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)=lim So if \mu(A)>0, there is a N such that \mu(A_N)>0 Since X is \sigma-finite X=\cup_m\{X_m\} where \mu(X_m)< \infty. If \mu(A_N)>0 then there must exist an M s.t. \mu(\cap{A_N,X_M})>0. and hence integral_intersection\{A_N,X_M\} {f} \ge integral_intersection\{A_N,X_M\} {g} + 1/N\mu(\cap\{A_N,X_M\}), a contradiction unless the left-hand side is infinite.




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



Choose M s.t. \mu(A_M,X_M,C_M) > 0, now integrate over this set instead to arrive at the desired contradiction. Hence \mu(A)=0



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

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