I have this problem:
Let $(\Omega, \mathscr{F}, \mathbb{P}$ be a probability space, $\mathscr{G}$ be a sub-$\sigma$-algebra. Let $X,Y$ be two real random variables such that for every bounded Borel-measurable function $g: \mathbb{R} \rightarrow \mathbb{R}^+$ we have $$\mathbb{E}\left(g(X)\mid\mathscr{G}\right) = g(Y) \text{ a.s.}$$ The question is if that implies $X=Y \text{ a.s}$ .
I guess the answer is yes, so I tried to prove that $\{X<Y\}$ is a nullset. If this set is in $\mathscr{G}$ it is obviously a nullset. But otherwise i get stuck. Does anyone have an idea?
First, integrating the equality given in the assumption gives that $\mathbb E\left[g(X)\right]= \mathbb E\left[g(Y)\right]$ for any bounded function $g$, hence considering $g$ as the indicator function of $(-\infty,t]$ where $t\in\mathbb R$, we derive that $X$ and $Y$ have the same distribution.
Therefore, for any bounded function $g$, $g(X)=g(Y)$ in distribution and by assumption, $\mathbb E\left[g(X)\mid\mathcal G\right]$ has the same distribution as $g(X)$, which entails that $\mathbb E\left[g(X)\mid\mathcal G\right]=g(X)$ a.s.. Going back to the initial assumption, we derive that for any bounded and non-negative function $g$, $g(X)=g(Y)$ a.s. From this, we can deduce equality of the positive parts, and by a similar reasoning that of the negative parts.