Is the following statement true on $L^0$ spaces? Let $(\Omega,\mathcal{F},P)$ be a probability space. Let $X,Y\in L^0(\Omega;\mathbb{R})$ two random variables taking values in $\mathbb{R}$. Is it true that:
$$\int_{A} f(X(\omega)) P(d\omega) = \int_{A} f(Y(\omega)) dP(\omega)$$
for all $A\in \mathcal{F}$ and all Lipschitz continuous functions $f:\mathbb{R}\rightarrow \mathbb{R}$ then
$$X=Y, \quad P-a.s.?$$
In other notation this is like saying:
$$E[f(X)] = E[f(Y)] \quad \forall f\in Lip(\mathbb{R}) \Rightarrow X=Y,\quad P-a.s.$$
Could we also relax it to all $f$ continuous and bounded (not necessarily Lipschitz)?
Thank you for the help :)
 A: Under the further hypothesis that $(\Omega, \mathcal{F},P)$ is a standard probability space, I believe that this is true even if we restrict to only one $f$, namely $f(x) = x$.
Recall the Lebesgue Differentiation Theorem. To apply that here, assume that $\Omega =[0, 1]$, $\mathcal{F}$ is the completed $\sigma$-algebra, and $P$ is Lebesgue measure. Equivalently, we can assume that $(\Omega, \mathcal{F},P)$ is mod $0$ isomorphic to that space, which is equivalent to the hypothesis that the space is standard.
From there, we have that $$ X(\omega) = \lim_{\epsilon \to 0} \frac{1}{2\epsilon} \int_{B_\epsilon (\omega)} X dP = \lim_{\epsilon \to 0} \frac{1}{2\epsilon} \int_{B_\epsilon (\omega)} Y dP = Y(\omega) $$ where the first and third equality are almost surely true by Lebesgue differentiation, and the second by the hypothesis.
So, with the probability space standard, this is true, but for more unusual situations, you'd need some way to get from integrals to function values, and Lebesgue Differentiation is the only method to do that that I am aware of.
