# 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 $(X, \cal{M}, \mu)$ 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)? –  Davide Giraudo Jul 18 '12 at 18:51
Yes, i did. But i think there was a gap in my solution, i just fixed it. –  srsly Jul 18 '12 at 18:53
You might want to consider (1) using mathjax/latex and (2) posting the problem (as opposed to a link to the problem). –  copper.hat Jul 18 '12 at 18:56
I LaTeX your solution. Please fix "integral-intersection" and the other odd things. Someone double-check please? –  user2468 Jul 18 '12 at 19:05
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 $g$. –  copper.hat Jul 18 '12 at 19:18

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.
Which $g$ and which set are you referring to? –  Zarrax Jul 18 '12 at 19:24
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. –  srsly Jul 18 '12 at 19:25