# Levy-Prokhorov metric question

I have a question relating to the Lévy-Prokhorov metric and its description on wikipedia.

The metric is defined for measures on $\mathbb R^d$ and is defined by $$d(\mu,\nu)=\inf\{\epsilon\colon \mu(A)\le \nu(B_\epsilon(A))+\epsilon,\ \nu(A)\le \mu(B_\epsilon(A))+\epsilon\text{ for all Borel A}\},$$ where $B_\epsilon(A)$ denotes the (open) $\epsilon$-ball around $A$.

Wikipedia tells us that the metric changes if Borel sets are replaced by closed sets. This seems false to me. I'll write out the simple argument. I'd be very grateful if someone can spot any error, or indicate that they agree.

Claim: $\inf\{\epsilon:\mu(A)\le \nu(B_\epsilon(A))+\epsilon,\ \nu(A)\le \mu(B_\epsilon(A))+\epsilon\text{ for all Borel$A$}\}$ = $\inf\{\epsilon:\mu(A)\le \nu(B_\epsilon(A))+\epsilon,\ \nu(A)\le \mu(B_\epsilon(A))+\epsilon\text{ for all$A$closed}\}$

Proof: Let the set over which the infimum is taken on the left side be $S_1$ and the set on the right side be $S_2$. Since Borel sets are closed, $S_1\subset S_2$. On the other hand, let $\epsilon\in S_2$ and let $\eta>\epsilon$. Then for any Borel set $A$, we have $B_\eta(A)\supset B_\epsilon(\bar A)$, so that $$\mu(A)\le \mu(\bar A)\le \nu(B_\epsilon(\bar A))+\epsilon\le \nu(B_\eta(A))+\epsilon,$$ with a similar statement in which $\mu$ and $\nu$ are reversed. It follows that $\eta\in S_1$. $\square$

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Should not the clause "Since Borel sets are closed" read "Since closed sets are Borel" in anthonyquas proof? – user72118 Apr 13 '13 at 15:40

You are right, in fact $B_\epsilon(A)= B_\epsilon(\bar A)$ to make the argument a little easier.