My Math Forum f a.e. equals zero

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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 $f\in L^1([0,1]), 0, and for any measuable subset $E\subset [0,1]$ with $m(E)=c$, we have $\int_E f(t)dt=0$. Here $m$ is the Lebesgue measure on $[0,1]$. Show that $f\equiv 0,\ a.e. [m]$.
 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 $0 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.

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