Hausdorff distance is a lower (or upper bound) for what probability metric?

Question

In a metric space $X=(X, d)$, given a probability measure $\mu$ and two subsets $A$ and $B$ of positive measure, it's not hard to prove that

$$ d(A, B) \le W(\mu|_A, \mu|_B), $$ where

• $d(A, B):= \inf_{a \in A,\;b \in B}d(x,y)$ is the distance between $A$ and $B$.
• $\mu|_A$ defined by $\mu|_A(C):= \mu(A\cap C)/\mu(A)$ is the conditional measure induced by $A$
• $W(\mu|_A,\mu|_B)$ is the Wasserstein distance between $\mu|_A$ and $\mu|_B$.

Now, consider the Hausdorff distance between $A$ and $B$, defined by $$ d_H(A,B) := \max\left(\sup_{b \in B}\inf_{a \in A}d(a,b),\sup_{a \in A}\inf_{b \in B}d(a,b)\right) = \inf\{\epsilon \ge 0 | A \subseteq B^\epsilon \text{ and } B\subseteq A^\epsilon\}, $$ where $A^\epsilon := \{x \in X | \exists a \in A\text{ with }d(a,x)\le \epsilon\} = \cup_{a \in A}\operatorname{Ball}_\epsilon(a)$ is the $\epsilon$-blowup of $A$.

Question

Is $d_H(A,B)$ a lower bound (or upper bound) for some probability metric ? Which one ? Between what and what ?

Edit

A user mentioned "$L^\infty$ Wasserstein distance" in the comments. This comment seems to have gone unnoticed until now. Indeed, Exercise 36 of OTAM asks to prove that $d_H(supp(\mu),supp(\nu)) \le W_\infty(\mu,\nu)$ for every pair of measures $\mu$ and $\nu$ on a Polish space. This solves one half of my question. The other half has been solved (in the negative) by another user.

Answer 1

A general note is that the answer depends heavily on the properties of $\mu$.

First a note that in general $d_H(A,B) \not \le C \cdot W_p(\mu|_A,\mu|_B)$ for $p\in[1,\infty)$ and some $C>0$. Though it's true for the case $p = \infty$. Here the example: Let $\mu_\lambda = (1-\lambda) \delta_x + \lambda \delta_y$ for distinct $x,y \in X$. Choose $A=\{x\}$ and $B=\{x,y\}$. Then $\mu_\lambda|_A$ is $\delta_x$ and $\mu_\lambda|_B$ is $\mu_\lambda$. Thus $d_H(A,B) = d(x,y)$ and $W_p(\mu|_A,\mu|_B) = \lambda^{\frac{1}{p}}d(x,y)$.

For the case $p=\infty$, i.e. $W_\infty(\mu,\nu) = \inf \|d\|_{L^\infty(\pi)} $, it is true.

But there is no lower bound for the Hausdorff distance $d_H(\cdot,\cdot)$ w.r.t. to any Wasserstein distance $W_p$ if there is at least one non-isolated point (as $W_p \le W_\infty$ it suffices to get counterexamples for $p=\infty$).

Choose $x_n \to x$ and $\mu = \frac{1}{2}\delta_x + \sum_{n\in \mathbb{B}} \lambda_n \delta_{x_n}$ for some $\lambda_n \ge 0$ with $\sum_{n\in \mathbb{B}} \lambda_n = \frac{1}{2}$. Assume, in addition, $\lambda_1 > 2 \lambda_n$ for $n>1$. Now Choose $A = \{x, x_1\}$ and $B_n = \{x_n,x_1\}$. Then $d_H(A,B_n) = d(x,x_n) \to 0$. However, $W_\infty(\mu|_A, \mu|_{B_n}) \ge \inf \{d(x,x_1),d(x_n,x_1)\}$ because $\mu|_A(\{x\}) > \frac{1}{2} > \mu|_{B_n}(\{x_n\})$.