I am trying to understand the proof behind Levy's theorem. The statement is as follows
Theorem: Let $(X_n)_{n\in \mathbb N}$ be a family of random variables, $(\mu_n)_{n\in \mathbb N}$ their distributions and $(\varphi_n)_{n\in \mathbb N}$ their characteristic functions.
Suppose that for every $t \in \mathbb R$
$$\lim_{n\to \infty}\varphi_n(t)= \varphi(t)$$
and that $\varphi$ is continuous in $0$. Then there exists a distribution $\mu$ such that $\varphi$ is $\mu$'s characteristic function and $(\mu_n)_{n \in \mathbb N}$ converges weakly to $\mu$.
The proof that I have been provided with shows that the family $(\mu_n)_{n\in \mathbb N}$ is tight through an integral argument.
From my (limited) knowledge in probability, Prokhorov's theorem states that if the family $(\mu_n)_{n\in \mathbb N}$ is tight then there exists a distribution $\mu$ and a subsequence of the given family $(\mu_{\kappa_n})_{n\in \mathbb N}$ that converges weakly to $\mu$, but there is no general result about the convergence of $(\mu_n)_{n \in \mathbb N}$.
Supposing that we have shown that $(\mu_n)_{n\in \mathbb N}$ is tight, we can conclude that there exists a sequence $\kappa_n$ such that
$$\lim_{n\to \infty}\varphi_{\kappa_n}(t)=\varphi(t)$$
but
How can we conclude that $(\mu_n)_{n \in \mathbb N}$ converges weakly to $\mu$?
Best Answer
Here are the pieces you are missing, I think:
That means limits are unique, and the sequence $(\mu_n)_{n \in \mathbb{N}}$ converges weakly to a probability measure $\mu$ if and only if each subsequence has a further subsequence converging weakly to that measure --- true because of steps 1, 2.