I'm solving the following homework problem. Given a sequence of random variables $(X_n)_{n \in \mathbb{N}}$. Show that $X_n \to X$ in probability $\iff$ $\lim_{n \to \infty} \mathbb{E}[\min\{|X_n - X|, 1\}] = 0.$
Let $Y_n := \min\{|X_n - X|, 1\} = |X_n - X| \cdot \mathbb{1}_{\{|X_n - X| < 1\}} + \mathbb{1}_{\{|X_n - X| > 1\}}$. If $X_n \xrightarrow{\mathbb{P}} X$ then we have $\lim_{n \to \infty}\mathbb{P}[|X_n - X| > 1] = 0$ and $$\lim_{n \to \infty}\mathbb{E}[Y_n] = \lim_{n \to \infty} \mathbb{E}[|X_n - X| \cdot \mathbb{1}_{\{|X_n - X| < 1\}}] + \mathbb{E}[\mathbb{1}_{\{|X_n - X| > 1\}}]$$ The second term goes to zero as $n \to \infty$, so $\lim_{n \to \infty}\mathbb{E}[Y_n] = \lim_{n \to \infty} |X_n - X| \mathbb{P}[|X_n - X| < 1]$ (not sure if that's correct, the probability term should go to $1$ but then $|X_n - X| \xrightarrow{\mathbb{P}} 0$ doesn't imply that the limit will be zero). I got stuck here and couldn't figure out how to prove the other direction as well. Any hint/help would be great!
This is one way to look at the problem.
Given $\varepsilon>0$ and (finite) random variables $X$ and $Y$ $$ \varepsilon\mathbb{1}_{\{|X-Y|>\epsilon\}}\leq |X-Y|\wedge 1 \leq \varepsilon +\mathbb{1}_{\{|X-Y|>\varepsilon\}} $$
Take expectations to get
$$ \varepsilon\mathbb{P}[|X-Y|>\varepsilon]\leq \mathbb{E}[|X-Y|\wedge1]\leq \varepsilon +\mathbb{P}[|X-Y|>\varepsilon]\tag{1}\label{one} $$
From this, you can see that $X_n\xrightarrow{n\rightarrow\infty}X$ in probability iff $\lim_n\mathbb{E}[|X_n-X|\wedge1]=0$.
Notes:
The metric $d(x,y)=1\wedge|x-y|$ (on $\mathbb{R}$) may be substituted by any other bounded metric $\rho(x,y)$ in $\mathbb{R}$ (equivalent to the standard metric $\ell(x,)=|x-y|$). For instance $\rho(x,y)=\frac{|x-y|}{1+|x-y|}$
A similar inequality to $\eqref{one}$ with $\rho$ in place of $d$ can be used to prove the metrizability of the topology of convergence in probability.