Here's a statement I've come across multiple times but have never seen a proof of:
Two random variables $X$ and $Y$ are independent, if for all continuous and bounded funtions $f, g: \mathbb R\to\mathbb R$ it holds that $$E[f(X)g(Y)]=E[f(X)]E[g(Y)].\tag{1}$$
I found this answer but I'm not sure that I'm filling in the details correctly:
Suppose $X$ and $Y$ satisfy the condition in $(1).$ We want to show that $X$ and $Y$ are independent. Since the closed intervals generate the Borel sigma algebra, it suffices to show that $$P(X\in I_1, Y\in I_2)=P(X\in I_1)P(Y\in I_2)$$ for all closed intervals $I_1, I_2\subset\mathbb R.$ Given two such intervals let $f_n, g_n\ge0$ be sequences of continuous and bounded functions with $$f_n(\cdot)\uparrow 1(\cdot\in I_1), \quad g_n(\cdot)\uparrow 1(\cdot\in I_2). \tag{2}$$
Then \begin{align*} P(X\in I_1, Y\in I_2) &= E[1(X\in I_1, Y\in I_2)]\\ &= E[1(X\in I_1)1(Y\in I_2)]\\ &= E[\lim_{n\to\infty}f_n(X)\lim_{m\to\infty}g_m(Y)]\\ &= \lim_{n\to\infty}E[f_n(X)\lim_{m\to\infty}g_m(Y)]\quad \text{(by monotone convergence)}\\ &= \lim_{n\to\infty}\lim_{m\to\infty}E[f_n(X)g_m(Y)]\quad \text{(by m.c.)}\\ &= \lim_{n\to\infty}\lim_{m\to\infty}E[f_n(X)]E[g_m(Y)]\quad \text{(by assumption)}\\ &= E[1(X\in I_1)]E[1(Y\in I_2)]\quad \text{(by m.c.)}\\ &=P(X\in I_1)P(Y\in I_2),\\ \end{align*} hence the claim.
Question: Is my above proof correct?
