8.1: Elementary and Measurable Functions
( \newcommand{\kernel}{\mathrm{null}\,}\)
From set functions, we now return to point functions
f:S→(T,ρ′)
whose domain Df consists of points of a set S. The range space T will mostly be E, i.e., E1,E∗,C,En, or another normed space. We assume f(x)=0 unless defined otherwise. (In a general metric space T, we may take some fixed element q for 0. ) Thus Df is all of S, always.
We also adopt a convenient notation for sets:
"A(P)" for "{x∈A|P(x)}."
Thus
A(f≠a)={x∈A|f(x)≠a},A(f=g)={x∈A|f(x)=g(x)},A(f>g)={x∈A|f(x)>g(x)}, etc.
A measurable space is a set S≠∅ together with a set ring M of subsets of S, denoted (S,M).
Henceforth, (S,M) is fixed.
An M-partition of a set A is a countable set family P={Ai} such that
A=⋃iAi(disjoint),
with A,Ai∈M.
We briefly say "the partition A=⋃Ai."
An M-partition P′={Bik} is a refinement of P={Ai}( or P′ refines P, or P′ is finer than P) iff
(∀i)Ai=⋃kBik
i.e., each Bik is contained in some Ai.
The intersection P′∩P′′ of P′={Ai} and P′′={Bk} is understood to be the family of all sets of the form
Ai∩Bk,i,k=1,2,…
It is an M -partition that refines both P′ and P′′.
A map (function) f:S→T is elementary, or M-elementary, on a set A∈M iff there is an M-partition P={Ai} of A such that f is constant (f=ai) on each Ai.
If P={A1,…,Aq} is finite, we say that f is simple, or M-simple, on A.
If the Ai are intervals in En, we call f a step function; it is a simple step function if P is finite.
The function values ai are elements of T (possibly vectors). They may be infinite if T=E∗. Any simple map is also elementary, of course.
A map f:S→(T,ρ′) is said to be measurable (or M -measurable ) on a setA in (S,M) iff
f=limm→∞fm( pointwise ) on A
for some sequence of functions fm:S→T, all elementary on A. (See Chapter 4, §12 for "pointwise.")
Note 1. This implies A∈M, as follows from Definitions 2 and 3.(Why ? )
If f:S→(T,ρ′) is elementary on A, it is measurable on A.
- Proof
-
Set fm=f,m=1,2,…, in Definition 4. Then clearly fm→f on A. square
If f is simple, elementary, or measurable on A in (S,M), it has the same property on any subset B⊆A with B∈M.
- Proof
-
Let f be simple on A; so f=ai on Ai,i=1,2,…,n, for some finite M -partition, A=⋃ni=1Ai.
If A⊇B∈M, then
{B∩Ai},i=1,2,…,n,
is a finite M -partition of B( why? ), and f=ai on B∩Ai; so f is simple on B.
For elementary maps, use countable partitions.
Now let f be measurable on A, i.e.,
f=limm→∞fm
for some elementary maps fm on A. As shown above, the fm are elementary on B, too, and fm→f on B; so f is measurable on B.◻
If f is elementary or measurable on each of the (countably many ) sets An in (S,M), it has the same property on their union A=⋃nAn.
- Proof
-
Let f be elementary on each An (so An∈M by Note 1).
By Corollary 1 of Chapter 7, §1,
A=⋃An=⋃Bn
for some disjoint sets Bn⊆An(Bn∈M).
By Corollary 2,f is elementary on each Bn; i.e., constant on sets of some M -partition {Bni} of Bi.
All Bni combined (for all n and all i) form an M-partition of A,
A=⋃nBn=⋃n,iBni.
As f is constant on each Bni, it is elementary on A.
For measurable functions f, slightly modify the method used in Corollary 2.◻
If f:S→(T,ρ′) is measurable on A in (S,M), so is the composite map g∘f, provided g:T→(U,ρ′′) is relatively continuous on f[A].
- Proof
-
By assumption,
f=limm→∞fm (pointwise)
for some elementary maps fm on A.
Hence by the continuity of g,
g(fm(x))→g(f(x)),
i.e., g∘fm→g∘f (pointwise) on A.
Moreover, all g∘fm are elementary on A (for g∘fm is constant on any partition set, if fm is).
Thus g∘f is measurable on A, as claimed. ◻
If the maps f,g,h:S→E1(C) are simple, elementary, or measurable on A in (S,M), so are f±g,fh,|f|a (for real a≠0) and f/h (if h≠0 on A).
Similarly for vector-valued f and g and scalar-valued h.
- Proof
-
First, let f and g be elementary on A. Then there are two M-partitions,
A=⋃Ai=⋃Bk,
such that f=ai on Ai and g=bk on Bk, say.
The sets Ai∩Bk (for all i and k) then form a new M -partition of A( why? ), such that both f and g are constant on each Ai∩Bk( why?); hence so is f±g.
Thus f±g is elementary on A. Similarly for simple functions.
Next, let f and g be measurable on A; so
f=limfm and g=limgm (pointwise) on A
for some elementary maps fm,gm.
By what was shown above, fm±gm is elementary for each m. Also,
fm±gm→f±g( pointwise ) on A,
Thus f±g is measurable on A.
The rest of the theorem follows quite similarly. ◻
If the range space is En( or Cn), then f has n real (complex) components f1,…,fn, as in Chapter 4,§3 (Part II). This yields the following theorem.
A function f:S→En(Cn) is simple, elementary, or measurable on a set A in (S,M) iff all its n component functions f1,f2,…,fn are.
- Proof
-
For simplicity, consider f:S→E2,f=(f1,f2).
If f1 and f2 are simple or elementary on A then (exactly as in Theorem 1), one can achieve that both are constant on sets Ai∩Bk of one and the same M-partition of A. Hence f=(f1,f2), too, is constant on each Ai∩Bk, as required.
Conversely, let
f=¯ci=(ai,bi) on Ci
for some M-partition
A=⋃Ci.
Then by definition, f1=ai and f2=bi on Ci; so both are elementary (or simple) on A.
In the general case (En or Cn), the proof is analogous.
For measurable functions, the proof reduces to limits of elementary maps (using Theorem 2 of Chapter 3, §15). The details are left to the reader. ◻
Note 2. As C=E2, a complex function f:S→C is simple, elementary, or measurable on A iff its real and imaginary parts are.
By Definition 4, a measurable function is a pointwise limit of elementary maps. However, if M is a σ-ring, one can make the limit uniform. Indeed, we have the following theorem.
If M is a σ-ring, and f:S→(T,ρ′) is M-measurable on A, then
f=limm→∞gm (uniformly) on A
for some finite elementary maps gm.
- Proof
-
Thus given ε>0, there is a finite elementary map g such that ρ′(f,g)<ε
on A.
If M is a σ-ring in S, if
fm→f(pointwise) on A
(fm:S→(T,ρ′)), and if all fm are M -measurable on A, so also is f.
Briefly: A pointwise limit of measurable maps is measurable (unlike continuous maps; cf. Chapter 4, §12).
- Proof
-
By the second clause of Theorem 3, each fm is uniformly approximated by some elementary map gm on A, so that, taking ε=1/m,m=1,2,…,
ρ′(fm(x),gm(x))<1m for all x∈A and all m.
Fixing such a gm for each m, we show that gm→f( pointwise ) on A, as required in Definition 4.
Indeed, fix any x∈A. By assumption, fm(x)→f(x). Hence, given δ>0,
(∃k)(∀m>k)ρ′(f(x),fm(x))<δ.
Take k so large that, in addition,
(∀m>k)1m<δ.
Then by the triangle law and by (1), we obtain for m>k that
ρ′(f(x),gm(x))≤ρ′(f(x),fm(x))+ρ′(fm(x),gm(x))<δ+1m<2δ.
As δ is arbitrary, this implies ρ′(f(x),gm(x))→0, i.e., gm(x)→f(x) for any (fixed) x∈A, thus proving the measurability of f.◻
Note 3. If
M=B(= Borel field in S),
we often say "Borel measurable" for M-measurable. If
M={ Lebesgue measurable sets in En},
we say "Lebesgue (L) measurable" instead. Similarly for "Lebesgue-Stieltjes (LS) measurable."