In optimal transport, I have encountered an integral in which the integrand is the quantile function. Could you confirm if my below attempt is fine?
Let $\mu$ be a Borel probability measure on $\mathbb R$ and $F: \mathbb R \to [0, 1]$ its cumulative distribution function. Then $F$ is right-continuous and non-decreasing. So $F$ is Borel measurable. We define its quantile function $F^{-1}: [0, 1] \to \mathbb R \cup \{\pm \infty\}$ by $$ F^{-1} (t) := \inf \{x \in \mathbb R \mid F(x) \ge t\} \quad \forall t \in [0, 1]. $$ Then $F^{-1}$ is measurable (w.r.t. the Borel $\sigma$-algebra generated by the order topology of extended real line). We adopt the convention that $\inf \emptyset = +\infty$.
My attempt: Fix $x \in \mathbb R$. Let $$ A_x := \{t \in [0, 1] \mid F^{-1} (t) < x\} \quad \text{and} \quad B_x := \{t \in [0, 1] \mid F^{-1} (t) > x\}. $$
We have the equivalence $F^{-1} (t) < x \iff t < F(x)$. Then $$ A_x = \{ t \in [0, 1] \mid t < F(x)\} \quad \text{and} \quad B_x = \{t \in [0, 1] \mid t > F(x)\}. $$
Clearly, $A_x$ is either empty (in case $F(x) = 0$) or $A_x = [0, F(x))$. As such, $A_x$ is Borel. Similarly, $B$ is Borel. It follows that $$ \{t \in [0, 1] \mid F^{-1} (t) = +\infty\} = \bigcap_{n \in \mathbb N} B_n $$ is Borel. Similarly, $$ \{t \in [0, 1] \mid F^{-1} (t) = -\infty\} = \bigcap_{n \in \mathbb N} A_n $$ is Borel. This completes the proof.
