Equivalent conditions for two functions to be equal almost everywhere in terms of integral

Question

My question is.... if two measurable functions $f$ and $g$ on a measure space $(X, M, \mu)$ are such that for all $A \in M$, $$ \int_{A}f\,d\mu = \int_{A}g\,d\mu, $$ does this imply that $f=g$ a.e. $\mu?$ If this isn't true, then are there any restrictions we can impose (positivity or $\sigma$-finiteness of $\mu$, etc.) to salvage the statement?

I've tried defining $E$ to be the set such that $f \neq g$ and show that it must have measure zero, but I can't find the right manipulations of the integral expressions to squeeze that out.... any suggestions?

Answer 1

Let $$A_\epsilon=\newcommand{\set}[1]{\left\{{#1}\right\}}\set{x\in X: f(x) \ge g(x)+\epsilon},$$ which is measurable because $f$ and $g$ are. Observe that $$0=\int_{A_\epsilon} f-g \,d\mu \ge \epsilon \mu(A_\epsilon),$$ so $\mu(A_\epsilon)=0$ for all $\epsilon$. Can you finish the proof from here?

Answer 2

Lemma: Suppose $\mu(C)>0$ and $h(x)>0$ for every $x\in C.$ Then $\displaystyle \int_C h\,d\mu>0.$

To prove this, consider the inverse-images under $h$ of $[1/(n+1),\, 1/n)$ for $n=1,2,3,\ldots\,.$ At least one of them has positive measure and you get a positive lower bound

Let $A=\{x\in X : f(x)>g(x)\}$ and $B=\{x\in X: f(x)< g(x)\}.$

By the Lemma, if $\mu(A)>0$ then $\displaystyle \int_A f\, d\mu > \int_A g\,d\mu.$ If $\mu(B)>0$ then $\displaystyle \int_B f\,d\mu < \int_B g\, d\mu.$

So if $\displaystyle\int_C f\,d\mu = \int_C g\,d\mu$ for every measurable set $C$, then $\mu(A) = \mu(B)=0,$ so $f=g$ almost everywhere.