I have the following proposition:
Let $ (\Omega, \mathcal{F}, \mathbb{P} ) $ be a probability space and $(X,d)$ a metric space. Let $ \{ f_h \}_{h >0} $ be s.t. $f_h : \Omega \to X$ and $f: \Omega \to X$. Then $$f_h \overset{\mathbb{P}}{\to} f \Leftrightarrow \int_{\Omega} [1 \wedge d(f_h, f)] d \mathbb{P} \overset{h \to 0}{\longrightarrow} 0$$
In the proof I have this equality
$$\int_{\Omega} [1 \wedge d(f_h, f)] d \mathbb{P} = \int_0^1 \mathbb{P}(d(f_h,f)>s)ds $$
Why is it true?
For any non-negative measurable function $X$ we have $\int_{\Omega} XdP=\int_0^{\infty} P\{X>t\} dt$ $\,\,$ (1). In this case the integrand on RHS is $0$ for $t>1$. In fact $P\{1\wedge d(f_h,f) >s\}=P\{ d(f_h,f) >s\}$ or $0$ according as $s <1$ or $s >1$.
Proof of (1): by Fubini's Theorem $\int_0^{\infty} P\{X>t\} dt=\int_0^{\infty} \int_{\Omega} I_{\{t<X\}} dP= \int_{\Omega} \int_0^{\infty}I_{\{t<X\}} dP=\int_{\Omega} XdP$.