204Lecture132006 - Economics 204 Lecture 13Wednesday,...

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Unformatted text preview: Economics 204 Lecture 13Wednesday, August 16, 2006 Revised 8/16/06, Revisions indicated by ** Sections 5.2 (Cont.), Transversality and Genericity Definition 1 Suppose A R n . A has Lebesgue measure zero if, for every > 0, there is a countable collection of rectangles I 1 , I 2 , . . . such that X k =1 Vol( I k ) < and A k =1 I k Notice that this defines Lebesgue measure zero without defining Lebesgue measure(!) This is a natural formulation of the notion that A is a small set: If you choose x R n at random, the probability that x A is zero. It is easy to show that A n has Lebesgue measure zero n N A n has Lebesgue measure zero In particular, Q and every countable set has Lebesgue measure zero. A function may have many critical points; for example, if a func- tion is constant on an interval, then every element of the interval is a critical point. But it cant have many critical values . Theorem 2 (2.4, Sards Theorem) Let X R n be open, f : X R m , f is C r with r 1 + max { , n m } . Then the set of all critical values of f has Lebesgue measure zero. Proof: First, we give a false proof that conveys the essential idea as to why the theorem is true; it can be turned into a correct proof. 1 Suppose m = n . Let C be the set of critical points of f , V the set of critical values. Then Vol( V ) = Vol( f ( C )) Z C | det Df ( x ) | dx (equality if f is one-to-one) = Z C dx = 0 Now, we outline how to turn this into a proof. First, show that we can write X = j N X j , where each X j is a compact subset of [ j, j ] n . Let C j = C X j . Fix j for now. Since f is C 1 , x k x det Df ( x k ) det Df ( x ) { x k } C j , x k x det Df ( x ) = 0 x C j so C j is closed, hence compact. Since X is open and C j is compact, there exists 1 > 0 such that B [ C j , 1 ] = x C j B [ x, 1 ] X B [ C j , 1 ] is bounded, and, using the compactness of C j , one can show it is closed, so it is compact. det Df ( x ) is continuous on B [ C j , 1 ], so it is uniformly continuous on B [ C j , 1 ], so given > 0, we can find 1 such that B [ C j , ] [ 2 j, 2 j ] n and x B [ C, ] det | Df ( x ) | 2 4 n j n Then f ( C j ) f ( B [ C j , ]) Vol( f ( B [ C j , ])) Z B [ C j , ] | det Df ( x ) | dx Z [ 2 j, 2 j ] n 2 4 n j n dx = 2 2 Since f is C 1 , show that f ( C j ) can be covered by a countable collection of rectangles of total volume less than . Since > 0 is arbitrary, f ( C j ) has Lebesgue measure zero. Then f ( C ) = f ( j N C j ) = n N f ( C j ) is a countable union of sets of Lebesgue measure zero, so f ( C ) has Lebesgue measure zero....
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This note was uploaded on 03/06/2009 for the course ECON 0204 taught by Professor Staff during the Summer '08 term at University of California, Berkeley.

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204Lecture132006 - Economics 204 Lecture 13Wednesday,...

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