Convergence in probability is metrized by $d_0$

The displayed integrand is measurable, nonnegative and bounded by $1$, so its expectation is finite. Almost-sure replacement of either representative leaves the integrand unchanged almost surely and hence leaves its expectation unchanged by [L1]. Symmetry is immediate. The real triangle inequality and $\min (1,a+b)\le \min (1,a)+\min (1,b)$ for $a,b\ge 0$ give $\min (1,\mathrm{\mid }X-Z\mathrm{\mid })\le \min (1,\mathrm{\mid }X-Y\mathrm{\mid })+\min (1,\mathrm{\mid }Y-Z\mathrm{\mid })$ pointwise. Taking expectations gives the triangle inequality.

For $0<\epsilon \le 1$, on $\{\mathrm{\mid }{X}_n-X\mathrm{\mid }>\epsilon \}$ the integrand is at least $\epsilon$, while on its complement it is at most $\epsilon$ and everywhere it is at most $1$. Splitting the expectation over those events yields $\epsilon \mathbb{P}(\mathrm{\mid }{X}_n-X\mathrm{\mid }>\epsilon )\le d_0([X_n],[X])\le \epsilon +\mathbb{P}(\mathrm{\mid }{X}_n-X\mathrm{\mid }>\epsilon )$.

If $d_0([X],[Y])=0$, [L2] makes $\min (1,\mathrm{\mid }X-Y\mathrm{\mid })=0$ almost surely, hence $X=Y$ almost surely. Conversely, almost-sure equality makes the expectation zero. Thus $d_0$ is a metric on almost-sure classes.

If $d_0([X_n],[X])\to 0$, the first inequality gives convergence in probability for each $0<\epsilon \le 1$; for a larger threshold, bound its tail event by the tail event at threshold $1$. Conversely, convergence in probability and the second inequality give ${\mathrm{lim\hspace{0.17em}sup}}_n{d}_0([X_n],[X])\le \epsilon$ for every $0<\epsilon \le 1$, hence $d_0([X_n],[X])\to 0$.