Suppose that $f, g$ are two nonnegative measurable functions that integrate to one over $\mathbf{R}$, with respect to Lebesgue measure and are such that the probability measure $\mu(A) = \int_A f d\lambda = \int_A g d\lambda$ for all Borels $A$. Our goal is to show that $f = g$ a.e.
I think the proof can go like this. Let $E = \{x : f(x) > g(x)\}$, and suppose that $\lambda(E) > 0$. Then for some $n$, $E_n = \{x : f(x) - g(x) \geq 1/n\}$ is such that $\lambda(E_n) > 0$ (by countable subadditivity this must be true), in which case $\int_{E_n} (f - g) d\lambda > 0$, which is a contradiction. We use here that $f$ and $g$ are measurable so that $E_n$ is Borel. Reversing the roles of $f$ and $g$, this means $\{x : f(x) > g(x)\} \cup \{x : g(x) > f(x)\} = \{x : f(x)\neq g(x)\}$ is null as desired.
By the way, it looks like the more general statement is that if $f$ and $g$ are two measurable functions agreeing (by integration) over all Borels, then $f = g$ a.e.
My only question is if this argument looks right.