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 $\mu$?


1 Answer 1


Here are the pieces you are missing, I think:

  1. Prokhorov's theorem actually says that $\textit{every}$ subsequence of your $(\mu_n)_{n \in \mathbb{N}}$ has a sub-subsequence converging in the weak topology to some probability measure.
  2. By your condition on the sequence $(\phi_n)_{n \in \mathbb{N}}$ converging to $\phi$, every one of the sub-subsequences of $(\mu_n)_{n \in \mathbb{N}}$ must converge to the measure $\mu$ whose characteristic function is $\phi$.
  3. In a Polish space, the topology of weak convergence is metrizable.

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.


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