A convolution of prob. measures is also a prob. measure. Let $\mu,\nu$ be two probability measures on $(\Omega=\mathbb{Z},\sigma=2^{\mathbb{Z}})$. Then the convolution of $\mu *\nu$ is defined as the probability measure on $(\Omega=\mathbb{Z},\sigma=2^{\mathbb{Z}})$,  
$$ (\mu *\nu)(\{n\})=\sum_{m \in \mathbb{Z}}\mu(\{m\})\nu(\{n-m\})$$.
How to prove that this defines in fact a prob. measure?
For that purpose I'm thinking of using the following theorem:

(Extension theorem) Let $\mathcal{A}$ be a semi-ring, and let
  $\mu:\mathcal{A}\rightarrow [0,\infty]$ be an additive,
  $\sigma$-subadditive, and $\sigma$-finite set function with
  $\mu(\emptyset)=0$. Then there is a unique $\sigma$-finite measure
  $\tilde \mu :\sigma (\mathcal{A}) \rightarrow [0,\infty]$ such that
  $\tilde \mu(A)=\mu(A) \forall_{A \in \mathcal{A}}$.

$\mathcal{A}=\{\{n\}:n\in \mathbb{Z}\}$ is a semi-ring, and with sigma-algebra $2^\mathbb{Z}$.
I don't see how to prove that $\mu *\nu$ is additive... since it's not defined for $\{n,k\}$. Maybe it's additive precisely because only sets $\{n\} \in \mathcal{A}$
Also, after applying the theorem I need to prove that $(\mu *\nu)(\mathbb{Z})=1$. For this I've done $(\mu *\nu)(\mathbb{Z})=\sum_{n\in \mathbb{Z}} (\mu *\nu)(\{n\})=\sum_{n\in \mathbb{Z}} \sum_{m\in \mathbb{Z}}\mu(\{m\})\nu(\{n-m\})$ $=\sum_{m\in \mathbb{Z}}\mu(\{m\}) \sum_{n\in \mathbb{Z}}\nu(\{n-m\})=\sum_{m\in \mathbb{Z}}\mu(\{m\})=1$
The problem here, is why could I interchange the infinite sums? If I had $\sum_n \sum_m \mu(\{m\})\nu(\{n-m\})<\infty$ then I could interchange them...
 A: For probability measures on $\mathbb{Z}$, it's not necessary to use the extension theorem you quote.  Any function $p : \mathbb{Z} \to \mathbb{R}_{\geq 0}$ such that $\sum_{n \in \mathbb{Z}} p(n) = 1$ defines a probability measure via $p(A) = \sum_{j \in A} p(j)$.  That this is well-defined, first of all, follows from the fact that bounded non-decreasing sequences of real numbers always have finite limits.  
Next, I'll check that this construction is, in fact, countably additive.  Suppose $\{A_{1},A_{2},\dots,\}$ is a disjoint collection of subsets of $\mathbb{Z}$ and $A = \bigcup_{n = 1}^{\infty} A_{n}$.  Fix $N \in \mathbb{N}$. 
 Then if we let $(b_{k})_{k \in \mathbb{N}}$ denote an enumeration of $A_{1} \cup \dots \cup A_{N}$, then
\begin{align*}
\sum_{j \in A} p(j) &\geq \sum_{k =1}^{\infty} p(b_{k}) \\
&= \sum_{k = 1}^{\infty} p(b_{k}) \left(1_{A_{1}}(b_{k}) + 1_{A_{2}}(b_{k}) + \dots + 1_{A_{N}}(b_{k})\right) \\
&= \sum_{n = 1}^{N} \sum_{k = 1}^{\infty} p(b_{k}) 1_{A_{n}}(b_{k}) \\
&= \sum_{n = 1}^{N} \sum_{j \in A_{n}} p(j)
\end{align*}
and, thus, $\sum_{j \in A} p(j) \geq \sum_{n = 1}^{\infty} p(A_{n})$.  Note that here I've used the fact that series of non-negative terms are unconditionally convergent: the limit is independent of the order of summation.  
On the other hand, arranging the elements of $A$ in a sequence $(a_{k})_{k \in \mathbb{N}}$, we have
$$\sum_{j \in A} p(j) = \lim_{L \to \infty} \sum_{k = 1}^{L} p(a_{k})$$
and $\sum_{k = 1}^{L} p(a_{k}) \leq \sum_{n = 1}^{M} p(A_{n})$ for some $M = M(L)$ depending on $L$.  Since $\sum_{n = 1}^{M} p(A_{n}) \leq \sum_{n = 1}^{\infty} p(A_{n})$, we find
$$\sum_{j \in A} p(j) \leq \sum_{n = 1}^{\infty} p(A_{n}),$$
proving countable additivity.
If you have seen a bit more measure theory, then an alternative way to construct the measure $p$ is to distinguish between the function $p : \mathbb{Z} \to \mathbb{R}_{\geq 0}$ and the measure, let's call it $\lambda_{p}$.  For this construction, you need to know about the counting measure $c$ on $\mathbb{Z}$.  Note that $\sum_{j \in \mathbb{Z}} p(j) = 1$ is just the same thing as $p \in L^{1}(\mathbb{Z},c)$ (by non-negativity of $p$).  Thus, the formula 
$$\lambda_{p}(A) = \int_{A} p(\xi) c(d \xi) = \sum_{j \in A} p(j)$$
defines a measure on $2^{\mathbb{Z}}$.  I only mention this because you may have seen this construction before and, at any rate, it comes up quite a lot (esp. in probability theory).  
As a commenter already mentioned, that $p = \mu * \nu$ satisfies $\sum_{n \in \mathbb{Z}} p(n) = 1$ follows from the Fubini-Tonelli Theorem (whether you call it Fubini or Tonelli depending on taste/culture).  
