We solved the following problem in class and I do not understand what happens.
Definition: A sequence of random variables $W_1, W_2, \ldots$ converges in distribution to random variable $W$ if for every $w \in \mathbb{R}$ s.t. $P(W = w) = 0$, it is true that $\lim_{n \to \infty} P(W_n \leq w) = P(W \leq w)$.
Claim: Sequence of random variables $W_n$ converges weakly to $W$ iff for every $w \in \mathbb{R}$, s.t. $P(W = w) = 0$, it is true that $\lim_{n \to \infty} P(W_n < w) = P(W < w)$.
I do not understand the proof that we did. We only argued $(\Rightarrow)$, as the converse is similar. We used the fact that $ \{ W_n \leq w \} = \cap_k \{ W_n < w + \frac{1}{k} \}$. Then we wrote:
\begin{multline*} P(W \leq w) = \lim_{n \to \infty} P(W_n \leq w) = \lim_{n \to \infty} P \left( \cap_k \left( W_n < w + \frac{1}{k} \right) \right) = \lim_{n \to \infty} \left( \lim_{k \to \infty} P \left( W_n < w + \frac{1}{k} \right) \right) = \\ = \lim_{k \to \infty} \left( \lim_{n \to \infty} P \left( W_n < w + \frac{1}{k} \right) \right) = \lim_{k \to \infty} P \left(W < n + \frac{1}{k} \right) \end{multline*}
I have no idea why, first of all, $\lim_{k \to \infty} \left( \lim_{n \to \infty} P \left( W_n < w + \frac{1}{k} \right) \right) = \lim_{k \to \infty} P \left(W < n + \frac{1}{k} \right)$. Then I also do not understand why what we did is enough; how does the claim (at least $(\Rightarrow)$) follow from what we have shown?
We also stated that $F_X(x)$ has at most finitely many points of discontinuity and that this is crucial to our argument. Which makes me even more confused.
I think there is a typo in the final probability $P(W<n+\frac1k)$; it should be $P(W<w+\frac1k)$. Even so, this does not seem to be a proof of the claimed implication, because $\lim_{k\to\infty}P(W<w+\frac1k)=P(W\le w)$. So the proof as written is a roundabout way of saying that $P(W\le w)=P(W\le w)$. (An additional critique of the claimed proof: It's not clear why it is legal to interchange $\lim_{n\to\infty}$ and $\lim_{k\to\infty}$. And it's not obvious that $\lim_{n\to\infty}P(W_n<w+\frac1k)=P(W<w+\frac1k)$; this is what $(\Rightarrow)$ is trying to prove, which btw would require $P(W=w+\frac1k)=0$.)
Here's one way to prove $(\Rightarrow)$. Suppose that $P(W=w)=0$. To show that $P(W_n<w)\to P(W<w)$, it is enough to prove that $P(W_n=w)\to 0$, because we already have $P(W_n\le w)\to P(W\le w)$ by weak convergence. Let $(w_k)$ be a sequence of reals increasing to $w$ such that $P(W=w_k)=0$ for every $k$. It is possible to find such a sequence because $F_W$ has only a countable number of discontinuities. We have for every $k$ and $n$: $$P(W_n=w)\le P(W_n\in (w_k, w])= P(W_n\le w) - P(W_n\le w_k).\tag1$$ By weak convergence $$\lim_{n\to\infty}\left[P(W_n\le w) - P(W_n\le w_k)\right]= P(W\le w)-P(W\le w_k)\tag2$$ and so $$\limsup_n P(W_n=w)\le P(W\in (w_k,w]).\tag3$$ But as $k\to\infty$ the RHS of (3) converges to $P(W=w)$ (by continuity from above of $P$). Since $P(W=w)=0$, the LHS of (3) must be zero, completing the proof.