Question about convergence in probability implies convergence in distribution of sequence of random variables

Question

We have sequence of random variable $\{X_n:n\geq 1\}$ in probability space $(\Omega,F,P)$ then if $X_n$ is converge in probability that is $\forall \epsilon>0$ $$lim_{n\to \infty}P({\omega: |(X_n-X)(\omega)|>\epsilon})=0$$ then we have to show that $$lim_{n\to \infty}F_{X_n}(x)=F_{X}(x)$$
the proof goes like this
Proof: fix $\epsilon>0$ $$F_{X_n}(x)=P(X_n<x)$$ $$=P(X_n\leq x,X\leq x+\epsilon)+P(X_n\leq x,X>x+\epsilon)...(1)$$
$$=P(X\leq x+\epsilon)+P(X_n<X-\epsilon)..(2)$$ can someone explain how we got $(1)$ and $(2)$.I am confused in this part of the proof.Thanks. can

Answer 1

(1) is the result of the general fact that $P(A) = P(A \land B) + P(A \land \neg B)$; since exactly one of $X \leq x + \epsilon$ and $X > x + \epsilon$ is true, it must be that $$ P(X_n < x) = P(X_n \leq x \land X \leq x + \epsilon) + P(X_n \leq x \land X > x + \epsilon). $$

(2) I think should be in fact a $\leq$ and not and equality, since certainly $$P(X \leq x + \epsilon) \leq P(X_n \leq x, X \leq x + \epsilon)$$ and on the other part of the sum we know that $$P(X_n \leq x \land X > x + \epsilon) = P(X_n \leq x \land x < X - \epsilon) \leq P(X_n \leq X - \epsilon).$$

To conclude what you want, note that $$ P(X_n < X - \epsilon) = P(X - X_n< \epsilon) $$

So we may conclude $$ \lim_{n \to \infty}F_{X_n}(x) \leq P(X \leq x + \epsilon) $$ But when you do everything with the sign of $\epsilon$ flipped, then we can see that $$ P(X \leq x - \epsilon) \leq \lim_{n \to \infty} F_{X_n}(x) \leq P(X \leq x + \epsilon) $$

Convergence in distribution means convergence at points of continuity of $F$, so we are done, since $\epsilon$ was arbitrary.