F n 2 f n 1 k p 1 2 then continue in this fashion put

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f n 2 - f n 1 k p < 1 2 . Then continue in this fashion.) Put g k = k X i =1 | f n i - 1 - f n i | and g = 1 X i =1 | f n i - 1 - f n i | . By the triangle inequality, k g k k p k X i =1 k f n i +1 - f n i k p k X i =1 1 2 i < 1 8 k. Notice that g k % g , so g p k % g p , so by the Monotone Convergence theorem, Z g p = lim k !1 Z g p k = lim k !1 k g k k p p 1 . Thus g 2 L p and k g k p 1. We now claim that since R g p < 1 , it must be the case that g < 1 a.e. . Suppose g < 1 except on A where m ( A ) = 0. So for all x / 2 A , 1 X i =1 | f n i +1 - f n i ( x ) | < 1 so for all x / 2 A , P 1 i =1 f n i +1 - f n i ( x ) is absolutely convergent. Define f ( x ) = f n 1 ( x ) + 1 X i =1 f n i +1 - f n i ( x ) for x / 2 A and f ( x ) = 0 for x 2 A i.e., f ( x ) = lim k !1 f n 1 ( x ) + k X i =1 f n i +1 ( x ) - f n i ( x ) = lim k !1 f n k +1 ( x ) = lim k !1 f n k ( x ) for x / 2 A so f is measurable and | f ( x ) | | f n 1 | + 1 X i =1 | f n i +1 ( x ) - f n i ( x ) | = | f n 1 | + g for x / 2 A so since f n 1 , g 2 L p , we have f 2 L p .
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2 INTRODUCTION TO LEBESGUE INTEGRATION 23 Lecture 15: June 8 . We now have to prove that f n k ! f in the L p -norm, i.e. that k f n k - f k p ! 0. We know that | f n k - f | p ! 0 pointwise on A C . Notice that | f n k - f | p = 1 X i = k f n i +1 ( x ) - f n i ( x ) p 1 X i = k | f n i +1 ( x ) - f n i ( x ) | ! p g p 2 L 1 . By the dominated convergence theorem, R A C | f n k - f | p ! R A C lim n !1 | f n k - f | p = 0. Then k f n k - f k p p = Z A C | f n k - f | p + Z A | f n k - f | p ! 0 . Thus f n k ! f in L p -norm. Remark. The proof actually shows that if f n ! f in L p -norm, there exists a subsequence f n k such that f n k ( x ) ! f ( x ) a.e. . Since ( f n ) is convergent in L p -norm, it is Cauchy in L p . Hence by our proof there exists some function F and a subsequence ( f n k ) such that f n k ! F a.e. pointwise and f n k ! F in L p -norm. (See box in proof.) But limits are unique, so f = F a.e. . We now consider the p = 1 case. Let ( f n ) be Cauchy on L 1 . Let A k = { x : | f k ( x ) | > k f k k 1 } and B n,m = { x : | f n ( x ) - f m ( x ) | > k f n - f m k 1 } Let E = ( S 1 k =1 A k ) [ ( S 1 n =1 S 1 m =1 B n,m ), so m ( E ) = 0. On E C , we have sup e 2 E C | f n ( x ) - f m ( x ) |  k f n - f m k 1 ! 0 So on E C , ( f n ( x )) is uniformly Cauchy, so converges uniformly (i.e. w.r.t. the supremum norm) to some f , i.e. sup x 2 E C | f n ( x ) - f ( x ) | ! 0 as n ! 1 . Define f = 0 on E . Then k f n - f k 1 sup x 2 E C | f n ( x ) - f ( x ) | ! 0 . Therefore f n ! f in L 1 -norm. Furthermore pick N such that k f N - f k 1 1, so k f k 1  k f - f N k 1 + k f N k 1 1 + k f N k 1 < 1 so f 2 L 1 . Proposition (From Assignment 2) . If f : [ a, b ] ! R is measurable and δ > 0, then there is a continuous function h such that m { x : | f ( x ) - h ( x ) | δ } < δ . Furthermore, if m f M , then h can be chosen such that m h M . Theorem (Lusin) . C [ a, b ] is dense in L p [ a, b ] for all 1 p < 1 and C [ a, b ] are not dense in L 1 [ a, b ]. Proof. Let f 2 L p [ a, b ], then | f | < 1 a.e. . So f = g a.e. where g : [ a, b ] ! C . (In particular, g ( x ) 6 = 1 8 x .) Thus we assume WLOG that Range( f ) C . Further, assume WLOG that f is real-valued (since we can approximate
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2 INTRODUCTION TO LEBESGUE INTEGRATION 24 Re( f ) and Im( f ) and then put them back together). First, suppose f is bounded, say | f | N . Let " > 0. Take δ = min( " (2 N ) p , " b - a , 1). Take h such that m { x : | f ( x ) - h ( x ) | δ } < δ . Look at k h - f k p p = Z { x : | h ( x ) - f ( x ) | δ } | f - h | p + Z { x : | f ( x ) - h ( x ) | < δ } | f - h | p (2 N ) p m { x : | f ( x ) - h ( x ) | δ } + δ p Z [ a,b ] 1 δ (2 N ) p + δ p ( b - a )
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