Solutions9 - Mathematics 105 — Spring 2004 — M Christ Problem Set 9 — Solutions to Selecta IX.A Consider the measure space R 1 B R 1 where

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Unformatted text preview: Mathematics 105 — Spring 2004 — M. Christ Problem Set 9 — Solutions to Selecta IX.A Consider the measure space ( R 1 , B R 1 , λ ) where λ denotes Lebesgue measure. Consider the measurable functions f n ( x ) = 1 n χ [0 ,n ] . Show that f n → 0 uniformly on R . Show that R f n dλ → 1. Explain why this does not contradict any of our three basic convergence theorems. Solution. That f n → 0 uniformly is obvious: Given ε > 0, choose an integer N > 1 /ε . Then whenever n ≥ N , for any x ∈ R , either | f n ( x )- | = n- 1 < ε , or | f n ( x )- | = |- | < ε . Since N does not depend on x , this establishes uniform convergence. It’s equally obvious that R f n dλ = n- 1 λ ([0 , n ]) = n/n = 1. On the other hand, R lim n →∞ f n dλ = R dλ = 0, which is not equal to 1 = lim n →∞ R f n dλ . This doesn’t contradict the monotone convergence theorem, because the hypoth- esis that f n ≤ f n +1 for all n is violated. There’s no violation of the dominated convergence theorem. If f n ( x ) ≤ g ( x ) for all x, n , then of course g ( x ) ≥ sup n f n ( x ), and if k- 1 < x < k for some positive integer k then sup n f n ( x ) = k- 1 . Thus g ( x ) ≥ k- 1 for all x ∈ ( k- 1 , k ). Therefore for any “Lebesgue dominator” g , R g dλ ≥ ∑ ∞ k =1 k- 1 λ (( k- 1 , k )) = ∑ ∞ k =1 k- 1 = + ∞ . Therefore there is no integrable Lebesgue dominator, so the Dominated Convergence Theorem doesn’t apply. The hypotheses of Fatou’s lemma are satisfied — but there’s no contradiction since Fatou’s lemma doesn’t assert equality, only the inequality R dλ ≤ lim n →∞ 1 = 1. IX.B Let ( E, A , μ ) be any measure space. Suppose that f : E → R 1 is a measurable function (which never takes on the values ±∞ ). Recall that for any Borel measurable set S ⊂ R 1 , the set f- 1 ( S ) belongs to A . (You need not prove this.). Define ν ( S ) = μ ( f- 1 ( S )) for all S ∈ B R 1 Show that ν is a measure on the Borel σ-algebra B R ....
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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.

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Solutions9 - Mathematics 105 — Spring 2004 — M Christ Problem Set 9 — Solutions to Selecta IX.A Consider the measure space R 1 B R 1 where

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