Tuesday, 9 January 2018

analysis - problem in measure theory

I would a appreciate if someone could take the time to check if my solution to the following problem is correct:





From http://www.math.chalmers.se/~borell/MeasureTheory.pdf, page 64, ex.6.



Let (X,M,μ) be a positive measure space and suppose f and g are non-negative measurable functions such that Afdμ=Agdμ,all AM.



(a). Prove that f=g a.e. [μ] if μ is σ-finite.



(b). Prove that the conclusion of Part (a) may fail if μ is not σ-finite.




Consider the set where f>g. Denote this set A. But A=n[An] where An={x:f(x)>g(x)+1/n},

a strictly increasing sequence hence μ(A)=limnμ(An). So if μ(A)>0, there is a N such that μ(AN)>0 Since X is σ-finite X=m{Xm} where μ(Xm)<. If μ(AN)>0 then there must exist an M s.t. μ(AN,XM)>0. and hence integral_intersection{AN,XM}f integral_intersection{AN,XM}g+1/Nμ({AN,XM}), a contradiction unless the left-hand side is infinite.




In that case consider Cn={x:g(x)<n}.



Choose M s.t. μ(AM,XM,CM)>0, now integrate over this set instead to arrive at the desired contradiction. Hence μ(A)=0



The same applies to the set B={x:g(x)>f(x)},μ(B)=0.

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