Let X and Y be B-measurable and integrable r.v.s. We further assume that (B X dP (
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By setting Z = Y - X, we have (B Z dP ( 0 for all B ( B, and we wish to conclude that Z ( 0 a.s. Set C = (Z ( 0) and D = (Z > 0) (= Cc). Then it suffices to show that P(D) = 0. By taking B = D, we have (D Z dP ( 0 and (D Z dP = ((Z ID) dP ( 0 since Z > 0 on D, so that (D Z dP = 0. Thus, it suffices to show that, if for a r.v. Z with D = (Z > 0) it holds that (D Z dP = 0, then P(D) = 0. This can be done through the four familiar steps?
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B YdP B XdP or equivalently B Y XdP 0 or B ZdP 0 where Z Y X Thus to show B ZdP 0 for every B implie...View the full answer
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Related Book For 
An Introduction to Measure Theoretic Probability
ISBN: 978-0128000427
2nd edition
Authors: George G. Roussas
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