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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?

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1 Answer 1

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You cannot take closed intervals in this argument. The argument works if you take open intervals.

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  • $\begingroup$ Is that because otherwise I would not be able to find $f_n$ and $g_n$ with $$f_n(\cdot)\uparrow 1(\cdot\in I_1), \quad g_n(\cdot)\uparrow 1(\cdot\in I_2)?$$ $\endgroup$
    – Epiousios
    Aug 16, 2018 at 9:40
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    $\begingroup$ Yes, you can do this for open intervals but not for closed intervals. $\endgroup$ Aug 16, 2018 at 9:47

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