convergence in probability induced by a metric Let $M$ be the set of all random variables from a fixed probability space to $\mathbb R$ with its borel sets.
Let's define a metric on $M$ by $d(X,Y)=E(\frac{|X-Y|}{1+|X-Y|})$
I want to prove that $d$ it's in fact a metric. The only difficult part it's to prove that $d(X,Y)=0$ if and only if $X=Y$ almost surely (a.e) and also that $X_n \to X$ in probability if and only iff $d(X_n,X)\to 0$  Thus this metric induces the probability convergence.
 A: This is can be seen as part of a general result:

Theorem: Suppose $(\Omega,\mathscr{F},\mu)$ is a finite measure space and $(S,\rho)$ a separable metric space. Assume  $\{f_n,\, f:n\in\mathbb{N}\}\subset S^\Omega$ is a sequence of measurable functions with respect to the outer measure $\mu^*$. Let $F:[0,\infty)\rightarrow[0,\infty)$ a bounded  continuous nondecreasing function with $F(t)=0$ iff $t=0$. Then,
$\rho(f_n,f)\in\mathscr{M}_{\mathbb{R}}(\mu^*)$ for all $n\in\mathbb{N}$, and
$f_n$ converges in measure to $f$ if and only if $\lim_n\int F(\rho(f_n,f))\,d\mu=0$.

Here is a short proof:
Let $\varepsilon>0$ arbitrary and $\|F\|_\infty:=M$. Notice that
$$
F(\varepsilon)\mathbb{1}_{\{\rho(f_n,f)>\varepsilon\}} \leq F(\rho(f_n,f))\leq F(\varepsilon) + M\mathbb{1}_{\{\rho(f_n,f)>\varepsilon\}}
$$
and denote by $D(f_n,f)=\int F(\rho(f_n,f))\,d\mu$. Then
$$
\begin{align}
F(\varepsilon)\mu(\{\rho(f_n,f)>\varepsilon\})\leq  D(f_n,f) \leq F(\varepsilon)\mu(\Omega)+ M \mu(\{\rho(f_n,f)>\varepsilon\})\tag{1}\label{mean-metric}
\end{align}
$$
Necessity follows by letting $n\nearrow\infty$ and then $\varepsilon\searrow0$. Sufficiency follows by letting $n\nearrow\infty$.

Corollary: Suppose $(\Omega,\mathscr{F},\mu)$ is a finite measure space. Assume  $F:[0,\infty)\rightarrow[0,\infty)$ is a bounded nondecreasing subadditive function such that $F(x)=0$ iff $x=0$, then
$$
D_F(f,g)=\int_\Omega F(|f-g|)\,d\mu
$$
is a complete metric on $L_0$ which results in the same topology as in $(L_0,d_0)$.
Here is a short proof:
It is easy to check that $D_F$ is a metric on $L_0$. By Theorem above it is enough to show that $(L_0,D_F)$ is complete.  If $\{f_n:n\in\mathbb{N}\}$ is a Cauchy sequence w.r.t. $D_F$,  then
$$
\lim_{M\rightarrow\infty} \sup_{n,m\geq M} \mu(|f_n-f_m|>\varepsilon)\leq \tfrac{1}{F(\varepsilon)}\lim_{M\rightarrow\infty}\sup_{n,m\geq  M}D_F(f_n,f_m)=0
$$
by $\eqref{mean-metric}$. Hence there are integers  $n_k< n_{k+1}$  such that $\sup_{n,m\geq n_k}\mu(|f_n-f_m|>2^{-k})< 2^{-k}$ and so, $\sum_k\mu(|f_{n_{k+1}}-f_{n_k}|>2^{-k})<\infty$.  By the Borel--Cantelli lemma, the set
$$
A=\{|f_{n_{k+1}}-f_{n_k}|>2^{-k},\,i.o\}
$$
has $\mu$--measure zero. It follows that    $\{f_{n_k}\}$ is $\mu$--a.s. a Cauchy sequence in $\mathbb{R}$; thus,   $f_{n_k}$ converges $\mu$--a.s to some $f\in\mathscr{M}_{\mathbb{R}}(\mu^*)$. By dominated convergence,  $\lim_kD_F(f_{n_k},f)\rightarrow0$.  Therefore $\lim_nD_F(f_n,f)=0$.

Examples

*

*In $\mathbb{R}$, take $\rho(x,y)=|x-y|$ and $F(t)=\min(t,1)$

*In $\mathbb{R}$, take $\rho(x,y)=|x-y|$, and $F(t)=\frac{t}{1+t}$
A: If $d(X,Y)=0$, then $Z=\frac{|X-Y|}{1+|X-Y|}$ is a nonnegative random variable whose expectation is zero, so it must be zero a.e., so $|X-Y|=0$ a.e., so $X=Y$ a.e.
Suppose that $(X_n) \to X$ in probability. Let $\varepsilon >0$. Then for large
enough $n$ we have $P(|X_n-X| > \varepsilon ) \leq \varepsilon$. For those
$n$ we have
$$
\begin{array}{lcl}
d(X_n,X) &=& {\bf E}(\frac{|X_n-X|}{1+|X_n-X|}{\bf 1}_{|X_n-X| \leq \varepsilon})
+{\bf E}(\frac{|X_n-X|}{1+|X_n-X|}{\bf 1}_{|X_n-X| > \varepsilon}) \\
&\leq& {\bf E}(|X_n-X|{\bf 1}_{|X_n-X| \leq \varepsilon})
+{\bf E}({\bf 1}_{|X_n-X| > \varepsilon}) \\
&\leq& {\bf E}(\varepsilon{\bf 1}_{|X_n-X| \leq \varepsilon})
+P(|X_n-X| > \varepsilon)=2\varepsilon
\end{array}
$$
So $d(X_n,X) \to 0$.
Conversely, suppose that $d(X_n,X) \to 0$. Let $\varepsilon >0$. Then for large
enough $n$ we have $d(X_n,X) \leq \frac{\varepsilon^2}{1+\varepsilon}$. For those
$n$ we have
$$
\frac{\varepsilon^2}{1+\varepsilon} \geq {\bf E}(\frac{|X_n-X|}{1+|X_n-X|}{\bf 1}_{|X_n-X| > \varepsilon})
\geq {\bf E}(\frac{\varepsilon}{1+\varepsilon}{\bf 1}_{|X_n-X| > \varepsilon})=
\frac{\varepsilon}{1+\varepsilon} P(|X_n-X|  > \varepsilon )
$$
So $P(|X_n-X| > \varepsilon) \leq \varepsilon$. This shows that $(X_n) \to X$ in probability.
A: Here is another proof that $d(X_n,X)\to 0$ implies $X_n\to X$ in probability. By subtracting $X$, we can assume $X_n \ge 0$, $d(X_n,0)\to 0$ and prove that $X_n\to 0$ in probability. Using the "layer cake" representation of the integral, the fact that $\frac{X_n}{1+X_n} \le 1$ a.s., and the change of variable $x=\frac{\lambda}{1-\lambda}$ our assumption $d(X_n,0) \to 0$ can be rewritten as
\begin{align*}
0 = \lim_{n\to\infty}E\bigg[\frac{X_n}{1+X_n}\bigg] &= \int_0^1 P\bigg(\frac{X_n}{1+X_n}>\lambda\bigg)\,d\lambda\\
&= \int_0^\infty P(X_n>x)(1+x)^{-2}\,dx.
\end{align*}
(By the dominated convergence theorem, it is now easy to see that $X_n\to 0$ in probability implies that $d(X_n,0)\to 0$.)
Suppose to the contrary that $x_0 > 0$ and $\limsup_{n\to\infty}P(X_n > x_0) = c > 0$. By the definition of the limsup of a numerical sequence, there is a subsequence $\{n_k\}_{k=1}^\infty$ so that $P(X_{n_k} > x_0) \to c$. Because the cdf $F_k(x) \equiv P(X_{n_k} > x)$ is nonincreasing and nonnegative, we have $$F(x) \equiv \liminf_{k\to\infty}F_k(x) \ge \liminf_{k\to\infty}F_k(x_0) \ge c\,\mathbf 1_{[0,x_0]}(x).^{\text{[1]}}$$
By Fatou's lemma, and our assumption that $d(X_n,0)\to 0$,
\begin{align*}
0 = \liminf_{k\to\infty}\int_0^\infty F_k(x)(1+x)^{-2}\,dx &\ge \int_0^\infty F(x)(1+x)^{-2}\,dx \\
&\ge \int_0^{x_0}c\,(1+x)^{-2}\,dx\\
&= c\,\frac{x_0}{1+x_0} > 0,
\end{align*}
a contradiction.

[1]: For a Borel set $A\subseteq\mathbf R$, $\mathbf 1_A(x) = 1$ if $x\in A$ and $\mathbf 1_A(x) = 0$ if $x\in \mathbf R\smallsetminus A$. $\mathbf 1_A$ is called the indicator function of $A$.
