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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 lefthand 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
 MichaelChrist
 Math

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