[Math] Total Variation Distance Triangle Inequality

Question

I have the following definition for the total variation distance between two probability measures on a metric space $X$. $d(\mu,\nu) = \inf\{\mathbb{P}(V\neq W) : \text{$V:\Omega\rightarrow X$ has distribution $\mu$, $W:\Omega\rightarrow X$ has distribution $\nu$}\}.$ How do I check that the triangle inequality holds for this function?

Answer 1

The total variation distance is a special case of minimal metric. So, triangle inequality follows from this fact.

For example, in the paper "Metric distances in spaces of random variables and their distributions", V. M. Zolotarev, section 1.4, there is a proof.