Convolution of discrete measures

Question

Exercise (Billingsley's Probability and Measure (3e), p.271, Exr.20.13): Given discrete measures, $\mu$ and $\nu$, consisting of masses $\alpha_n$ and $\beta_n$ for $n = 0, 1,2,\ldots$ Show that $\mu * \nu$ consists of a mass of $\sum_{k=0}^{n} \alpha_{k} \beta_{n-k}$ at $n, n=0,1,2, \ldots$

What I thought about doing first was pushing symbols together. First, since $\sum \alpha_n = 1 = \sum \beta_n$. Multiplying would result in $\sum_{k=0}^n \sum_{j=0}^n \alpha_j \beta_k = 1$. But I don't get the formula that I want. Especially not if I want to find the mass for an exact $n$.

The formula for the convolution of two (possibly arbitrary finite) measures (in Billingsley) is as follows:

$$(\mu * \nu) (H) = \int_{- \infty}^{+ \infty} \nu(H -x)\mu(dx), H \in \mathcal{R}^1,$$

where $\mathcal{R}^1$ is the $\sigma$-field generated by intervals in $\mathbb{R}$. Blindly pushing ahead with the current problem, we see that

$$\int_{- \infty}^{+ \infty} \nu(H -x)\mu(dx) = \sum \nu(H-x) \mu(x)$$

How do I manipulate $H$ and my equation here? Since after $H$ is measured, it is going to be $0$ almost everywhere except for some points that correspond to the masses of our measure.

A technical explanation of what we want to do in this problem and sort of verbalizing exactly what we do would be extremely helpful. But suggestions are also helpful.

Answer 1

Let $a:\mathbb{R}\times\mathbb{R}\to\mathbb{R}, (x,y)\mapsto x+y$ be addition (which is continuous hence Borel measurable). Then for $\mu,\nu$ two anonymous measures on $\mathbb{R}$, their convolution $\mu\ast\nu$ is by definition the pushforward of the product measure $\mu\otimes \nu$ via $a$; thus for $\phi:\mathbb{R}\to \mathbb{R}$ bounded measurable we have

\begin{align*} \int_\mathbb{R} \phi(z) \, d\mu\ast\nu(z) &=\int_\mathbb{R} \phi(z) \, da_\ast(\mu\otimes\nu)(z)\\ &=\int_{\mathbb{R}\times \mathbb{R}} \phi\circ a(x,y) \, d\mu\otimes \nu(x,y)\\ &=\int_{\mathbb{R}\times \mathbb{R}} \phi(x+y) \, d\mu\otimes \nu(x,y)\\ &=\int_{\mathbb{R}}\int_{\mathbb{R}} \phi(x+y) \, d\mu(x)\, d\nu(y). \end{align*}

In particular, if $\mu$ and $\nu$ are purely atomic with support nonnegative integers, so that

$$\mu = \sum_{n\in\mathbb{Z}_{\geq0}} \alpha_n\, \delta_n,\,\, \nu= \sum_{m\in\mathbb{Z}_{\geq0}} \beta_m\,\delta_m,$$

the above formula reads:

\begin{align*} \int_\mathbb{R} \phi(z) \,d\mu\ast\nu(z) &= \sum_{n\in\mathbb{Z}_{\geq0}}\sum_{m\in\mathbb{Z}_{\geq0}} \alpha_n\beta_m\phi(n+m)\\ &= \sum_{k\in\mathbb{Z}_{\geq0}}\sum_{\substack{p,q\in\mathbb{Z}_{\geq0}\\ p+q=k}} \alpha_p\beta_q\phi(k). \end{align*}

Here one can think of the rearrangement as follows: consider the integer points $(n,m)$ in the first quadrant of the plane. The first sum corresponds to summing first over verticals and then over horizontals, whereas the second sum corresponds to first summing over the diagonals $p+q=k$ and then summing the results. (Note that this rearrangement is valid as these are integrals in the sense of Lebesgue.)

Thus we have

$$\left(\sum_{n\in\mathbb{Z}_{\geq0}} \alpha_n\, \delta_n\right)\ast\left(\sum_{m\in\mathbb{Z}_{\geq0}} \beta_m\,\delta_m\right) = \sum_{k\in\mathbb{Z}_{\geq0}}\left(\sum_{\substack{p,q\in\mathbb{Z}_{\geq0}\\ p+q=k}} \alpha_p\beta_q\right)\, \delta_k.$$