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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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    It seems ok. Did you solve part b)?2012-07-18
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    Yes, i did. But i think there was a gap in my solution, i just fixed it.2012-07-18
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    You might want to consider (1) using mathjax/latex and (2) posting the problem (as opposed to a link to the problem).2012-07-18
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    I LaTeX your solution. Please fix "integral-intersection" and the other odd things. **Someone double-check please?**2012-07-18
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    For b), let $\mu$ be the measure on $\mathbb{R}$ that is $+\infty$ on any non-empty set. Then let $f,g$ be *any* two positive functions. Then the integral equality criterion is satisfied, but $f$ need not equal $g$.2012-07-18
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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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    how do you know for sure that g has finite integral on this set?2012-07-18
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    Which $g$ and which set are you referring to?2012-07-18
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    Sorry, I might have missinerpreted you. Sorry, I might have missinerpreted you. I thought at first you meant that you would just integrate over $A \cap X_n$ and this would give a contradiction right away.2012-07-18
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    ok I corrected my answer to deal with infinite integrals, thanks for pointing this out2012-07-18
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    "You do have to worry about f or g having infinite integral however". That is why I introduced the sets $C_n?2012-07-18
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    I wasn't sure what you were doing there with the $C_n$.. but yes that's the idea, and you also must consider where one or both functions are infinite.2012-07-18
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    I belive my solution handles the case when one side is infinite. Both sides cant be infinite when restricting to the intersection of A_M, X_M and C_M2012-07-18
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    You might have $f = \infty$ where $g < \infty$ over a set where both integrate to $\infty$ for example. You might be doing something up there to deal with it correctly, but since you didn't define $\mu(A_m, X_m, C_m)$ I wasn't exactly sure. Also, "a contradiction unless the left-hand side is infinite", both sides can be infinite (and maybe you knew that). Again what you are doing might work fine, it's just not that easy for me to know what you mean.2012-07-18
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    someone edited it, it was supposed to say the intersection of these three sets. One set assures that g is bounded, another that the measure is finite and one makes sure f is strictly larger than g on this set.2012-07-18
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    I think your answer is correct then.2012-07-18