# Does convolution of Borel measures have the cancellation property?

The convolution of two Borel measures $$\mu$$ and $$\nu$$ is given by $$(\mu * \nu)(E) = \int_{-\infty}^\infty \nu(E - x) \; \mu(d x).$$

I have been trying to figure out whether the following cancellation law holds for convolutions: $$\text{if } \nu \neq 0 \text{ and } \mu_1 * \nu = \mu_2 * \nu, \text{ then also } \mu_1 = \mu_2$$ Here I use $$\mu = \nu$$ to mean that $$\mu(E) = \nu(E)$$ for all Borel measurable $$E$$.

Intuitively, I feel that the cancellation property should hold, but I have seen examples of the cancellation property failing for very similar kinds of convolution, so I am unsure.

Does the cancellation property above hold for convolution of measures? And if not, can we recover it by restricting for example to probability measures and/or continuous measures?

• $\mu$ and $\nu$ are only defined on measurable sets, so you can't really interpret it any way else – Jakobian Jul 16 '19 at 13:18
• This is definitely not true for complex measures (even continuous): take any continuous compactly supported, nonzero functions $f,g$ with disjoint supports, and look at the (continuous) measures $\mu,\nu_1,\nu_2$ such that $\hat\mu=f$, $\hat\nu_1=g$ and $\hat \nu_2=2g$. Then $\mu*\nu_1=\mu*\nu_2=0$, but $\nu_1\neq2\nu_1=\nu_2$. – tomasz Jul 16 '19 at 13:32
• Even simpler, consider $\mu_1(E) = 0$ for every measurable $E$. We either need stronger conditions on $\mu_1$, or consider the situation in general, where equality holds for any $\mu_1$. – Jakobian Jul 16 '19 at 13:36
• @Jakobian: Yeah, but that is completely trivial, obviously the question only makes sense for nonzero measures. – tomasz Jul 16 '19 at 13:37
• @Jakobian Right, $\mu_1$ must of course be non-zero, I will add that. – mrp Jul 16 '19 at 13:38

## 1 Answer

Definition: We say that $$\mu$$ is infinitely divisible, if for any $$n$$ there exists a measure $$\mu_n$$ such that $$\mu = (\mu_n)^{*n}$$, the $$n$$-th convolution power. In other words, for any $$n$$ there exists $$n$$-th root of $$\mu$$.

Theorem: Suppose that $$\mu_1, \mu_2, \nu$$ are of finite measure and $$\nu\neq 0$$ is infinitely divisible. It holds that $$\mu_1*\nu = \mu_2*\nu \implies \mu_1 = \mu_2.$$

Proof: Without loss of generality we may assume that $$\mu_1, \mu_2, \nu$$ are probability measures. This is because for any Borel measure $$\mu$$, $$\mu \neq 0\iff \mu(R)>0$$, and $$(\mu_1*\nu)(R) = \mu_1(R)\nu(R) = \mu_2(R)\nu(R) = (\mu_2*\nu)(R).$$ From which $$\mu_1(R) = \mu_2(R)$$. If $$\mu_1(R) = 0$$, then $$\mu_1 = \mu_2 = 0$$ as desired. If $$\mu_1(R) \neq 0$$, then we can divide both sides by $$\mu_1(R)\nu(R) = \mu_2(R)\nu(R)$$, and take new (this time probability) measures $$\mu_1' = \frac{\mu_1}{\mu_1(R)}, \mu_2' = \frac{\mu_2}{\mu_2(R)}, \nu' = \frac{\nu}{\nu(R)}.$$ Those new measures satisfy $$\mu_1'*\nu' = \mu_2'*\nu'$$ since convolution is bilinear. We will keep refering to those as $$\mu_1, \mu_2, \nu$$.

Consider $$\varphi_1, \varphi_2, \phi$$, the characteristic functions of $$\mu_1, \mu_2,\nu$$. We have $$\varphi_{\mu_1*\nu} = \varphi_1\phi = \varphi_2\phi = \varphi_{\mu_2*\nu},$$ hence $$\forall_{x\in R}\ (\varphi_1(x) = \varphi_2(x)\ \lor \phi(x) = 0). \tag{1}$$ It is because for any probability measures $$\mu, \hat{\mu}$$ we can consider independent random variables $$X\sim \mu, Y\sim \hat{\mu}$$, and it is well known that $$X+Y\sim \mu*\hat{\mu}$$, hence $$\varphi_{\mu*\hat{\mu}} = \varphi_{\mu}\varphi_{\hat{\mu}}$$.

From [1] it follows that $$\phi$$ has no zeros, since it's a characteristic function of infinitely divisible measure. Hence from $$(1)$$ we have that $$\varphi_1 = \varphi_2$$, and so $$\mu_1 = \mu_2$$, as had to be shown.

Examples of infinitely divisible measures are Normal distribution or Cauchy distribution. Distributions of the form $$\delta_a$$ are also infinitely divisible. Another example would be Poisson distribution.

It can also be shown that it is false in general for probability measures. Consider functions $$\phi(x) = (-|x|+1)1_{[0, 1]}(|x|),\ \varphi_1(x) = e^{-|x|},\\ \varphi_2(x) = e^{-|x|}1_{[0, 1)}(|x|)+e^{-1}(-|x|+2)1_{[1, 2]}(|x|)$$ We have that those are characteristic functions from Pólya criterion, and they satisfy $$(1)$$, which is equivalent to $$\varphi_1\phi = \varphi_2\phi$$, and so $$\mu_1*\nu = \mu_2*\nu$$ where $$\mu_1, \mu_2, \nu$$ are probability measures corresponding to $$\varphi_1, \varphi_2,\phi$$, and $$\mu_1 \neq \mu_2$$ because $$\varphi_1\neq \varphi_2$$.

There can be doubt about if $$\varphi_2$$ is convex for positive values because of the point $$x = 1$$, but it can be easily checked that $$(e^{-x})'_{x=1} = -e^{-1}$$, so it is indeed a convex function.

In fact, those characteristic functions are all integrable, hence they correspond to absolutely continuous random variables (in addition, symmetric).