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 A∈M.
(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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