Convergence in Probability implies Weak Convergence Proof Question I'm trying to follow a proof for showing $\displaystyle \lim_{n\rightarrow \infty} P[|X_n-X|>\epsilon] = 0 \Rightarrow X_n \rightarrow_p X$
The first step of the proof says:
$P[X \leq x-\epsilon] - P[|X_n - X| \geq \epsilon] \leq P[X_n \leq x] \leq P[X\leq x+\epsilon] + P[|X_n - x| \geq \epsilon]$
The first inequality is true because any $\omega$ such that $X(\omega) \leq x-\epsilon$ and $X_n - X \leq \epsilon$, will certainly satisfy $X_n(\omega) \leq x$. 
But I cannot think of similar reasoning to see why the latter inequality is true.
 A: I believe there is a significant typo in the final term making the second inequality impossible to prove.

Claim: $P[|X_n - x| \geq \epsilon]$ should be $P[|X_n - X| \geq \epsilon]$.

Proof of claim:
Let $\Omega$ have only one element: $\omega$. In that case,
$\mu(\omega) =1$, and $X_n = X_n(\omega)$ is just a sequence of real numbers. Suppose it is defined by:
$$
X_n = 
\begin{cases}
n & n \leq 10 \\
10 & n > 10
\end{cases}
$$
Also, let $x = 5$, $n = 10$, and $\epsilon = 1$. Then 
$$P[X_5 \leq 5] = 1$$
because in fact $P[X_5 = 5] = 1$. Therefore,
$$P[|X_5 - 5| \geq 1] = 0.$$ 
But we know that since the sequence $X_n \to 10$, we have the fact that $X = 10$. And hence, 
$$P[X \leq 5+1] =0.$$
This contradicts your second inequality because
$$ 1> 0 +0.$$
Proof with typo fixed:
Let $A = \{\omega : X_n(\omega) \leq x\}$, $B = \{\omega :X(\omega) \leq x+\epsilon\}$, and $C = \{\omega : |X_n(\omega) - X| \geq \epsilon\}$.
What we want to show is $P(A) \leq P(B) + P(C)$. It suffices to show $A \subseteq B \cup C$. So let $\omega \in A$, and assume that $\omega \notin C$. We have these two inequalities:


*

*$X_n(\omega) \leq x$ 

*$X(\omega) - X_n(\omega) < \epsilon$


By adding them we see that they imply that $X(\omega) \leq x + \epsilon$. Therefore $\omega \in B$. We have just shown that $A \subseteq B \cup C$, completing our proof of the second inequality.
