332 FOR F&Auml;&deg;NAL

332 FOR F&Auml;&deg;NAL - T Liggett Mathematics...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: T. Liggett Mathematics 131C – Final Exam Solutions June 7, 2010 (25) 1. (a) State Fatou’s Lemma. See Royden, page 86. (b) State the Bounded Convergence Theorem. See Royden, page 84. (c) Use Fatou’s Lemma to prove the Bounded Convergence Theorem. Suppose | f n | ≤ M for each n . Then M + f n and M- f n are nonnegative functions. By Fatou, Mm ( E )- Z E f = Z E ( M- f ) ≤ lim inf n →∞ Z E ( M- f n ) = Mm ( E )- lim sup n →∞ Z E f n and Mm ( E )+ Z E f = Z E ( M + f ) ≤ lim inf n →∞ Z E ( M + f n ) = Mm ( E )+lim inf n →∞ Z E f n . Since m ( E ) < ∞ , lim sup n →∞ Z E f n ≤ Z E f ≤ lim inf n →∞ Z E f n , so Z E f = lim n →∞ Z E f n . (20) 2. Evaluate lim n →∞ Z n 1- x n n e ax dx for each real a . Justify each of your steps. The integrand converges to e ( a- 1) x 1 [0 , ∞ ) ( x ) for each x . Since 1- t ≤ e- t for all t , the integrand is dominated by e ( a- 1) x 1 [0 , ∞ ) ( x ), which is integrable for a < 1. By the Dominated Convergence Theorem, lim n →∞ Z n 1- x n n e ax dx = Z ∞ e ( a- 1) x dx = 1 1- a for a < 1. Since Z n 1- x n n e ax dx is increasing in a , it follows that lim n →∞ Z n 1- x n n e ax dx = ∞ for a ≥ 1....
View Full Document

This note was uploaded on 11/08/2011 for the course MATH 332 taught by Professor Feritöztürk during the Spring '05 term at Boğaziçi University.

Page1 / 5

332 FOR F&Auml;&deg;NAL - T Liggett Mathematics...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online