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solutions9 - Mathematics 105 Spring 2004 M Christ Problem...

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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 f n 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 . Then whenever n N , for any x R , either | f n ( x ) - 0 | = n - 1 < ε , or | f n ( x ) - 0 | = | 0 - 0 | < ε . Since N does not depend on x , this establishes uniform convergence. It’s equally obvious that f n = n - 1 λ ([0 , n ]) = n/n = 1. On the other hand, lim n →∞ f n = 0 = 0, which is not equal to 1 = lim n →∞ f n . 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 , 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 0 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 . More generally, suppose that f : E R n is a measurable mapping, in the sense that f - 1 ( S ) ∈ A for all S ∈ B R n . Show that the above recipe defines a measure on B R n .
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