Q:

Equivalence definition for convergence in probability

Let $(X_n)$ be a sequnce of random variables in the probability space $(\Omega, \mathscr{F}, \mathbb{P}).$ Then $X_n \rightarrow X$ in probability if and only if $\mathbb{E} \min(|X_n - X|, 1) \rightarrow 0$ as $n \uparrow \infty.$
($\rightarrow)$ Let $Y_n = \min(|X_n - X|,1).$ Then $|Y_n| \leq 1$. So by Dominated convergence thm, $$\lim \mathbb{E}Y_n = \mathbb{E} \lim Y_n.$$ Since $X_n \rightarrow X$ in probability, almost every $x$,
$$Y_n(x) = |X_n-X|(x)$$ for $n$ large. So it should be like $$Y(x) = \lim |X_n - X|(x)$$ almost every $x$. I feel that $\lim |X_n - X|$ should be $0$, but I cannot find a good reason to support this (I know just $X_n \rightarrow X$ in probability, but not $X_n \rightarrow X$ in usual sequence sence.)
Any help for this direction ?

A:

Note that convergence in probiability does not give you that for almost every $x$ that $|X_n - X|$ is small almost surely. It just guarantees that the set where $|X_n - X|$ is large has small (not zero!) probiability.
Recall that $X_n - X$ in probiability means 
$$ \forall \epsilon > 0: \def\P{\mathbf P}\def\E{\mathbf E}\P\bigl(|X_n - X| > \epsilon\bigr) \to 0 $$
Suppose this is true and we want to prove $\E(Y_n) \to 0$. Let $\epsilon \in(0, 1)$. Choose $N \in \mathbf N$ such that 
$$ \P\bigl(|X_n - X| > \epsilon\bigr)  < \epsilon, \qquad n \ge N.$$
We have
\begin{align*}
  Y_n &= \min(1, |X_n - X|)\\
      &\le \epsilon \chi_{\{|X_n - X|\le \epsilon\}} + 1\chi_{\{|X_n - X| > \epsilon\}}
\end{align*}
Taking the expected value, we have
\begin{align*}
  \E Y_n &\le \epsilon \P(|X_n - X| \le \epsilon) + \P(|X_n - X| > \epsilon)\\
       &\le \epsilon + \epsilon\\
       &= 2\epsilon
\end{align*}
Hence $\E Y_n \to 0$.

For the other direction suppose $\E Y_n \to 0$, let $\epsilon \in (0,1)$. We have by Markov
\begin{align*}
  \P(|X_n - X| > \epsilon) &= \P(\min\{1,|X_n - X|\}> \epsilon)\\
         &= \P(Y_n > \epsilon)\\
         &\le \frac 1\epsilon \E Y_n\\
         &\to 0.
\end{align*}

