$\mathbb E [f(X)] = \mathbb E [f(Y)]$ for all $f \in \mathcal C_b (\mathbb R) \implies \mathbb P (X = Y) = 1.$

Question

Let $X$ and $Y$ be two random variables on a probability measure space $(\Omega, \mathcal F, \mathbb P)$ such that $\mathbb E [f(X)] = \mathbb E [f(Y)]$ for all $f \in \mathcal C_b (\mathbb R)$ (bounded continuous functions). Can we conclude that $\mathbb P (X = Y) = 1$ i.e. $X = Y$ almost surely?

Answer 1

No. You can conclude they have the same distribution but not that they are almost surely equal.

There are plenty counter-examples. For instance $X$ and $-X$ for any centred gaussian variable, or $U$ and $1-U$ for $U$ uniformly distributed on $(0,1)$.