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204Lecture132006

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

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Unformatted text preview: Economics 204 Lecture 13–Wednesday, 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 ε > 0, there is a countable collection of rectangles I 1 , I 2 , . . . such that ∞ X k =1 Vol( I k ) < ε 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 can’t have many critical values . Theorem 2 (2.4, Sard’s 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 δ 1 > 0 such that B [ C j , δ 1 ] = ∪ x ∈ C j B [ x, δ 1 ] ⊆ X B [ C j , δ 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 , δ 1 ], so it is uniformly continuous on B [ C j , δ 1 ], so given ε > 0, we can find δ ≤ δ 1 such that B [ C j , δ ] ⊆ [ − 2 j, 2 j ] n and x ∈ B [ C, δ ] ⇒ det | Df ( x ) | ≤ ε 2 · 4 n j n Then f ( C j ) ⊆ f ( B [ C j , δ ]) Vol( f ( B [ C j , δ ])) ≤ Z B [ C j ,δ ] | det Df ( x ) | dx ≤ Z [ − 2 j, 2 j ] n ε 2 · 4 n j n dx = ε 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 ε . Since ε > 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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204Lecture132006 - Economics 204 Lecture 13–Wednesday...

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