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: 6,764 Thanks: 697  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  
Newbie Joined: May 2012 Posts: 14 Thanks: 0  Re: f a.e. equals zero Quote:
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: 6,764 Thanks: 697  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  
Newbie Joined: May 2012 Posts: 14 Thanks: 0  Re: f a.e. equals zero Quote:
 
May 6th, 2012, 12:51 PM  #6 
Global Moderator Joined: May 2007 Posts: 6,764 Thanks: 697  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.


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