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Topic: Measure Spaces and Integral. Prelim. Problem
Replies: 2   Last Post: Aug 1, 2012 1:42 PM

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G. A. Edgar

Posts: 2,504
Registered: 12/8/04
Re: Measure Spaces and Integral. Prelim. Problem
Posted: Aug 1, 2012 7:40 AM
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In article <fdec5e28-8fb4-4f70-85fa-d525322e367d@googlegroups.com>,
<baclesback@gmail.com> wrote:

> Hi, Again:
> We have a measure triple (X, mu, sigma), and f: X-->[0,oo] measurable.
> We want to show that if Int_X f=0 , then f==0 . I think mu(X)>0 , though
> this is not stated on the problem.

Perhaps the conclusion should be f=0 almost everywhere (according to

> If we knew, e.g. f is continuous and X compact, we could conclude, e.g.,
> m<f< M, and bound the integral by m*mu(X); unfortunately, we don't know this.
> Unfortunately, we cannot use version of Luzin, to approximate f with g
> continuous.
> My idea is to assume f>k>0 on a subset S< X. Then Int_X f >= k.m(S),
> and reach the contradiction that k=0 . But I'm not sure we can conclude this.
> One thing we can do is use the fact that f is measurable, so that
> f^{-1}[a,oo]:=V is measurable in X (assuming the Lebesgue measure on [0,oo] )
> then Int_x f > a*mu(V) . But how do I conclude mu(V)>0 ?
> Any Ideas?
> Thanks in Advance.

G. A. Edgar http://www.math.ohio-state.edu/~edgar/

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