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Unformatted text preview: Analysis Midterm 1 Solutions 1. Let ( , F , ) be a measure space on which f, f 1 , f 2 , . . . are nonnegative integrable functions. Show that if f n f almost everywhere and R f n d R fd then R  f n f  d 0. Solution Notice that 0 6 ( f n f ) 6 f since f, f n are nonnegative. The Dominated Convergence Theorem therefore implies that R ( f n f ) d since f n f ae and the dominating function f is integrable. Now observe that Z  f n f  d = Z ( f n f ) d + 2 Z ( f n f ) d tends to 0 since on the right the first summand does so (given) and the second summand does so (shown). [This result is often called Sheff es lemma .] 2. Let ( , F , ) be a probability space. For Z define * ( Z ) = sup { ( A ) : F 3 A Z } * ( Z ) = inf { ( B ) : Z B F} define G = { Z : * ( Z ) = * ( Z ) } and for Z G write ( Z ) = * ( Z ) = * ( Z ). Prove that: (i) Z G if and only if for each > 0 there exist A, B F such that A Z B and ( B A ) < ; (ii) G is a algebra (containing F ) on which is a measure (extending ). Solution (i) Let > 0. ( ) If Z G then there exist F 3 A Z such that ( Z ) / 2 < ( A ) and Z B F such that ( B ) < ( Z ) + / 2; now ( B A ) = ( B ) ( A ) < . ( ) Let A, B F satisfy A Z B and ( B A ) < : then * ( Z ) > ( A ) > ( B ) > * ( Z ) so that * ( Z...
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This note was uploaded on 11/09/2011 for the course MAA 5228 taught by Professor Robinson during the Fall '09 term at University of Florida.
 Fall '09
 Robinson

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