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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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    This may be a little pedantic (the need for $\sigma$ finiteness frequently escapes me), but the crux of the argument is that to obtain a contradiction one must show the existence of a non-null set of *finite* measure on which $f$ and $g$ differ. The finite part is essential.2012-07-18

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Your answer looks like the right idea to me. You do have to worry about $f$ or $g$ having infinite integral however. One technique for doing this would be to replacing $X$ by $X_{lmn} = \{x \in A \cap X_n: f(x) < l, g(x) < m\}$, then doing what you did to show $f = g$ a.e. on $X_{lmn}$. Since the countable union of sets of measure zero has measure zero, you can then take the union over $l$, $m$, and $n$ to show that $f = g$ a.e. where both are finite. This is obviously true where both are infinite, so it remains to worry about where one is finite and the other is infinite. For that you can let $A = A_{ln} = \{x \in A \cap X_n: f(x) < l, g(x) = \infty \}$ or $B_{ln} = \{x \in A \cap X_n: g(x) < l, f(x) = \infty \}$, use the given condition and get that these sets are of measure zero. Taking the union over all $l$ and $n$ then finishes the proof.

For part b) you can just let $X$ be a measure space consisting of one point, of infinite measure.

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    I think your answer is correct then.2012-07-18