652 CHAPTER 32. MEASURES AND INTEGRALS

Example 32.11.4 Suppose you have (U,F ) and (V,S ) , two measurable spaces. LetK ⊆ U ×V consist of all sets of the form A×B where A ∈ F and B ∈ S . This is easilyseen to be a π system. When this is done, σ (K ) is denoted as F ×S .

Definition 32.11.5 When K is the open sets of Rp, the Borel sets, denoted asB (Rp) , are defined as B (Rp)≡ σ (K ).

Don’t try to describe a typical Borel set. Just use the definition that these are those setsin the smallest σ algebra that contains the open sets. However, if you wanted to give this atry, see Hewitt and Stromberg [20] who do something like this in showing the existence ofLebesgue measurable sets which are not Borel measurable.

For example, here is a useful result about the product of Borel sets.

Lemma 32.11.6 If Ak is a Borel set in R, then ∏pi=1 Ak is a Borel set in Rp.

Proof: Let πk : Rp → R be defined by πk (x)≡ xk, the kth entry of x. Then

π−1k (U) = R×·· ·×R×U ×R×·· ·×R

Let G be those Borel sets B such that π−1k (B) is Borel in Rp. Then from the above, this

is true if B is open. However, it follows from the definition of inverse image that G is a σ

algebra. Therefore, by definition G = B (Rp). Now note that

p

∏i=1

Ak = ∩pk=1π

−1k (Ak) ∈ B (Rp) .

32.12 Product MeasuresFirst of all is a definition.

Definition 32.12.1 Let (X ,F ,µ) be a measure space. Then it is called σ finite ifthere exists an increasing sequence of sets Rn ∈ F such that µ (Rn)< ∞ for all n and alsoX = ∪∞

n=1Rn.

Now I will show how to define a measure on ∏pi=1 Xi given that (Xi,Fi,µ i) is a σ finite

measure space. The main example I have in mind is the case where each Xi = R and ameasure µ = m to be described a little later, yielding p dimensional Lebesgue measure.However, there is no good reason not to do this in general. It is no harder, so this is what Iam doing here.

Let K denote all subsets of X ≡ ∏pi=1 Xi which are the form ∏

pi=1 Ei where Ei ∈ Fi.

These are called measurable rectangles. Let {Rni }

∞

n=1 be the sequence of sets in Fi whoseunion is all of Xi, Rn

i ⊆ Rn+1i , and µ i (R

ni ) < ∞. Thus if Rn ≡ ∏

pi=1 Rn

i , and E ≡ ∏pi=1 Ei,

then

Rn ∩E =p

∏i=1

Rni ∩Ei

Let I ≡ (i1, · · · , ip) where (i1, · · · , ip) is a permutation of {1, · · · , p}. Also, to save on space,let F be a subset of ∏

pi=1 Xi ≡X and denote the iterated integral∫

X11

· · ·∫

Xip

XF (x1, · · · ,xp)dµ i1 · · ·dµ ip