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by: Mittie Ernser


Mittie Ernser
GPA 3.59


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Class Notes
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This 2 page Class Notes was uploaded by Mittie Ernser on Friday September 18, 2015. The Class Notes belongs to MAA 4212 at University of Florida taught by Staff in Fall. Since its upload, it has received 11 views. For similar materials see /class/206866/maa-4212-university-of-florida in Mathematics (M) at University of Florida.

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Popular in Mathematics (M)




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Date Created: 09/18/15
MAA 42127Some lecture notes for 12109 Proposition Let ab E Ra lt b A function f ab 7 R is integrable on ab if and only if for all 6 gt 0 there em39st step functions f1f2 ab 7 R such that f1z 3 fx 3 f2z for allx 6 ab and it fjfg 7 fjfl lt 6 Proof partial The if77 direction was proven cleanly in class We prove just the only if77 direction below Assume f is integrable on ab and let 6 gt 0 Then there exists 6 gt 0 such that U5f L5f lt 6 1 Select such a 6 and let 73 xiBio be a partition of 1 b of width less than 6 Since f is integrable on ab it is bounded on ab by a previously proven proposition and hence is bounded on each interval zj1zj1 Sj S N For each such j let M and m be the supremum and in mum respectively of t z E zj1zj De ne f1 f2 ab 7 R by f 7 M if 711 lt z lt Q for some j 2 7 y if z 7 for some j ha 771739 if 711 lt z lt Q for some j fj If x 7 for some j Then f1 f2 are step functions on ab and f1z 3 fx 3 f2z for all z 6 ab It remains to show only that f2 7 f1 lt 6 In previous work we calculated the integral of a general step function Applying this calculation to f1 and f2 we have b N f2 A23 ZMJ39W 9571 2 a j1 b N and f1 A1 3 Z7777ltj 7 jil a j1 In case you havent seen it before 77 is used to indicate that we are de ning the notation to the left of the 77 to mean whatever7s to the right of the 77 Now let S be the set of Riemann sums of f associated to 73 On each interval xi1 zj we have 771 S f S M Hence for any pointing T Of 73 A1 3 Sf73T 3 A2 Thus A1 and A2 are respectively lower and upper bounds for S We claim that A2 is the least upper bound of S For suppose not Then for some 6 gt 0 A2 7 6 is an upper bound for S For 1 S j S N since M7 LUBfx z E zj1zj there exists t E xi1le such that ft gt M 7 b 7 a The collection T tiBil is a pointing Of 73 and N 6 503731 gt Z rbiax j jil j1 N 6 N EMA j i 95H i b 7 a 95 9571 71 71 A2 67 a contradiction since by our de nition A2 7 6 is an upper bound for 8 Therefore A2 LUBS which by de nition is the number SUf73 A similar argument shows that A1 SLf73 Then since L5f S SLf73 S SUf73 S U5f inequality 1 implies that A2 A1 SUf73 SLf73 S U5f L6f lt 6 But by equations 273 A f for 239 12 so we are done I


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