Let $\mu_n$ be Borel probability measures on $[0, 1]$. I'm trying to prove a special case of Carleman's theorem, i.e.,
Theorem If the sequence $(\int_0^1 x^k \mathrm d \mu_n (x), n\in \mathbb N)$ converges for all $k \in \mathbb N$. Then there is a Borel probability measures on $[0, 1]$ such that $\mu_n \to \mu$ weakly.
Could you confirm if my below attempt is fine? Suggestion for simpler proof is very welcome.
Proof Let $\mathcal P([0, 1])$ be the space of Borel probability measures on $[0, 1])$. Because $[0, 1]$ is compact, $\mathcal P([0, 1])$ is compact in topology of weak convergence. So there is a sub-sequence $(n_k)$ and $\mu \in \mathcal P([0, 1])$ such that $\mu_{n_k} \underset{k \to \infty}{\longrightarrow} \mu$ weakly. It suffices to prove that $(\mu_n)$ converges weakly to $\mu$.
Fix a continuous bounded function $f:[0, 1] \to \mathbb R$. It suffices to prove $\int_0^1 f \mathrm d \mu_n \to \int_0^1 f \mathrm d \mu$. Fix $\varepsilon>0$. By Weierstrass approximation theorem, there is a polynomial function $g$ such that $\sup_{x \in [0, 1]} |f(x)-g(x)| < \varepsilon$. So it suffices to prove that $\int_0^1 g \mathrm d \mu_n \to \int_0^1 g \mathrm d \mu$.
Because $\mu_{n_k} \underset{k \to \infty}{\longrightarrow} \mu$ weakly, we have $$ \int_0^1 g \mathrm d \mu_{n_k} \underset{k \to \infty}{\longrightarrow} \int_0^1 g \mathrm d \mu. $$
WLOG, we assume $g (x) = \sum_{i=0}^p a_i x^i$ for some $a_0, \ldots, a_p \in \mathbb R$. Then $$ \int_0^1 g \mathrm d \mu_n = \sum_{i=0}^p a_i \int_0^1 x^i \mathrm d \mu_n (x). $$
Because $(\int_0^1 x^i \mathrm d \mu_n (x), n\in \mathbb N)$ converges for all $i=0, \ldots, p$, we get $(\int_0^1 g \mathrm d \mu_n, n\in \mathbb N)$ converges and thus $$ \lim_n \int_0^1 g \mathrm d \mu_n = \lim_k \int_0^1 g \mathrm d \mu_{n_k} = \int_0^1 g \mathrm d \mu. $$
This completes the proof.
