Lévy and Lévy Prokhorov metric

Question

I was reading "On choosing and bounding probability metrics" (https://www.math.hmc.edu/~su/papers.dir/metrics.pdf) and encountered the Lévy distance,defined by

$ d_L(\mu,\nu):=\inf\lbrace{}\varepsilon>0\,:\, \mu((-\infty,x-\varepsilon])\leq\nu((-\infty,x])\leq\mu((-\infty,x+\varepsilon]), \forall x\in\mathbb{R}\rbrace $

as a special case (on $\mathbb{R}$) of the Lévy-Prokhorov distance, defined by

$ d_P(\mu,\nu):=\inf\lbrace\varepsilon>0\,:\,\mu(B)\leq\nu(B^{\varepsilon})+\varepsilon,\forall \text{ measurable }B\subset\mathcal{X}\rbrace $

In the transcript it said that $d_P$ metrized weak convergence. It is clear that $d_L\leq{}d_P.$

Additionally it mentioned that $d_L$ metrizes weak convergence too, which I don't quite understand since it is only bounded from above by Lévy-Prokhorov. For it to metrize weak convergence it should have to be equivalent to $d_P$, which I don't believe it is. Or maybe I'm mistaken. Can anyone explain?

Thanks a lot.

Answer 1

I think you may be confusing «topologically equivalent» and «uniformly equivalent», which is an easy thing to do since both notions coincide for norms.

Convergence in the Lévy metric is equivalent to convergence of the cumulative distribution functions at all continuity points of the limit, which is the usual definition of convergence in distribution in the real line, i.e. weak convergence.