Consider the probability space $(\Omega, {\cal B}, \lambda)$ where $\Omega=(0,1)$, ${\cal B}$ is the Borel sets, and $\lambda$ is Lebesgue measure.

For random variables $W,Z$ on this space, we define the Ky Fan metric by

$$\alpha(W,Z) = \inf \lbrace \epsilon > 0: \lambda(|W-Z| \geq \epsilon) \leq \epsilon\rbrace.$$

Convergence in this metric coincides with convergence in probability.

Fix the random variable $X(\omega)=\omega$, so the law of $X$ is Lebesgue measure, that is, ${\cal L}(X)=\lambda$.

Question:For any probability measure $\mu$ on $\mathbb R$, does there exist a random variable $Y$ on $(\Omega, {\cal B}, \lambda)$ with law $\mu$ so that $\alpha(X,Y) = \inf \lbrace \alpha(X,Z) : {\cal L}(Z) = \mu\rbrace$ ?

*Notes:*

By Lemma 3.2 of Cortissoz, the infimum above is $d_P(\lambda,\mu)$: the Lévy-Prohorov distance between the two laws.

The infimum is achieved if we allowed to choose both random variables. That is, there exist $X_1$ and $Y_1$ on $(\Omega, {\cal B}, \lambda)$ with ${\cal L}(X_1) = \lambda$, ${\cal L}(Y_1) = \mu$, and $\alpha(X_1,Y_1) = d_P(\lambda,\mu)$. But in my problem, I want to fix the random variable $X$.

*Why the result may be true:*the space $L^0(\Omega, {\cal B}, \lambda)$ is huge. There are lots of random variables with law $\mu$. I can't think of any obstruction to finding such a random variable.*Why the result may be false:*the space $L^0(\Omega, {\cal B}, \lambda)$ is huge. A compactness argument seems hopeless to me. I can't think of any construction for finding such a random variable.