Pointwise convergence vs weak convergence

I just have a short question:

I’ve prove the fact that given densities fn(x)f_n(x) of the tnt_n distribution, then they converge pointwise to the density Φ(x)\Phi(x) of a standard normal distributed random variable, i.e.
limnfn(x)=Φ(x)\lim_n f_n(x) = \Phi(x)

How can I show the weak convergence t_n \rightarrow ^D N(0,1)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?

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1 Answer
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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\{X_n\}_n with distribution function F_nF_n, these converge to a r.v. XX having distribution function FF if \lim_{n\to\infty}F_n(x)=F(x)\lim_{n\to\infty}F_n(x)=F(x) for every xx point of continuity of FF.

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)\;dxF_n(x)=P(X_n\leq x)=\int_{-\infty}^xf_n(x)\;dx and F(x)=P(X\leq x)=\int_{-\infty}^xf(x)\; dxF(x)=P(X\leq x)=\int_{-\infty}^xf(x)\; dx.

You know that f_n(x)f_n(x) tends pointwise to f(x)f(x), i.e. the limit holds for every x\in\mathbb{R}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, Scheffأ©’s Lemma, see https://en.wikipedia.org/wiki/Scheff%C3%A9%E2%80%99s_lemma). Hence, if FF is continuous in xx, 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).

\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).