Suppose that the rvs $X \cong F$(*) and $X_{n} \cong$ $F_{n}$ satisfy $X_{n} \rightarrow_{p} X .$ Then $X_{n} \rightarrow_{d} X$.
(*)$X \cong F$ denotes the induced distribution $P_{X}(\cdot)$ of the $\mathrm{rv} X$ has distribution function(df) $F$.
The proof is given but I don't get some parts of it (I have pointed them):
There exists an integer $n_{\varepsilon}$ such that $$ \begin{aligned} F_{n}(t) &=P\left(X_{n} \leq t\right) \qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\quad\quad\text{(By the definition df)} \\ &\leq P(X \leq t+\varepsilon)+P\left(\left|X_{n}-X\right| \geq \varepsilon\right) \qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\text{(Q1)}\\ & \leq F(t+\varepsilon)+\varepsilon\quad\quad\quad\quad\quad\quad\quad\quad\text{(Definition of df and convergence in measure)} \end{aligned} $$
Also, there exists an integer $n_{\varepsilon}^{\prime}$ such that $$ \begin{aligned} F_{n}(t) &=P\left(X_{n} \leq t\right) \\ & \geq P\left(X \leq t-\varepsilon \text { and }\left|X_{n}-X\right| \leq \varepsilon\right) \equiv P(A B) \quad\quad\qquad\qquad\qquad\qquad\qquad\text{(Q2)}\\ & \geq P(A)-P\left(B^{c}\right)=F(t-\varepsilon)-P\left(\left|X_{n}-X\right|>\varepsilon\right) \\ & \geq F(t-\varepsilon)-\varepsilon, \text { for } n \geq n_{\varepsilon}^{\prime} \end{aligned} $$
Therefore, we have for $n \geq n_{\varepsilon} \vee n_{\varepsilon}^{\prime}$ $$ F(t-\varepsilon)-\varepsilon \leq \underline{\lim } F_{n}(t) \leq \overline{\lim } F_{n}(t) \leq F(t+\varepsilon)+\varepsilon\qquad\qquad\qquad\qquad\qquad\qquad(Q3) $$ If $t$ is a continuity point of $F,$ then letting $\varepsilon \rightarrow 0$ in above relation gives $F_{n}(t) \rightarrow F(t) .$ Thus $F_{n} \rightarrow_{d} F$
Q1
I understand that suppose $\omega\notin[X\leq t+\epsilon]$ then the distance between $X$ and $X_n$ is greater than $\epsilon$. Therefore set $[X\leq t+\epsilon]$ and $[|X_n-X|\geq\epsilon]$ are disjoint. But how exactly $P(X_n\leq t)$ is connected with $P(X\leq t+\epsilon)+P(|X_n-X|\geq\epsilon)$? The only sensible explanation is the distance between $X$ and $X_n$ can either be less than $\epsilon$ or greater than $\epsilon$. But this leads me to my second question:
Q2
If $|X_n(\omega)-X(\omega)|\leq\epsilon$,$X_n-\epsilon\leq X\leq\epsilon+X_n$, and since $X\leq t$, $t-\epsilon \leq X$. Then how can $A$ and $B$ happen simultaneously?
Q3
It seems the proof is done by showing the upper bound is the upper limit and the lower limit of the sequence is the same and equal to the distribution at the limit. However, it is hard for me to picture the intuitive meaning of the lower and upper bound of $F_n(t)$ as pointed in Q1 and Q2. Could someone please explain?
Q1: Note that event $\{X_n \le t\}$ can be rewritten in trivial way as
$ \{ (X_n \le t \cap |X-X_n| < \varepsilon ) \cup (X_n \le t \cap |X-X_n| \ge \varepsilon) \} $
Now, since those two events are disjoint, we get:
$$ \mathbb P(X_n \le t) = \mathbb P(X_n \le t, |X-X_n|<\varepsilon) + \mathbb P(X_n \le t, |X-X_n| \ge \varepsilon)$$
Use easy inequality of type $\mathbb P(A \cap B) \le \mathbb P(B)$ to get $\mathbb P(X_n \le t,|X-X_n| \ge \varepsilon) \le \mathbb P(|X-X_n| \ge \varepsilon)$
As for the first one, note that $\{X_n \le t, |X-X_n| < \varepsilon\} = \{X_n \le t, |X-X_n| < \varepsilon, X \le t+\varepsilon\}$ (indeed, if $X > t+\varepsilon$, then $\{X_n \le t, |X-X_n| < \varepsilon\}$ cannot happen. So again using $\mathbb P(A \cap B \cap C) \le \mathbb P(C)$ we arrive at $$ \mathbb P(X_n \le t) \le \mathbb P(X \le t+\varepsilon) + \mathbb P(|X-X_n| \ge \varepsilon)$$
Q2: Similarly write $\{X_n \le t\}$ in a way we did to arrive at $$ \mathbb P(X_n \le t) = \mathbb P(X_n \le t, |X-X_n|<\varepsilon) + \mathbb P(X_n \le t, |X-X_n| \ge \varepsilon) \ge \mathbb P(X_n \le t, |X-X_n| < \varepsilon)$$
note that (again, similarly as above) $\{X_n \le t, |X-X_n| < \varepsilon \} = \{X_n \le t, |X-X_n| < \varepsilon, X < t+\varepsilon \} \supset \{X_n \le t, |X-X_n| < \varepsilon, X \le t-\varepsilon \} = \{|X_n -X| < \varepsilon , X \le t- \varepsilon \}$
(Last equality due to fact, that if both $|X-X_n| < \varepsilon, X \le t-\varepsilon$ holds, then trivially $X_n \le t$, too)
Hence $$ \mathbb P(X_n \le t) \ge \mathbb P(X \le t-\varepsilon , |X-X_n| < \varepsilon)$$
Now, $\mathbb P(A \cap B) = \mathbb P(A) - (\mathbb P(B \cup A) - \mathbb P(B)) \ge \mathbb P(A) - (1 - \mathbb P(B)) = \mathbb P(A) - \mathbb P(B^c)$ so it follows that:
$$ \mathbb P(X_n \le t) \ge \mathbb P(X \le t - \varepsilon) - \mathbb P(|X_n-X| \ge \varepsilon) $$
Q3: OKay, from $(1),(2)$ we get that for given $t$ and $n$ sufficiently large (depending on $t$) we have inequality:
$$ F(t-\varepsilon)-\varepsilon \le F_n(t) \le F(t+\varepsilon) + \varepsilon $$
Since both lower and upper bounds hold for sufficiently large $n$, it means that $\liminf,\limsup$ of sequence $(F_n(t))_{n}$ is bounded below/above by those bounds we get from $(2),(1)$. Hence the inequality (because obviously $\liminf a_n \le \limsup a_n$) $$ F(t-\varepsilon)-\varepsilon \le \liminf F_n(t) \le \limsup F_n(t) \le F(t+\varepsilon) + \varepsilon $$
Now, if $t$ is a continuity point of $F$, then taking $\varepsilon \to 0^+$, we get that both lower and upper bound converges to $F(t)$. Hence $\liminf F_n(t) = \limsup F_n(t) = F(t)$ so that $\lim F_n(t) = F(t)$ (for those $t$ - continuity points of $F$).