Link between expected value and probability $\{X_n\}_{n\geq 1}$ is a sequence of random variables. Want to show that 
\begin{align*}
X_n\xrightarrow[]{P}X \text{  if and only if  } \lim_{n\rightarrow\infty}E[1-\exp\{-\vert X-X_n\vert\}]=0
\end{align*}
$\Rightarrow$
Convergence in probability gives that there exist a $\epsilon$ such that $\lim_{n\rightarrow\infty}P(\vert X_n-X\vert < \epsilon)=1$ or equivalent $\lim_{n\rightarrow\infty}P(\vert X_n-X\vert > \epsilon)=0$.
Hence 
\begin{align*}
\lim_{n\rightarrow\infty}E[1-\exp\{-\vert X-X_n\vert\}]&= 1-\lim_{n\rightarrow\infty}E[\exp\{-\vert X-X_n\vert\}] \\
&= 1-\lim_{n\rightarrow\infty}\int^\infty_{-\infty} \exp\{-\vert X-X_n\vert\}P(\vert X-X_n\vert) d\vert X-X_n\vert
\end{align*}
Is this correct or do I have to proceed in another way? 
Since exp is continuous I guess somehow it should be simple. I can also see the final argument and way this is try but the relation between expected values $E$ and probability $P$  gives me problems. 
 A: $\{X_n\}_{n\geq 1}$ is a sequence of random variables. We should show that 
\begin{align*}
X_n\xrightarrow[]{P}X \text{  if and only if  } \lim_{n\rightarrow\infty}E[\exp\{-\vert X-X_n\vert\}]=1
\end{align*}
$\Rightarrow$:
$X_n\xrightarrow[]{P}X$ means that there exist a $\epsilon$ such that $\lim_{n\rightarrow\infty}P(\vert X_n-X\vert > \epsilon)=0$ or $\lim_{n\rightarrow\infty}P(\vert X_n-X\vert \leq \epsilon)=1$
Hence 
\begin{align*}
&\lim_{n\rightarrow\infty}E[\exp\{-\vert X-X_n\vert\}]\\ 
=& \lim_{n\rightarrow\infty}\left(E[\exp\{-\vert X-X_n\vert\}\textbf{1}\{\vert X-X_n\vert>\varepsilon\}] + E[\exp\{-\vert X-X_n\vert\}\textbf{1}\{\vert X-X_n\vert\leq \varepsilon\}] \right)\\
\geq& \exp(-\varepsilon)\lim_{n\rightarrow\infty}P(\vert X-X_n\vert\leq\varepsilon) = \exp(-\varepsilon)
\end{align*}
Now since $\varepsilon$ can be arbitrarily small, we got $\lim_{n\rightarrow\infty}E[\exp\{-\vert X-X_n\vert\}]\geq 1$. But $E[\exp\{-\vert X-X_n\vert\}]\leq 1$ as $\vert X-X_n\vert\geq 0$, thus $\lim_{n\rightarrow\infty}E[\exp\{-\vert X-X_n\vert\}]= 1$.
$\Leftarrow$: 
For any $\varepsilon>0$ we have
\begin{align*}
&E[\exp\{-\vert X-X_n\vert\}]\\ 
=& \left(E[\exp\{-\vert X-X_n\vert\}\textbf{1}\{\vert X-X_n\vert>\varepsilon\}] + E[\exp\{-\vert X-X_n\vert\}\textbf{1}\{\vert X-X_n\vert\leq \varepsilon\}] \right)\\
\leq& \exp(-\varepsilon)P(\vert X-X_n\vert > \varepsilon)+ P(\vert X-X_n\vert\leq\varepsilon) \\
=& \exp(-\varepsilon)[1-P(\vert X-X_n\vert \leq \varepsilon)]+ P(\vert X-X_n\vert\leq\varepsilon) \\
=& \exp(-\varepsilon)+ [1-\exp(-\varepsilon)]P(\vert X-X_n\vert\leq\varepsilon)
\end{align*}
Thus 
$$[1-\exp(-\varepsilon)]P(\vert X-X_n\vert\leq\varepsilon)\geq E[\exp\{-\vert X-X_n\vert\}] - \exp(-\varepsilon).$$
Taking $\liminf_{n\rightarrow \infty}$ on both sides gives us
$$[1-\exp(-\varepsilon)]\liminf_{n\rightarrow \infty}P(\vert X-X_n\vert\leq\varepsilon)\geq 1 - \exp(-\varepsilon),$$
which means $\liminf_{n\rightarrow \infty}P(\vert X-X_n\vert\leq\varepsilon)\geq 1$. But on the other hand $\limsup_{n\rightarrow \infty}P(\vert X-X_n\vert\leq\varepsilon)\leq 1$, thus we get
$$\lim_{n\rightarrow \infty}P(\vert X-X_n\vert\leq\varepsilon)= 1,\ {\rm or}\ X_n\xrightarrow[]{P}X.$$
