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Let $(\Omega, \mathcal{F}, P)$ be a probability space and $X, Y:\Omega \to [0,1]$ be random variables. Prove that if

$$\mathbb{E}[f(X)]=\mathbb{E}[f(Y)] \text{ for all continuous }f:[0,1]\to\mathbb{R},$$

then $X$ and $Y$ have the same distribution.

My first idea was to prove that $P(X\leq \alpha)=P(Y\leq\alpha)$ for every $\alpha \in [0,1]$, which can be written as $\mathbb{E}[1_{\{X\leq\alpha\}}]=\mathbb{E}[1_{\{Y\leq\alpha\}}]$. Since $1_{\{X\leq\alpha\}}$ and $1_{\{Y\leq\alpha\}}$ may not be continuous, I thought about approximating them by continuous functions, but I'm having trouble to formalize this (not even sure it's possible).

Is there a better way? Thanks!

Community
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rmdmc89
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1 Answers1

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For any integer $n\ge 1$, consider the function \begin{align*} f_n(x) = \begin{cases} 1, & x \le \alpha-\frac{1}{n},\\ -n(x-\alpha), & \alpha-\frac{1}{n} < x \le \alpha,\\ 0, & x > \alpha. \end{cases} \end{align*} Then $f_n$ is continuous and $f_n(x) \nearrow 1_{x \le \alpha}$, and \begin{align*} E(f_n(X)) \rightarrow P(X \le \alpha). \end{align*}

Gordon
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