The norm of Borel measures I was looking on at the Convolution page on wikipedia and saw that it stated that we can define the convolution of two Borel measures of bounded variation on $\mathbb{R}^d$, $\mu$ and $\nu$, to be
$$\int_{\mathbb{R}^d} f(x) d(\mu\ast \nu)(x) = \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} f(x+y) d\mu(x) d\nu(y),$$
and that we have the result
$$||\mu \ast \nu || \leq ||\mu|| ||\nu||.$$
I'm not sure what the second statement means though. Is there a natural norm for the space of Borel measures with bounded variation? 
 A: 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 $|f| \leq 1$.
A: Yes.  Generally, $\| \mu\|$ is defined as
$$
\| \mu \| = |\mu|(\mathbb{R}^d),
$$
where $|\mu |$ is the total variation.
