Lévy–Prokhorov metric

Remarks

The function $\pi$ is a metric on $\mathcal{P}(X)$, and $\pi \le 1$ because every $\varepsilon \ge 1$ satisfies both inequalities. If $X$ is separable, then $\pi(\mu_n, \mu) \to 0$ if and only if $\mu_n$ converges weakly to $\mu$. If $X$ is separable and complete, then so is $(\mathcal{P}(X), \pi)$.

Example: let $\delta_x$ be the Dirac measure at $x$, and let $x, y \in X$. If $\varepsilon > d(x, y)$, then every Borel set $A$ containing $x$ has $y \in A^\varepsilon$, so $\delta_x(A) \le \delta_y(A^\varepsilon)$, and symmetrically. Hence $\pi(\delta_x, \delta_y) = \min(d(x, y), 1)$.

Non-example: for $\varepsilon < \min(d(x, y), 1)$ the value $\varepsilon$ is not admissible. Take $A = \set{x}$: then $y \notin A^\varepsilon$, and $1 \le 0 + \varepsilon$ fails.

On $\R$, the points $1/n$ and $0$ give $\pi(\delta_{1/n}, \delta_0) = 1/n$, while the total variation distance stays equal to $1$.