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 May 2nd, 2012, 10:49 PM #1 Newbie   Joined: May 2012 Posts: 14 Thanks: 0 f a.e. equals zero Let , and for any measuable subset with , we have . Here is the Lebesgue measure on . Show that . May 3rd, 2012, 03:13 PM #2 Global Moderator   Joined: May 2007 Posts: 6,834 Thanks: 733 Re: f a.e. equals zero General idea: Proof by contradiction. Assume f ? 0 on a set A of positive measure. A can be divided into 2 disjoint sets A+ and A- where f > 0 on A+ and f < 0 on A-. Since they are disjoint and the union has positive measure, at least one of them has positive measure. The integral of f over the set with positive measure cannot be 0 (possibly both). Added note: I am not sure I understood your question. Is there one value for c or for every value of c? May 3rd, 2012, 08:11 PM   #3
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Re: f a.e. equals zero

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 Originally Posted by mathman General idea: Proof by contradiction. Assume f ? 0 on a set A of positive measure. A can be divided into 2 disjoint sets A+ and A- where f > 0 on A+ and f < 0 on A-. Since they are disjoint and the union has positive measure, at least one of them has positive measure. The integral of f over the set with positive measure cannot be 0 (possibly both). Added note: I am not sure I understood your question. Is there one value for c or for every value of c?

Year, the is fixed...So your argument does not apply to this problem. Anyway, thank you very much. May 4th, 2012, 04:50 PM #4 Global Moderator   Joined: May 2007 Posts: 6,834 Thanks: 733 Re: f a.e. equals zero I haven't worked it out in detail, but it looks like you could assume that the interval can be divided into three sets A+ (f > 0), A- (f < 0), and A0 (f = 0). If either m(A+) or m(A-) > c, you can truncate that one to get m = c with an integral ? 0. If both m(A+) and m(A-) < c and m(A0) > 0, A+ can be supplemented by including part of A- only or by including part of A0 as well (I am expressing myself poorly), to get sets of measure c, but with different integrals. Finally in case m(A0) = 0, while both m(A+) and m(A-) < c, you can start with A- and add part of A+ or start with A+ and add part of A- until you get to something with a measure of c. The integrals should be different. May 5th, 2012, 10:53 PM   #5
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Re: f a.e. equals zero

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 Originally Posted by mathman I haven't worked it out in detail, but it looks like you could assume that the interval can be divided into three sets A+ (f > 0), A- (f < 0), and A0 (f = 0). If either m(A+) or m(A-) > c, you can truncate that one to get m = c with an integral ? 0. If both m(A+) and m(A-) < c and m(A0) > 0, A+ can be supplemented by including part of A- only or by including part of A0 as well (I am expressing myself poorly), to get sets of measure c, but with different integrals. Finally in case m(A0) = 0, while both m(A+) and m(A-) < c, you can start with A- and add part of A+ or start with A+ and add part of A- until you get to something with a measure of c. The integrals should be different.
I'm not sure you are right...So many poor proof. May 6th, 2012, 12:51 PM #6 Global Moderator   Joined: May 2007 Posts: 6,834 Thanks: 733 Re: f a.e. equals zero I agree it is poorly expressed. However the general idea is correct. Since c < 1, you can construct various combinations from A+ and A- to get sets of measure c, but with different integrals. Tags equals Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post yogazen2013 Calculus 2 October 5th, 2013 08:44 AM Anthony.R.Brown Number Theory 1 August 24th, 2013 08:56 AM maximus101 Calculus 3 March 8th, 2011 10:51 AM honzik Applied Math 1 June 10th, 2009 04:20 PM knowledgegain Calculus 2 May 2nd, 2009 01:32 AM

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