Let $X_1,X_2,...,X$ be random variables taking values in $[0,1]$. Show that $X_n \rightarrow X$ converge in distribution if and only if $E(X^k_n) \rightarrow E(X^k)$ for all $k=0,1,2,...$
The hint says I need to use Weierstrass’s polynomial approximation theorem. But I do not really see how to use this theorem to prove it. How should I start this question?
$(\Rightarrow)$. By definition, $X_{n}\longrightarrow X$ converges in distribution if and only if $\mathcal{P}_{X_{n}}$ converges weakly to $\mathcal{P}_{X}$ in $([0,1],\mathcal{B}([0,1]))$ where $\mathcal{P}_{X_{n}}$ and $\mathcal{P}_{X}$ are the distribution induced by $X_{n}$ and $X$, respectively. This happens if and only if for all continuous and bounded function $f:[0,1]\longrightarrow\mathbb{R}$, we have $\int fd\mathcal{P}_{X_{n}}\longrightarrow \int fd\mathcal{P}_{X}$, i.e. $\mathbb{E}(f(X))\longrightarrow\mathbb{E}(f(X_{n}))$. In particular, the function $x\mapsto x^{k}$ for any $k\in\mathbb{Z}_{\geq 1}$ is continuous and is bounded on $[0,1]$, and thus $\mathbb{E}X^{k}_{n}\longrightarrow\mathbb{E}X^{k}$ for any $k\in\mathbb{Z}_{\geq 1}$.
$(\Leftarrow)$. Suppose that $\mathbb{E}X^{k}_{n}\longrightarrow\mathbb{E}X^{k}$ for any $k\in\mathbb{Z}_{\geq 1}$. Then, by the linearity of the expectation, we have that $\mathbb{E}P(X_{n})\longrightarrow\mathbb{E}P(X)$ for any polynomial $P$. Hence, for any $\epsilon>0$, there exists $N=N(\epsilon)>0$ such that \begin{equation} \label{ex19eq1} |\mathbb{E}P(X_{n})-\mathbb{E}P(X)|<\epsilon,\forall n\geq N. \end{equation} Let $f:[0,1]\longrightarrow\mathbb{R}$ be a continuous function (and thus it is bounded), then there exists a polynomial $P$ such that $|f(x)-P(x)|<\epsilon$ for all $x\in [0,1]$, and thus \begin{equation} \label{ex19eq2} |\mathbb{E}P(X)-\mathbb{E}f(X)|=|\mathbb{E}(P(X)-f(X)|\leq \mathbb{E}|P(X)-f(X)|<\mathbb{E}\epsilon=\epsilon, \end{equation} and \begin{equation} \label{ex19eq3} |\mathbb{E}P(X_{n})-\mathbb{E}f(X_{n})|=|\mathbb{E}(P(X_{n})-f(X_{n})|\leq \mathbb{E}|P(X_{n})-f(X_{n})|<\mathbb{E}\epsilon=\epsilon, \forall n\in\mathbb{Z}_{\geq 1}. \end{equation} It then follows that \begin{align*} |\mathbb{E}f(X_{n})-\mathbb{E}f(X)|&=|\mathbb{E}f(X_{n})+\mathbb{E}P(X_{n})-\mathbb{E}P(X_{n})+\mathbb{E}P(X)-\mathbb{E}P(X)-\mathbb{E}f(X)|\\ &\leq |\mathbb{E}f(X_{n})-\mathbb{E}P(X_{n})|+|\mathbb{E}P(X_{n})-\mathbb{E}P(X)|+|\mathbb{E}P(X)-\mathbb{E}f(X)|\\ &<3\epsilon,\forall\ n\geq N. \end{align*} Hence, $\mathbb{E}f(X_{n})\longrightarrow \mathbb{E}f(X)$. Since $f$ was arbitrary, it follows that $\mathcal{P}_{X_{n}}$ converges to $\mathcal{P}_{X}$ weakly, and thus $X_{n}\longrightarrow X$ in distribution.