Convergence in distribution and characteristic function Convergence in distribution ( Two equivalent definitions)
I am referring to the above problem. There was one thing in the proof that I was not clear.
Once we have the $\textbf{existence of a}$ weakly convergent subsequence.  The distribution, call it $\mu$, this subsequence converges weakly to, must have characteristic function $\phi$
. Moreover, $\textbf{every subsequence has a further subsequence}$ that converges weakly to $\mu$. Using again the Portmanteau Theorem characterization mentioned above (and a general topological fact: if every subsequence has a further subsequence that converges to some point, then the whole sequence converges to that point) one can show that the whole sequence of distributions.
But we only find one subsequence that converges weakly, how can we get the conclusion that this is for $\textbf{every}$ subsequence converges weakly. I mean, we may just have one such subsequence.
 A: The following statement is proved in the linked answer:

Let $(\nu_k)_k$ be  a sequence of probability meaures with characteristic function $\phi_k$. If $\phi_k$ converges pointwise to a function $\phi$ which is continuous at $t=0$, then there exists a subsequence $(\nu_{k_\ell})_{\ell}$ such that $\nu_{k_{\ell}}$ converges weakly to a probability measure $\nu$ with characteristic function $\phi$.

Now let $(\mu_n)_{n \in \mathbb{N}}$ be a sequence of probability measures with characteristic function $\phi_n$ such that $\phi_n$ converges pointwise to a function $\phi$ which is continuous at $t=0$. Now choose an arbitrary subsequence $\nu_k := \mu_{n_k}$ of this (original) sequence. By the above lemma, we find a subsequence of $(\nu_k)_k$ which converges weakly to a probability measure $\nu$ with characteristic function $\phi$. Since distributions are uniquely determined by their characteristic function, this probability measure $\nu$ does not depend on the chosen subsequence $(\nu_k)_k$.
Therefore, we have shown that the sequence $(\mu_n)_n$ has the property that for any subsequence there exists a (further) subsequence converging weakly to a probability measure $\nu$ (not depending on the chosen subsequence). Invoking this general topological fact which was mentioned by the OP, it follows that $\mu_n$ converges weakly to $\nu$.
