DJC's answer is of course to the point. I'm adding a short sketch of a proof of the inequality $\|\mu \ast \nu\| \leq \|\mu\| \,\|\nu\|$ since you asked about it in a comment. In that comment you mentioned $\sigma$-finiteness. Note that bounded variation implies that all measures involved are in fact finite.
One of the most convenient ways of writing the total variation norms is as
$\|\mu\| = \sup_{|f| \leq 1}{\;\left|\int f\,d\mu\right|}$
where it doesn't matter much what kinds of measurable functions $f$ you allow in the supremum (continuous; simple; Borel; smooth; compactly supported or not). I'll take Borel here.
Given this, the inequality is then clear: By Fubini we have for every Borel function $f$ with $|f|\leq 1$ that the function $y \longmapsto \left|\int f(x+y)\,d\mu(x)\right|$ is Borel and bounded by $\|\mu\|$ hence the fact that $\|\nu\| = \| \,|\nu|\,\|$ gives $\left|\int f \, d(\mu \ast \nu) \right| = \left|\iint f(x+y)\,d\mu(x)\,d\nu(y)\right| \leq \int\left|\int f(x+y)\,d\mu(x)\right|\,d|\nu|(y) \leq \|\mu\| \,\|\nu\|.$ The norm of $\|\mu \ast \nu\|$ is by definition the supremum over the left hand side over $|f| \leq 1$.