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.
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.
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 $F(t)=\frac{t}{1+t}$