Topics In Analysis

10.9 Product Measures

Let

and

be two complete measure spaces. In this section consider the problem of defining a product measure, μ × ν which is defined on a σ algebra of sets of X ×Y such that

= μ

whenever E ∈S and F ∈T . I found the following approach to product measures in [?] and they say they got it from [?].

Definition 10.9.1 Let ℛ denote the set of countable unions of sets of the form A×B, where A ∈S and B ∈T (Sets of the form A × B are referred to as measurable rectangles) and also let

ρ (A \times B ) = μ(A) ν(B) (10.9.30)

(10.9.30)

More generally, define

\int \int ρ(E) \equiv XE (x,y)dμdν (10.9.31)

(10.9.31)

whenever E is such that

x \to XE (x,y) is μ measurable for all y (10.9.32)

(10.9.32)

and

\int y \to XE (x,y)dμ is ν measurable. (10.9.33)

(10.9.33)

Note that if E = A × B as above, then

\int \int \int \int XE (x,y)dμdν = XA \timesB(x,y)dμdν \int \int = XA (x)XB (y)dμdν = μ (A)ν(B ) = ρ(E )

and so there is no contradiction between 10.9.31 and 10.9.30.

The first goal is to show that for Q ∈ℛ, 10.9.32 and 10.9.33 both hold. That is, x →X_Q

is μ measurable for all y and y →∫ X_Q

dμ is ν measurable. This is done so that it is possible to speak of ρ

. The following lemma will be the fundamental result which will make this possible. First here is a picture.

Lemma 10.9.2 Given C × D and

_i=1ⁿ, there exist finitely many disjoint rectangles,

_i=1^p such that none of these sets intersect any of the A_i ×B_i, each set is contained in C × D and

(\cupni=1Ai \times Bi)\cup (\cuppk=1C'k \times D'k) = (C \times D )\cup(\cupni=1Ai \times Bi) .

Proof: From the above picture, you see that

(C \times D )∖(A1 \times B1) = C \times (D ∖B1) \cup(C ∖A1 )\times (D \capB1 )

and these last two sets are disjoint, have empty intersection with A₁ × B₁, and

(C \times (D ∖ B1)\cup (C ∖ A1)\times (D \cap B1))\cup(\cupni=1Ai \timesBi) = (C \times D )\cup (\cupni=1Ai \times Bi)

Now suppose disjoint sets,

_i=1^m have been obtained, each being a subset of C × D such that

( ) (\cupni=1Ai \times Bi)\cup \cupmk=1C^k \times ^Dk = (\cupni=1Ai \times Bi)\cup (C \timesD )

and for all k,

_k ×

_k has empty intersection with each set of

_i=1^p. Then using the same procedure, replace each of

_k ×

_k with finitely many disjoint rectangles such that none of these intersect A_p+1 × B_p+1 while preserving the union of all the sets involved. The process stops when you have gotten to n. This proves the lemma.

Lemma 10.9.3 If Q = ∪_i=1^∞A_i × B_i ∈ℛ, then there exist disjoint sets, of the form A_i^′×B_i^′ such that Q = ∪_i=1^∞A_i^′×B_i^′, each A_i^′×B_i^′ is a subset of some A_i ×B_i, and A_i^′∈S while B_i^′∈T . Also, the intersection of finitely many sets of ℛ is a set of ℛ. For ρ defined in 10.9.31, it follows that 10.9.32 and 10.9.33 hold for any element of ℛ. Furthermore,

\sum \sum ρ (Q ) = μ (A 'i) ν(B'i) = ρ(A 'i \times B'i). i i

Proof: Let Q be given as above. Let A₁^′× B₁^′ = A₁ × B₁. By Lemma 10.9.2, it is possible to replace A₂ ×B₂ with finitely many disjoint rectangles,

_i=2^m₂ such that none of these rectangles intersect A₁^′× B₁^′, each is a subset of A₂ × B₂, and

\infty m2 ' ' \infty \cupi=1Ai \times Bi = (\cup i=1A i \timesB i)\cup(\cupk=3Ak \times Bk)

Now suppose disjoint rectangles,

_i=1^m_p have been obtained such that each rectangle is a subset of A_k × B_k for some k ≤ p and

\cup\infty Ai \times Bi = (\cupmp A'\times B ')\cup (\cup\infty Ak \times Bk). i=1 i=1 i i k=p+1

By Lemma 10.9.2 again, there exist disjoint rectangles

_{i=m_p+1}^m_p+1 such that each is contained in A_p+1 × B_p+1, none have intersection with any of

_i=1^m_p and

\infty ( mp+1 ' ') ( \infty ) \cupi=1Ai \times Bi = \cupi=1 A i \times Bi \cup \cupk=p+2Ak \timesBk .

Note that no change is made in

_i=1^m_p. Continuing this way proves the existence of the desired sequence of disjoint rectangles, each of which is a subset of at least one of the original rectangles and such that

Q = \cup\inftyi=1A'i \times Bi'.

It remains to verify x →X_Q

is μ measurable for all y and

\int y \to X (x,y)dμ Q

is ν measurable whenever Q ∈ℛ. Let Q ≡∪_i=1^∞A_i ×B_i ∈ℛ. Then by the first part of this lemma, there exists

_i=1^∞ such that the sets are disjoint and ∪_i=1^∞A_i^′× B_i^′ = Q. Therefore, since the sets are disjoint,

\sum\infty \infty\sum XQ (x,y) = XA '\timesB '(x,y) = XA'(x)XB '(y). i=1 i i i=1 i i

It follows x →X_Q

is measurable. Now by the monotone convergence theorem,

\int \int \sum\infty XQ (x,y)dμ = XA 'i (x)XB 'i (y)dμ \infty i=1 \int = \sum X '(y) X '(x)dμ i=1 Bi Ai \infty\sum = XB'(y)μ (A 'i) . i=1 i

It follows y →∫ X_Q

dμ is measurable and so by the monotone convergence theorem again,

\int \int \int \infty\sum ' XQ (x,y)dμdν = XB'i (y)μ (A i)dν i=1\int \infty\sum ' = XB'i (y)μ (A i)dν i=\infty1 = \sum ν(B')μ(A'). (10.9.34) i=1 i i

This shows the measurability conditions, 10.9.32 and 10.9.33 hold for Q ∈ℛ and also establishes the formula for ρ

, 10.9.34.

If ∪_iA_i × B_i and ∪_jC_j × D_j are two sets of ℛ, then their intersection is

\cupi \cupj (Ai \capCj) \times(Bi \capDj )

a countable union of measurable rectangles. Thus finite intersections of sets of ℛ are in ℛ. This proves the lemma.

Now note that from the definition of ℛ if you have a sequence of elements of ℛ then their union is also in ℛ. The next lemma will enable the definition of an outer measure.

Lemma 10.9.4 Suppose

_i=1^∞ is a sequence of sets of ℛ then

\sum\infty ρ (\cup \inftyi=1Ri) \leq ρ(Ri). i=1

Proof: Let R_i = ∪_j=1^∞A_jⁱ ×B_jⁱ. Using Lemma 10.9.3, let

_m=1^∞ be a sequence of disjoint rectangles each of which is contained in some A_jⁱ ×B_jⁱ for some i,j such that

\cup \inftyi=1Ri = \cup\inftym=1A 'm \times B'm.

Now define

Si \equiv {m : A ' \timesB ' \subseteq Ai \timesBi for some j}. m m j j

It is not important to consider whether some m might be in more than one S_i. The important thing to notice is that

\cupm \inSiA'm \times B'm \subseteq \cup\inftyj=1Aij \times Bij = Ri.

Then by Lemma 10.9.3,

\sum ρ(\cup\inftyi=1Ri) = ρ(A'm \times B 'm) m \sum\infty \sum ' ' \leq ρ (A m \timesB m) i=1m\inSi \sum\infty ' ' \infty\sum \leq ρ(\cupm \inSiA m \times Bm) \leq ρ(Ri). i=1 i=1

This proves the lemma.

So far, there is no measure and no σ algebra. However, the next step is to define an outer measure which will lead to a measure on a σ algebra of measurable sets from the Caratheodory procedure. When this is done, it will be shown that this measure can be computed using ρ which implies the important Fubini theorem.

Now it is possible to define an outer measure.

Definition 10.9.5 For S ⊆ X × Y, define

(μ\times-ν)(S) \equiv inf{ρ(R ) : S \subseteq R, R \in ℛ }. (10.9.35)

(10.9.35)

The following proposition is interesting but is not needed in the development which follows. It gives a different description of

Proposition 10.9.6

= inf

Proof: Let λ

≡ inf

. Suppose S ⊆∪_i=1^∞A_i × B_i ≡ Q ∈ℛ. Then by Lemma 10.9.3, Q = ∪_iA_i^′× B_i^′ where these rectangles are disjoint. Thus by this lemma, ρ

= ∑ _i=1^∞μ

≥ λ

and so λ

≤

. If λ

= ∞, this shows λ

. Suppose then that λ

< ∞ and λ

+ ε > ∑ _i=1^∞μ

where Q = ∪_i=1^∞A_i ×B_i ⊇ S. Then by Lemma 10.9.3 again, ∪_i=1^∞A_i × B_i = ∪_i=1^∞A_i^′× B_i^′ where the primed rectangles are disjoint, each is a subset of some A_i × B_i and so

\sum\infty \sum\infty ' ' ----- λ (S )+ ε \geq μ (Ai)ν(Bi) \geq μ (A i) ν(Bi) = ρ(Q ) \geq (μ\times ν)(S). i=1 i=1

Since ε is arbitrary, this shows λ

≥

and this proves the proposition.

Lemma 10.9.7 μ × ν is an outer measure on X × Y and for R ∈ℛ

----- (μ\times ν)(R) = ρ(R). (10.9.36)

(10.9.36)

Proof: First consider 10.9.36. Since R ⊇ R, it follows ρ

≥

. On the other hand, if Q ∈ℛ and Q ⊇ R, then ρ

≥ ρ

and so, taking the infimum on the left yields

≥ ρ

. This shows 10.9.36.

It is necessary to show that if S ⊆ T, then

----- ----- (μ \timesν) (S ) \leq (μ\times ν)(T), (10.9.37)

(10.9.37)

----- \infty \sum\infty ----- (μ \times ν)(\cupi=1Si) \leq (μ \times ν)(Si). (10.9.38) i=1

(10.9.38)

To do this, note that 10.9.37 is obvious. To verify 10.9.38, note that it is obvious if

= ∞ for any i. Therefore, assume

< ∞. Then letting ε > 0 be given, there exist R_i ∈ℛ such that

(μ-\times-ν)(Si)+ ε-> ρ(Ri),Ri \supseteq Si. 2i

Then by Lemma 10.9.4, 10.9.36, and the observation that ∪_i=1^∞R_i ∈ℛ,

----- ----- (μ\times ν)(\cup\inftyi=1Si) \leq (μ \times ν)(\cup\inftyi=1Ri) \sum\infty = ρ(\cup\inftyi=1Ri) \leq ρ(Ri) i=1 \infty\sum ( ----- ε-) \leq (μ\times ν)(Si)+ 2i i=(1 ) \sum\infty ----- = (μ \times ν)(Si) + ε. i=1

Since ε is arbitrary, this proves the lemma.

By Caratheodory’s procedure, it follows there is a σ algebra of subsets of X × Y, denoted here by S×T such that

is a complete measure on this σ algebra. The first thing to note is that every rectangle is in this σ algebra.

Lemma 10.9.8 Every rectangle is

measurable.

Proof: Let S ⊆ X × Y. The following inequality must be established.

----- ----- ----- (μ \times ν)(S ) \geq (μ\times ν)(S \cap(A \times B))+ (μ \timesν )(S ∖(A \times B)). (10.9.39)

(10.9.39)

The following claim will be used to establish this inequality.

Claim: Let P,A × B ∈ℛ. Then

ρ(P \cap (A\times B ))+ ρ(P ∖(A \times B)) = ρ(P ).

Proof of the claim: From Lemma 10.9.3, P = ∪_i=1^∞A_i^′×B_i^′ where the A_i^′×B_i^′ are disjoint. Therefore,

\infty⋃ ' ' P \cap(A \times B ) = (A \cap Ai)\times (B \capB i) i=1

while

\infty⋃ ' ' \infty⋃ ' ' P ∖ (A \times B ) = (A i ∖A )\times Bi \cup (A \cap Ai)\times (Bi ∖ B). i=1 i=1

Since all of the sets in the above unions are disjoint,

ρ(P \cap(A \times B))+ ρ (P ∖ (A \times B )) =

\int \int \infty \int \int \infty \sum X ' (x) XB\capB'(y)dμdν + \sum X ' (x)XB '(y)dμdν i=1 (A\capAi) i i=1 (Ai∖A) i