Counterexample for a continuous function

Question

Continuity definition for expectation: For all $\epsilon>0$, there exists a $\delta>0$ such that the following holds $$\forall x~|F(x)-G(x)|\leq\delta \implies |E_F[X]-E_G[X]| \leq \epsilon. $$ Does it suffice to say that since Cauchy distribution doesn't have a mean, so $$|E_F[X]-E_G[X]|$$ is not bounded, and so this continuity doesn't hold for expectation? Even if this is true, can we have a more interesting counterexample for $F\neq G$? Any hints are appreciated.

Answer 1

Let $X_n \to 0$ almost surely with $EX_n$ not tending to $0$. [Eg: $nI_{(0,1/n)}$ on $(0,1)$ with Lebesgue measure]. Let $Y$ have standard normal distribution and assume that $Y$ is in dependent of the $X_n$'s. Then $X_n+Y \to Y$ almost surely, hence in distribution. By a standard theorem about convergence in distribution $F_{X_n+Y} \to F_Y$ uniformly on the real line. [ This is because $Y$ has a continuous distribution]. But $E(X_n+Y)$ does not tend to $EY$.