is m -measurable on A, then
f=limn→∞fn(uniformly) on A−Q(mQ=0),
for some maps fn, all elementary on A−Q.
- 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:S→E∗(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:S→E∗(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:S→En(Cn) is M-measurable on A iff
A∩f−1[B]∈M
for each Borel set (equivalently, open set) B in En(Cn).
- Proof
-
See Problems 8−10 in §2 for a sketch of the proof.
Lemma 8.3.2
Amapf:(S,ρ)→(T,ρ′) is relatively continuous on A⊆S iff for any open set B⊆(T,ρ′), the set A∩f−1[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:A→T) is continuous in the ordinary sense.
Now, by Problem 15(iv)(v) in Chapter 4, §2, with S replaced by A, this means that g−1[B] is open (closed) in(A,ρ) when B is so in (T,ρ′). But
g−1[B]={x∈A|f(x)∈B}=A∩f−1[B].
(Why?) Hence the result follows. ◻
Theorem 8.3.1
Let m:M→E∗ be a topological measure in (S,ρ). If f:S→ En(Cn) is relatively continuous on a set A∈M, it is M -measurable on A.
- Proof
-
Let B be open in En(Cn). By Lemma 2,
A∩f−1[B]
is open in(A,ρ). Hence by Theorem 4 of Chapter 3, §12,
A∩f−1[B]=A∩U
for some open set U in (S,ρ).
Now, by assumption, A is in M. So is U, as M is topological (M⊇G).
Hence
A∩f−1[B]=A∩U∈M
for any open B⊆En(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 D⊆T (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:M→E∗ 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 F⊆A(F∈M) such that
m(A−F)<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=A−Q,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=S∩f−1[Gk]=f−1[Gk],k=1,2,…
By Corollary 2 in §2, Sk∈M. As m is strongly regular, we find for each Sk an open set
Uk⊇Sk,
with Uk∈M and
m(Uk−Sk)<ε2k+1.
Let Bk=Uk−Sk,D=⋃kBk; so D∈M and
mD≤∑kmBk≤∑kε2k+1≤12ε
and
Uk−Bk=Sk=f−1[Gk].
As D=⋃Bk, we have
(∀k)Bk−D=Bk∩(−D)=∅.
Hence by (2′),
(∀k)f−1[Gk]∩(−D)=(Uk−Bk)∩(−D)=(Uk∩(−D))−(Bk∩(−D))=Uk∩(−D).
Combining this with (1), we have, for each open set B=⋃iGkiinT,
f−1[B]∩(−D)=⋃if−1[Gki]∩(−D)=⋃iUki∩(−D).
since the Uki are open in S (by construction), the set (3) is open in S−D as a subspace of S. By Lemma 2, then, f is relatively continuous on S−D, or rather on
H−D=A−Q−D
(since we actually substituted S for H in the course of the proof). As mQ=0 and mD<12ε by (2),
m(H−D)<mA−12ε.
Finally, as m is strongly regular and H−D∈M, there is a closed M-set
F⊆H−D⊆A
such that
m(H−D−F)<12ε.
since f is relatively continuous on H−D, it is surely so on F. Moreover,
A−F=(A−(H−D))∪(H−D−F);
so
m(A−F)≤m(A−(H−D))+m(H−D−F)<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+(b−a)ρ(x,A)ρ(x,A)+ρ(x,B).
As A is closed, ρ(x,A)=0 iff x∈A (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 a≤g≤b 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 F⊆S, there is a function g:S→E∗ 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(f≤13)=F∩f−1[[0,13]]
and
B=F(f≥23)=F∩f−1[[23,1]].
As F is closed inS, so are A and B by Lemma 2. (Why? )
As B∩A=∅, Lemma 3 yields a continuous map g1:S→[0,13], with g1=0 on A, and g1=13 on B. Set f1=f−g1 on F; so |f1|≤23, and f1 is continuous on F.
Applying the same steps to f1 (with suitable sets A1,B1⊆F), find a continuous map g2, with 0≤g2≤23⋅13 on S. Then f2=f1−g2 is continuous, and 0≤f2≤(23)2 on F.
Continuing, obtain two sequences {gn} and {fn} of real functions such that each gn is continuous on S,
0≤gn≤13(23)n−1,
and fn=fn−1−gn is defined and continuous on F, with
0≤fn≤(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,
|f−n∑k=1gk|≤(23)n→0
on F( why? ); so f=g on F. Moreover,
0≤g1≤g≤∞∑n=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,b∈E1)
Set
h(x)=f(x)−ab−a
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=f−ab−a
on F, inf g0[S]=0, and supg0[S]=1. Set
a+(b−a)g0=g.
Then g is the required function. (Verify!)
Finally, if a,b∈E∗(a<b), all reduces to the bounded case by considering H(x)=arctanf(x). ◻
*Theorem 8.3.3
(Fréchet). Let m:M→E∗ be a strongly regular measure in (S,ρ). If f:S→E∗(En,Cn) is m -measurable on A, then
f=limi→∞fi(a⋅e.(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:S→E∗ (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 Fi⊆A such that
m(A−Fi)<1i
and f is relatively continuous on each Fi. We may assume that Fi⊆Fi+1 (if not, replace Fi by ⋃ik=1Fk).
Now, Lemma 4 yields for each i a continuous map fi:S→E∗ such that fi=f on Fi. We complete the proof by showing that fi→f (pointwise) on the set
B=∞⋃i=1Fi
and that m(A−B)=0.
Indeed, fix any x∈B. Then x∈Fi 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 x∈B. As Fi⊆B, we get
m(A−B)≤m(A−Fi)<1i
for all i. Hence m(A−B)=0, and all is proved. ◻