Suppose $(S, \mathcal S, \mu)$ is a probability space. Let $f$ be a nonnegative measurable function on $S$. Let $v$ be a measure on $\mathcal B([0,\infty))$ such that $M(u) = v([0,u)) < \infty$, for all $u$ in $\mathbb R$.
We have $$M(f(x)) = \int_{[0,\infty)} I_{[0,f(x))} dv,$$
but the proof starts with
$$\int M(f(x)) \mu(dx) = \int_S \int_0^\infty I_{( u < f(x))} v(du)\mu(dx)$$
Why the integrand is $I_{( u < f(x))}$ but not $I_{[0,f(x))}$?
For clarity write $\int_{[0,\infty)} I_{[0,f(x))} dv$ as $\int_{[0,\infty)} I_{[0,f(x))}(u) dv(u)$. The inequality $0\leq u <f(x)$ holds iff $f(x) >u$. There is nothing more to it.