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# Because i n a n b n a we have 2 f b n f a n z b n a n

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Because I n = ( a n , b n ) A we have (2) | f ( b n ) - f ( a n ) | = | Z b n a n f 0 ( x ) dx | = Z b n a n f 0 ( x ) dx = Z b n a n | f 0 ( x ) | dx ; similarly, because J k B , (3) | f ( d k ) - f ( c k ) | = | Z d k c k f 0 ( x ) dx | = - Z d k c k f 0 ( x ) dx = Z d k c k | f 0 ( x ) | dx. Let ² > 0. Since (by disjointness) X n =1 m ( I n ) = m ( A ) m ([ a, b ]) < , there is N such that n = N +1 m ( I n ) < ²/ (2 C ) where C is a bound for | f 0 | ; | f 0 ( x ) | ≤ C for all x [ a, b ]. Similarly, there is K such that k = K +1 m ( J k ) < ²/ (2 C ). Now consider the intervals I 1 , I 2 , . . . , I N , J 1 , J 2 , . . . , J M . These are pairwise disjoint open subintervals of [ a, b ]. We can order them by their left endpoints and relabel them as K 1 , K 2 , . . . , K N + M , where K 1 is the left-most of these N + M intervals, K 2 the one following K 1 , etc. Let us also relabel the endpoints, so K j = ( σ j , τ j ) for j = 1 , . . . , N + M . Adding if needed, the endpoints a, b , the points a σ 1 < τ 1 < σ 2 < · · · < σ N + M < τ N + M b forma partition of [ a, b ] and by definition of V b a ( f ) we have V b a ( f ) | f ( σ 1 ) - f ( a ) | + | f ( τ 1 ) - f ( σ 1 ) | + | f ( σ 2 ) - f ( τ 1 ) | + | f ( τ 2 ) - f ( σ 2 ) | + · · · + | f ( τ N + M - f ( σ N + M ) | + | f ( b ) - f ( τ N + M ) | | f ( τ 1 ) - f ( σ 1 ) | + | f ( τ 2 ) - f ( σ 2 ) | + · · · + | f ( τ N + M - f ( σ N + M ) | . The last sum is the sum over the intervals K 1 , . . . , K N + M , which are just the intervals I 1 , . . . , I N , J 1 , . . . , J M in some order. That is, each f ( τ j ) - f ( σ j ) is either one of | f ( b n ) - f ( a n ) | or | f ( d k ) - f ( c k ) | ; that is, we proved V b a ( f ) N X n =1 | f ( b n ) - f ( a n ) | + M X k =1 | f ( d n ) - f ( c k ) | . In view of (2), (3), we proved (4) V b a ( f ) N X n =1 Z b n a n | f 0 ( x ) | dx + M X k =1 Z d k c k | f 0 ( x ) | dx = Z S N n =1 I n S M k =1 J k | f 0 | , the last equality holding because all the intervals ar disjoint. Now Z A | f 0 | = X n =1 Z I n | f 0 | = Z S N n =1 I n | f 0 | + Z S n = N +1 I n | f 0 | Since | f 0 | ≤ C , Z S n = N +1 I n | f 0 | ≤ Cm ( [ n = N +1 I n ) = C X n = N +1 m ( I n ) < ² 2 . Thus Z A | f 0 | < Z S N n =1 I n | f 0 | + ² 2 Similarly, Z B | f 0 | < Z S M k =1 J k | f 0 | + ² 2 , 7

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so that Z S N n =1 I n S M k =1 J k | f 0 | = Z S N n =1 I n | f 0 | + Z S M k =1 J k | f 0 | > Z A B | f 0 | - ². By (4), Z A B | f 0 | < V b a ( f ) + ². But f 0 = 0 in [ a, b ] \ ( A B ), thus Z b a | f 0 ( x ) | dx = Z A B | f 0 | < V b a ( f ) + ². Since ² > 0 is arbitrary, this proves Z b a | f 0 ( x ) | dx V b a ( f ) . 8
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• Spring '11
• Speinklo
• Empty set, Open set, Topological space, Dominated convergence theorem, Lebesgue integration, Non-measurable set

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