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Measure, Integration & Real Analysis, 2021a

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138 Chapter 5 Product <strong>Measure</strong>s<br />

The next result tells us that the collection of Borel sets from various dimensions<br />

fit together nicely.<br />

5.39 product of the Borel subsets of R m and the Borel subsets of R n<br />

B m ⊗B n = B m+n .<br />

Proof Suppose E is an open cube in R m+n . Thus E is the product of an open cube<br />

in R m and an open cube in R n . Hence E ∈B m ⊗B n . Thus the smallest σ-algebra<br />

containing all the open cubes in R m+n is contained in B m ⊗B n .Now5.38(b) implies<br />

that B m+n ⊂B m ⊗B n .<br />

To prove the set inclusion in the other direction, temporarily fix an open set G<br />

in R n . Let<br />

E = {A ⊂ R m : A × G ∈B m+n }.<br />

Then E contains every open subset of R m (as follows from 5.36). Also, E is closed<br />

under countable unions because<br />

( ⋃ ∞ ) ⋃ ∞<br />

A k × G = (A k × G).<br />

k=1<br />

Furthermore, E is closed under complementation because<br />

(<br />

R m \ A ) (<br />

)<br />

× G = (R m × R n ) \ (A × G) ∩ ( R m × G ) .<br />

Thus E is a σ-algebra on R m that contains all open subsets of R m , which implies that<br />

B m ⊂E. In other words, we have proved that if A ∈B m and G is an open subset of<br />

R n , then A × G ∈B m+n .<br />

Now temporarily fix a Borel subset A of R m . Let<br />

k=1<br />

F = {B ⊂ R n : A × B ∈B m+n }.<br />

The conclusion of the previous paragraph shows that F contains every open subset of<br />

R n . As in the previous paragraph, we also see that F is a σ-algebra. Hence B n ⊂F.<br />

In other words, we have proved that if A ∈B m and B ∈B n , then A × B ∈B m+n .<br />

Thus B m ⊗B n ⊂B m+n , completing the proof.<br />

The previous result implies a nice associative property. Specifically, if m, n, and<br />

p are positive integers, then two applications of 5.39 give<br />

(B m ⊗B n ) ⊗B p = B m+n ⊗B p = B m+n+p .<br />

Similarly, two more applications of 5.39 give<br />

B m ⊗ (B n ⊗B p )=B m ⊗B n+p = B m+n+p .<br />

Thus (B m ⊗B n ) ⊗B p = B m ⊗ (B n ⊗B p ); hence we can dispense with parentheses<br />

when taking products of more than two Borel σ-algebras. More generally, we could<br />

have defined B m ⊗B n ⊗B p directly as the smallest σ-algebra on R m+n+p containing<br />

{A × B × C : A ∈B m , B ∈B n , C ∈B p } and obtained the same σ-algebra (see<br />

Exercise 3 in this section).<br />

<strong>Measure</strong>, <strong>Integration</strong> & <strong>Real</strong> <strong>Analysis</strong>, by Sheldon Axler

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