# 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?

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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 $|f| \leq 1$.

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Thanks! The reason I mentioned $\sigma$-finiteness is because I wanted to verify that you couldn't get the inequality for just any two infinite measures (in particular ones that aren't $\sigma$-finite). Also, if the measures in question are positive, do you still need to use Fubini, or can you use some sort of product measure notion to say that $f \equiv 1$ would be the function we use and so the norm of the convolution would be equal to the product of each of the two measures on the whole space? – user1736 Aug 25 '11 at 16:20
Also, does that alternate definition only work for Borel measures? Is there an easy way to see why we can use any of those types of functions to define the norm? Thanks again for your help! – user1736 Aug 25 '11 at 16:23
A quick first reaction to the second comment: The usual definition of total variation (see e.g. Rudin's book on real and complex analysis) involves cutting up the domain of the measure in disjoint pieces $E_i$ and maximizing $|\sum a_i \mu E_i|$ where $|a_i| = 1$. Now if we're working with (regular) Borel measures we can approximate the $E_i$ with open sets $U_{i}$ from the outside and the characteristic functions of the $U_i$ with functions from any of those classes from below. Is this enough as a hint for that? – t.b. Aug 25 '11 at 16:33
Hm, yeah I think I get it. So does that mean the the definition you used only works for regular measures? – user1736 Aug 25 '11 at 16:41
Two things: Yes, I believe that it is fair to say that you only want to do it for (finite) Borel measures. However, every such is automatically regular (and tight) on complete separable metric spaces. Note also that a finite measure automatically has a certain integrability condition built in. As much as you can't convolve any two (non-negative) functions without getting something meaningless, you can't do it with any two measures. Note that in working on $\mathbb{R}^n$ we're working on a space with lot of structure. In particular it has a group structure and a translation invariant measure. – t.b. Aug 25 '11 at 17:09

Yes. Generally, $\| \mu\|$ is defined as

$$\| \mu \| = |\mu|(\mathbb{R}^d),$$

where $|\mu |$ is the total variation.

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That definition should be well-defined for every signed(not necessarily being Borel) measure right? Also, using that definition, does the convolution inequality then come from using applying Fubini's Theorem (like the way it would if the two measures were absolutely continuous with respect to Lebesgue measure)? You would also need the $\sigma$-finite property in that case then right? – user1736 Aug 24 '11 at 21:30