Orthonormal bases for product measures

Jordan Bell
October 22, 2015

1 Measure and integration theory

Let be the Borel σ-algebra of , and let ¯ be the Borel σ-algebra of [-,]={-,}: the elements of ¯ are those subsets of ¯ of the form B,B{-},B{},B{-,}, with B.

Let (X,𝒜,μ) be a measure space. It is a fact that if fn is a sequence of 𝒜¯ measurable functions then supnfn and infnfn are 𝒜¯ measurable, and thus if fn is a sequence of 𝒜¯ measurable functions that converge pointwise to a function f:X¯, then f is 𝒜¯ measurable.11 1 Heinz Bauer, Measure and Integration Theory, p. 52, Corollary 9.7. If f1,,fn are 𝒜¯ measurable, then so are f1fn and f1fn, and a function f:X¯ is 𝒜¯ measurable if and only if both f+=f0 and f-=-(f0) are 𝒜¯ measurable. In particular, if f is 𝒜¯ measurable then so is |f|=f++f-.

A simple function is a function f:X that is 𝒜 measurable and whose range is finite. Let E=E(𝒜) be the collection of nonnegative simple functions. It is straightforward to prove that

u,vE,α0αu,u+v,uv,uv,uvE.

Define Iμ:E[0,] by

Iμu=i=1naiμ(Ai),

where u has range {a1,,an} and Ai=u-1(ai). One proves that Iμ:E[0,] is positive homogeneous, additive, and order preserving.22 2 Heinz Bauer, Measure and Integration Theory, pp. 55–56, §10.

It is a fact33 3 Heinz Bauer, Measure and Integration Theory, p. 57, Theorem 11.1. that if un is a nondecreasing sequence in E and uE then

usupnunIμusupnIμun.

It follows that if un and vn are sequences in E then

supnun=supnvnsupnIμun=supnIμvn. (1)

Define E*=E*(𝒜) to be the set of all functions f:X[0,] for which there is a nondecreasing sequence un in E satisfying supnun=f, in other words, there is a sequence un in E satisfying unf. From (1), for fE* and sequences un,vnE with supnun=f and supnvn=f, it holds that supnIμun=supnIμvn. Also, if uE then un=u is a nondecreasing sequence in E with u=supnun, so uE*. Then it makes sense to extend Iμ from E[0,] to E*[0,] by defining Iμf=supnIμun. One proves44 4 Heinz Bauer, Measure and Integration Theory, pp. 58–59, §11. that

f,gE*,α0αf,f+g,fg,fg,fgE*

and that Iμ:E*[0,] is positive homogeneous, additive, and order preserving.

The monotone convergence theorem55 5 Heinz Bauer, Measure and Integration Theory, p. 59, Theorem 11.4. states that if fn is a sequence in E* then supnfnE* and

Iμ(supnfn)=supnIμfn.

We now prove a characterization of E*.66 6 Heinz Bauer, Measure and Integration Theory, p. 61, Theorem 11.6.

Theorem 1.

E* is equal to the set of functions X[0,] that are AB¯ measurable.

Proof.

If fE*, then there is a sequence un in E with unf. Because each un is measurable 𝒜¯, so is f.

Now suppose that f:X[0,] is 𝒜¯ measurable. For n1 and 0in2n-1 let

Ai,n={fi2-n}{f<(i+1)2-n}={i2-nf<(i+1)2-n},

and for i=n2n let

Ai,n={fn}.

Because f is 𝒜¯ measurable, the sets Ai,n belong to 𝒜. For each n, the sets A0,n,An2n-1,n,An2n,n are pairwise disjoint and their union is equal to X. It is apparent that

Ai,n=A2i,n+1A2i+1,n+1,0in2n-1. (2)

Define

un=i=0n2ni2-n1Ai,n,

which belongs to E. For xX, either f(x)= or 0f(x)<. In the first case, un(x)=n for all n1. In the second case, un(x)f(x)<un(x)+2-n for all n>f(x). Therefore un(x)f(x) as n, and because this is true for each xX, this means unf and so fE*. ∎

So far we have defined Iμ:E*[0,]. Suppose that f:X¯ is 𝒜¯ measurable. Then f+,f-:X[0,] are 𝒜¯ measurable so by Theorem 1, f+,f-E*. Then Iμf+,Iμf-[0,]. We say that a function f:X¯ is μ-integrable if it is 𝒜¯ measurable and Iμf+< and Iμf-<. One checks that a function f:X¯ is μ-integrable if and only if it is 𝒜¯ measurable and Iμ|f|<. If f:X¯ is μ-integrable, we now define Iμf by

Iμf=Iμf+-Iμf-.

For example, if μ(X)< and S is a subset of X that does not belong to 𝒜, define f:X by f=1S-1XS. Then f+=1S and f-=1XS, and thus f is not 𝒜¯ measurable, so it is not μ-integrable. But |f|=1 belongs to E, and Iμ|f|=μ(X)< by hypothesis, showing that |f| is μ-integrable while f is not.

One proves that if f,g:X¯ are μ-integrable and α then αf is μ-integrable and

Iμ(αf)=αIμf,

if f+g is defined on all X then f+g is μ-integrable and

Iμ(f+g)=Iμf+Iμg,

and fg,fg are μ-integrable.77 7 Heinz Bauer, Measure and Integration Theory, p. 65, Theorem 12.3. Furthermore, Iμ is order preserving.

Let f:X be a function and write f=u+iv. One proves that f is Borel measurable (i.e. 𝒜 measurable), if and only if u and v are measurable 𝒜. We define f to be μ-integrable if both u and v are μ-integrable, and define

Iμf=Iμu+iIμv.

2 ℒ²

Let (X,𝒜,μ) be a measure space and for 1p< let p(μ) be the collection of Borel measurable functions f:X such that |f|p is μ-integrable. For complex a,b, because xxp is convex we have by Jensen’s inequality

|a+b2|p(12|a|+12|b|)p12|a|p+12|b|p=12(|a|p+|b|p),

so |a+b|p2p-1(|a|p+|b|p). Thus if f,gp(μ) then

|f+g|p2p-1(|f|p+|g|p),

which implies that p(μ) is a linear space.

For Borel measurable f:X define

fLp=(X|f|p𝑑μ)1/p.

For f,gp(μ), by Hölder’s inequality, with 1p+1p=1 (for which p=pp-1),

f+gLpp X|f||f+g|p-1𝑑μ+X|g||f+g|p-1𝑑μ
fLp|f+g|p-1Lp+gLp|f+g|p-1Lp
=fLpf+gLpp-1+gLpf+gLpp-1,

which implies that f+gLpfLp+gLp, and hence Lp is a seminorm on p(μ).

Let 𝒩p(μ) be the set of those fp(μ) such that fLp=0. 𝒩p(μ) is a linear subspace of p(μ), and we define

Lp(μ)=p(μ)/𝒩p(μ)={f+𝒩p(μ):fp(μ)}.

Lp(μ) is a normed linear space with the norm Lp.

It is a fact that if V is a normed linear space then V is complete if and only if each absolutely convergent series in V converges in V. Suppose that fk is a sequence in p(μ) with k=1fLp<. For n1 let gn(x)=(k=1n|fk(x)|)p and define g:X[0,] by

g(x)=(k=1|fk(x)|)p=limngn(x),

which is 𝒜¯ measurable, being the pointwise limit of a sequence of functions each of which is 𝒜¯ measurable. Because g1g2, by the monotone convergence theorem,

Xg𝑑μ=limnXgn𝑑μ.

But

(Xgn𝑑μ)1/p=k=1n|fk|Lpk=1nfkLpk=1fkLp,

which implies that Xg𝑑μ<, meaning that g:X[0,] is integrable. The fact that g is integrable implies μ(E)=0, where E={xX:g(x)=}𝒜. For xXE, k=1|fk(x)|< and because is complete this implies that k=1fk(x), and so it makes sense to define f:X by

f(x)=1XE(x)k=1fk(x),

which is Borel measurable. Furthermore, |f|pg, and because g is integrable this implies that fp(μ). For xXE,

limn|k=1nfk(x)-f(x)|p=0

and

|k=1nfk(x)-f(x)|pg(x),

so by the dominated convergence theorem,88 8 Heinz Bauer, Measure and Integration Theory, p. 83, Theorem 15.6.

limnX|k=1nfk(x)-f(x)|p𝑑μ=0.

Because xx1/p is continuous this implies

limnk=1nfk-fLp=0.

Hence, if fk is a sequence in Lp(μ) such that k=1fkLp< then there is some fLp(μ) such that k=1nfkf in the norm Lp. This implies that Lp(μ) is a Banach space.

We say that the σ-algebra 𝒜 is countably generated if there is a countable subset 𝒞 of 𝒜 such that 𝒜=σ(𝒞) and we say that a topological space is separable if there exists a countable dense subset of it. It can be proved that if 𝒜 is countably generated and μ is σ-finite, then for 1p< there is a countable collection of simple functions that is dense in Lp(μ), showing that Lp(μ) is separable.99 9 Donald L. Cohn, Measure Theory, second ed., p. 102, Proposition 3.4.5.

Theorem 2.

Let (X,A,μ) be a measure space and let 1p<. Lp(μ) with the norm Lp is a Banach space, and if A is countably generated and μ is σ-finite then Lp(μ) is separable.

For f,g2(μ), let

f,gL2(μ)=Xfg¯𝑑μ.

This is an inner product on L2(μ), and thus L2(μ) is a Hilbert space.

3 Product measures

Let (X1,𝒜1,μ1) and (X1,𝒜1,μ1) be measure spaces and let 𝒜1𝒜2 be the product σ-algebra. For QX1×X2, write

Qx1={x2X2:(x1,x2)Q},Qx2={x1X1:(x1,x2)Q}.

One proves that if μ1 and μ2 are σ-finite, then for each Q𝒜1𝒜2 the function x1μ2(Qx1) is 𝒜1¯ measurable and the function x2μ1(Qx2) is 𝒜2¯ measurable.1010 10 Heinz Bauer, Measure and Integration Theory, p. 135, Lemma 23.2. If μ1 and μ2 are σ-finite, one proves1111 11 Heinz Bauer, Measure and Integration Theory, p. 136, Theorem 23.3. that there is a unique measure μ:𝒜1𝒜2[0,] that satisfies

μ(A1×A2)=μ1(A1)μ2(A2),A1𝒜1,A2𝒜2.

The measure μ satisfies

μ(Q)=X1μ2(Qx1)𝑑μ1(x1)=X2μ1(Qx2)𝑑μ2(x2)

for Q𝒜1𝒜2, and is itself σ-finite. We write μ=μ1μ2, and call μ the product measure of μ1 and μ2.

Let X be a set and let f:X1×X2X be a function. For x1X1, define fx1:X2X by

fx1(x2)=f(x1,x2),x2X2

and for x2X2, define fx2:X1X by

fx2(x1)=f(x1,x2),x1X1.

For QX1×X2,

(1Q)x1=1Qx1,(1Q)x2=1Qx2.

It is straightforward to prove that if (X,𝒜) is a measurable space and f:(X1×X2,𝒜1𝒜2)(X,𝒜) is measurable, then for each x1X1 the function fx1:X2X is measurable 𝒜2𝒜 and for each x2X2 the function fx2:X1X is measurable 𝒜1𝒜.1212 12 Heinz Bauer, Measure and Integration Theory, p. 138, Lemma 23.5.

Tonelli’s theorem1313 13 Heinz Bauer, Measure and Integration Theory, p. 138, Theorem 23.6. states that if (X1,𝒜1,μ1) and (X1,𝒜1,μ1) are σ-finite measure spaces and f:X1×X2[0,] is 𝒜1𝒜2¯ measurable, then the functions

x2X1fx2𝑑μ1,x1X2fx1𝑑μ2

are 𝒜2¯ measurable and 𝒜1¯ measurable respectively, and

X1×X2fd(μ1μ2)=X2(X1fx2𝑑μ1)𝑑μ2(x2)=X1(X2fx1𝑑μ2)𝑑μ1(x1). (3)

Fubini’s theorem1414 14 Heinz Bauer, Measure and Integration Theory, p. 139, Corollary 23.7. states that if (X1,𝒜1,μ1) and (X2,𝒜2,μ2) are σ-finite measure spaces and f:X1×X2¯ is μ1μ2-integrable then there is some A1𝒜1 with μ1(A1)=0 such that for x1X1A1 the function fx1:X2¯ is μ2-integrable, and there is some A2𝒜2 with μ2(A2)=0 such that for x2X2A2 the function fx2:X1¯ is μ1-integrable. Furthermore, define F1:X1 by F1(x1)=X2fx1𝑑μ2 for x1X1A1 and F1(x1)=0 for x1A1, and define F2:X2 by F2(x2)=X1fx2𝑑μ1 for x2X2A2 and F2(x2)=0 for x2A2. The functions F1 and F2 are μ1-integrable and μ2-integrable respectively, and

X1×X2fd(μ1μ2)=X1F1𝑑μ1=X2F2𝑑μ2.

Suppose that (X1,𝒜1,μ1) and (X2,𝒜2,μ2) are σ-finite measure spaces. For e:X1 and f:X2, define ef:X1×X2 by

(ef)(x1,x2)=e(x1)f(x2),

which is Borel measurable X1×X2 if e and f are Borel measurable. If e2(μ1) and f2(μ2), then by Tonelli’s theorem ef:X1×X2 belongs to 2(μ1μ2). For e,e2(μ1) and f,f2(μ2), by Fubini’s theorem,

ef,efL2(μ1μ2)=X1×X2e(x1)f(x2)e(x1)f(x2)¯d(μ1μ2)(x1,x2)=X2(X1e(x1)e(x1)¯𝑑μ1(x1))f(x2)f(x2)¯𝑑μ2(x2)=e,eL2(μ1)f,fL2(μ2).

Therefore, if E2(μ1) is an orthonormal set in L2(μ1) and F2(μ2) is an orthonormal set in L2(μ2), then {ef:eE,fF}2(μ1μ2) is an orthonormal set in L2(μ1μ2).

Theorem 3.

Let (X1,A1,μ1) and (X2,A2,μ2) be σ-finite measure spaces and suppose that L2(μ1) and L2(μ2) are separable. If EL2(μ1) is an orthonormal basis for L2(μ1) and FL2(μ2) is an orthonormal basis for L2(μ2), then Φ={ef:eE,fF}L2(μ1μ2) is an orthonormal basis for L2(μ1μ2).

Proof.

To show that Φ is an orthonormal basis for L2(μ1μ2) it suffices to prove that if h2(μ1μ2) belongs to the orthogonal complement of Φ then h𝒩2(μ1μ2). Thus, suppose that h2(μ1μ2) and that h,efL2(μ1μ2)=0 for all eE,fF. Using Fubini’s theorem,

X1e(x1)(X2hx1(x2)f(x2)𝑑μ2(x2))𝑑μ1(x1)=0.

Because this is true for all eE and E is dense in L2(μ1), it follows that there is some Af𝒜1 with μ1(Af)=0 such that X2hx1f𝑑μ2=0 for x1Af. Let A1=fFAf, for which μ1(A1)=0. If x1A1 then X2hx1f𝑑μ2=0 for all fF, and because F is dense in L2(μ2) this implies that hx1=0 μ2-almost everywhere. Then

X1×X2|h|2d(μ1μ2) =X1(X2|hx1|2𝑑μ2)𝑑μ1(x1)
=X1A1(X2|hx1|2𝑑μ2)𝑑μ1(x1)
=0,

which implies that h=0 μ1μ2-almost everywhere. ∎