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Unformatted text preview: Mathematics 105 Spring 2004 M. Christ A Supplementary Note, Selected Solutions for Problem Set 10, and Problem Set 11 Proposition. Denote Lebesgue measure in R k by k , and as usual let B R k , B R k denote the classes of Borel and Lebesgue measurable sets, respectively. For any m, n 1, B R m B R n = B R m + n . (1) For any set A B R m + n , m + n ( A ) = ( m n )( A ). Finally B R m B R n = B R m + n . (2) This proposition allows us to exploit Fubinis and Tonellis theorems in conjunc- tion with the product structure of R m + n ; without it we wouldnt know that the product measure and -algebra which arise in Fubinis theorem have any relevance to the real world of Lebesgue integration on Euclidean space. Proof. Ill simplify notation by systematically writing B k as an abbreviation for B R k . B m B n contains all products A 1 A 2 with A 1 B m , A 2 B n . In particular, it contains all such products in which both factors are open. Since any open set in R m + n can be expressed as a countable union of open squares, B m B n contains all open subsets of the product space; since its a -algebra, it therefore contains all Borel subsets. Thus B m B n B m + n . To show the reverse inclusion, it suffices to show that B m + n contains all products A 1 A 2 with A 1 B m , A 2 B n , since B m B n is by its definition contained in any -algebra which includes all these sets. All that is immediately obvious is that B m + n contains all products for which both sets A j are open . Fix an open set A 2 and consider the collection C of all sets B R m such that B A 2 B m + n . C contains all open sets B . C is a -algebra: it contains the open sets and R m ; its closed under countable unions since ( j B j ) A 2 = j ( B j A 2 ). To show that its closed under complementation let S C . Note that R m + n \ ( S A 2 ) = ( R m \ S ) A 2 R m ( R n \ A 2 ) , (3) and this is a disjoint union. The left-hand side is the complement of a set in B m + n , so belongs to B m + n . The second set on the right is the Cartesian product of two closed sets, so is closed, so belongs to B m + n . Therefore ( R m \ S ) A 2 B m + n , as desired. Since C is a -algebra which contains all open sets, it contains all of B m . Thus we conlude that if A 1 B m and if A 2 R n is open, then A 1 A 2 B m + n . Of course, the same conclusion then holds with the roles of R m , R n reversed. 1 Now fix any A 2 B n and redefine C to be the collection of all A 1 R m such that A 1 A 2 B m + n . By repeating the above reasoning we find that C is a -algebra; the only modification is that the set R m ( R n \ A 2 ) is no longer closed. However, it is the product of an open set with a Borel set, so by the preceding step belongs to B m + n ....
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This note was uploaded on 05/07/2008 for the course MATH 105 taught by Professor Michaelchrist during the Spring '04 term at University of California, Berkeley.
- Spring '04