$(X_n)_{n \geq 1}$ is said Cauchy in probability if $\forall \hspace{0.1cm} \delta > 0, \forall \hspace{0.1cm} \epsilon > 0 \hspace{0.1cm} \exists \hspace{0.1cm} n(\epsilon) \in \mathbb{N} : P(\lvert X_n - X_m \rvert > \epsilon) < \delta$
I'd like to prove the equivalence with $X_n$ being Cauchy for the distance $d(X,Y) = \mathbb{E}[\lvert X - Y \rvert \wedge 1]$
My effort : If $X_n$ is Cauchy for the distance then $\forall \hspace{0.1cm} \epsilon,m,n > 0$ we have that $\exists \hspace{0.1cm} n_0 : d(X_n,X_m) \leq \epsilon\hspace{0.1cm} \forall\hspace{0.1cm}m,n > n_0$. Si taking a ball $B$ centered in $n_{0}$ we have that $\forall \hspace{0.1cm} \delta > 0, \forall \hspace{0.1cm} \epsilon > 0 P(X_n \not\in B) = P(\lvert X_n - X_m \rvert > \epsilon)= 0 < \delta$.
But I'm stuck with the other implication, although the statement it seems true, since I already know that in general $d(X_n,X) \longmapsto 0 \iff X_n \overset{P}{\longmapsto} X$ so it makes sense to have the property on the cauchy sequences, but I'd like to see it explicitly.