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




Re: Measure Spaces and Integral. Prelim. Problem
Posted:
Aug 1, 2012 7:40 AM


In article <fdec5e288fb44f7085fad525322e367d@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 mu).
> > 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.ohiostate.edu/~edgar/



