Convergence in distribution ( Two equivalent definitions) I read that for convergence in distribution it is equivalent to have that either the characteristic functions of the random variables convergence pointwise or we have that $F_{X_n} \rightarrow F_{X}$ pointwise, where $F$(the distribution function) is continuous.
I could not find a proof of this, so I was wondering how hard it is to show? Does anybody here have a (Internet)-reference or could sketch the idea?
 A: Durrett's probability book appears to still be free (on author's page). Your subject is embedded in Chapter 3 Central Limit Theorems (Weak Convergence, Characteristic Functions etc., leading to Continuity Theorem 3.3.6).
Rick Durrett's Probability: Theory and Examples book Theorem 3.3.6. Levy's continuity theorem: 
Let $\mu_n$, $1\leq n \leq \infty$ be probability measures with characteristic functions $\phi_n$. 
(i) If $\mu_n$ converges weakly to $\mu_\infty$, then $\phi_n(t)$ converges pointwise to $\phi_\infty(t)$.
(ii) If $\phi_n(t)$ converges pointwise to a limit $\phi(t)$ that is continuous at $0$, then the associated sequence of distributions $\mu_n$ is tight and converges weakly to a measure $\mu$ with characteristic function $\phi$. 
Statement (i) follows from the Portmanteau Theorem characterization of weakly convergence that uses bounded and continuous functions, by noting that $\mathrm e^{itx}$ is bounded and continuous in $x$.
For (ii), Durrett first proves that sequence of distributions $\mu_n$ is tight (not easy, uses the continuity of $\phi$ at $0$; I'll try to update some details later), which in turn implies the existence of a weakly convergent subsequence. The distribution, call it $\mu$, this subsequence converges weakly to, must have characteristic function $\phi$. Moreover, 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 $\mu_n$ converges weakly to $\mu$. 
Edit: The tightness follows from the observation that:
$$ \mu_n\left(\{x\;|\;|x|>2u^{-1} \}\right) \leq u^{-1}\int_{-u}^u (1- \phi_n(t))dt,$$
using Fubini's theorem to re-write the integral as:
$$ u^{-1}\int_{-u}^u (1- \phi_n(t))dt = \int_{-\infty}^\infty \left(1- \frac{\sin(ux)}{ux}\right)\mu_n(dx),$$
and the fact that $|\sin x|\leq |x|$ for all $x$.
Edit2: For relationship of tightness and weak convergence topology see also Prokhorov's Theorem and its corollaries.
