Converge in probability and Expectation. Let $(\Omega, F , P)$ be probability space.
Let $X_i$ be sequence of random variable and $X$ be Random variable.

Claim. $X_n \to X$ in probability if and only if $E(\frac{|X_n-X|}{1+|X_n-X|})$ $\to 0$

I can not catch any hint.
Can I get some hints?
 A: We consider the probability space $(\Omega,\mathcal{F},P)$.
Firstly: I suppose you have to assume the existence of a mean of the random variables. So let that be the case.
$X_n \to X$ in probability if for all $\epsilon>0$,
$\lim_{n\to\infty} P(|X_n-X| \geq \epsilon) = 0$.
LH to RH:
Denote with $Y_n$ the difference $Y_n=X_n-X$. $Y_n$ is by assumption integrable. We then want to show that if $Y_n \to 0$ in probability, then 
$\lim_{n\to\infty}\mathbb{E}\left[\frac{|Y_n|}{1+|Y_n|}\right] = 0.$
As $Y_n$ converges to $0$ in probability, we know that for all $\epsilon>0$,
$\lim_{n\to\infty} P(|Y_n|\geq \epsilon) = 0$.
Now let $\epsilon>0$ be given. Then
$
0
\leq
\mathbb{E}\left[\frac{|Y_n|}{1+|Y_n|}\right]
=
\int_\Omega \frac{|Y_n|}{1+|Y_n|} \, \mathrm{d}P
=
\int_{(|Y_n|\geq\epsilon)} \frac{|Y_n|}{1+|Y_n|} \, \mathrm{d}P
+
\int_{(|Y_n|<\epsilon)} \frac{|Y_n|}{1+|Y_n|} \, \mathrm{d}P
$
$\leq$
$
\int_{(|Y_n|\geq\epsilon)} \frac{|Y_n|}{1+|Y_n|} \, \mathrm{d}P
+
P(|Y_n|< \epsilon) \frac{\epsilon}{1+\epsilon}
$
$\leq$
$
P(|Y_n|\geq\epsilon)
+
\frac{\epsilon}{1+\epsilon}
\to 0
$, as $n\to\infty$,
where the last convergence is due to the following: The first term converges to $0$ as $Y_n$ converges in probability to $0$. Furthermore, as $\epsilon>0$ was given, the last term can handled by choosing $\epsilon$ sufficiently small.
RH to LH:
See other answers.
A: Let $U_n=\frac{|X_n-X|}{1+|X_n-X|}$. Note that $U_n\in[0,1)$. First of all by Markov inequality:
$$
\Pr(U_n>\delta)\leq \frac{E(U_n)}{\delta}.
$$
If $E(U_n)\to 0$ as $n\to\infty$, $U_n$ converges to zero in probability. 
Secondly, if $U_n$ converges to zero in probability, for each $\delta$, $\Pr(U_n>\delta)\to 0$. We have:
$$
E(U_n)=E(U_n\mathbf 1_{U_n>\delta})+E(U_n\mathbf 1_{U_n\leq\delta})< \Pr(U_n>\delta)+\delta
$$
This inequality holds for each $\delta$. Therefore for any $\delta'$, there is $N_0$, such that for $n>N_0$, $E(U_n)\leq \delta'$. This means that $E(U_n)\to 0$. 
So we have proved that $E(U_n)\to 0$ if and only if $U_n$ converges to zero in probability. It is easy to check that $U_n$ converging to zero in probability is equivalent to $|X_n-X|$ converging to zero in probability.
