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 $(X, \cal{M}, \mu)$ be a positive measure space and suppose $f$ and $g$ are non-negative measurable functions such that $$ \int_{A} fd\mu = \int_A gd\mu,\quad \text{all } A \in \cal{M}. $$
$(a).$ Prove that $f = g$ a.e. $[\mu]$ if $\mu$ is $\sigma$-finite.
$(b).$ Prove that the conclusion of Part $(a)$ may fail if $\mu$ is not $\sigma$-finite.
Consider the set where $f > g.$ Denote this set $A.$ But $A=\cup_n[A_n]$ where $$A_n=\{x:f(x)>g(x)+1/n\},$$ a strictly increasing sequence hence $\mu(A)=\lim_n \mu(A_n).$ 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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