I just have a short question:
I've prove the fact that given densities $f_n(x)$ of the $t_n$ distribution, then they converge pointwise to the density $\Phi(x)$ of a standard normal distributed random variable, i.e. $$\lim_n f_n(x) = \Phi(x)$$
How can I show the weak convergence $t_n \rightarrow ^D N(0,1)$? I know, that pointwise convergence is stronger, so one implies the other (pointwise convergence is for every x, weak convergence for every continuity point). But I don't know how to write it down or prove it properly. Any ideas?
Let's make some definitions clear. Define weak convergence of r.v.s as follows: given a sequence of r.v.s $\{X_n\}_n$ with distribution function $F_n$, these converge to a r.v. $X$ having distribution function $F$ if $\lim_{n\to\infty}F_n(x)=F(x)$ for every $x$ point of continuity of $F$.
In our case, since the r.v.s have densities, we get: $F_n(x)=P(X_n\leq x)=\int_{-\infty}^xf_n(x)\;dx$ and $F(x)=P(X\leq x)=\int_{-\infty}^xf(x)\; dx$.
You know that $f_n(x)$ tends pointwise to $f(x)$, i.e. the limit holds for every $x\in\mathbb{R}$. In particular, it holds almost everywhere. Since the functions are all positive and Lebesgue-integrable, then you can pass the limit under the integral (use, for example, Scheffé's Lemma, see https://en.wikipedia.org/wiki/Scheff%C3%A9%E2%80%99s_lemma). Hence, if $F$ is continuous in $x$, we write: $$ \lim_{n\to\infty}F_n(x)=\lim_{n\to\infty}\int_{-\infty}^xf_n(x)\; dx=\int_{-\infty}^x\lim_{n\to\infty}f_n(x)\; dx=\int_{-\infty}^xf(x)\; dx=F(x). $$