Problems with infinite $\Omega$, when trying to define product spaces of discrete probability spaces Definitions In our course we defined a discrete probability
space as a tuple $\left(\Omega,P\right)$, where $P:\mathcal{P}(\Omega)\rightarrow\left[0,1\right]$
and $\Omega$ is at most countable, such that $P\left(\Omega\right)=1$
and for a finite our countable infinite set $I$ we have $P\left(\bigcup_{i\in I}A_{i}\right)=\sum_{i\in I}P\left(A_{i}\right)$
where all $A_{i}$ are pairwise disjoint.
Motivation Now we did several exercise that implicitly
involved product spaces. For all these exercises $\Omega$ was finite.
Problem To avoid having to construct for each example
separately a product space, I wanted to do it once and for all in
the abstract, by proving the following theorem: Let $\left(\Omega_{1},P_{1}\right),\ldots,\left(\Omega_{n},P_{n}\right)$
be discrete probability spaces. Then $\left(\Omega_{1}\times\ldots\times\Omega_{n},P\right)$
where $P:\Omega_{1}\times\ldots\times\Omega_{n}\rightarrow\left[0,1\right],\ P\left(\omega_{1}\ldots\omega_{n}\right)=P_{1}\left(\omega_{1}\right)\cdot\ldots\cdot P_{n}\left(\omega_{n}\right)$
is a discrete probability space as well.
But in trying to prove the second property ($\sigma$-addivity)
of discrete probability spaces, I ran into problems: How do I prove,
that 
$$
P\left(\bigcup_{i\in I}A_{i}\right)=P_{1}\left(\text{pr}_{1}\left[\bigcup_{i\in I}A_{i}\right]\right)\cdot\ldots\cdot P_{n}\left(\text{pr}_{n}\left[\bigcup_{i\in I}A_{i}\right]\right)=$$ $$=\sum_{i\in I}P_{1}\left(\text{pr}_{1}\left[A_{i}\right]\right)\cdot\ldots\cdot\sum_{i\in I}P_{n}\left(\text{pr}_{n}\left[A_{i}\right]\right)\quad\left(\star\right)
$$
equals $\sum_{i\in I}\left(P_{1}\left(\text{pr}_{1}\left[A_{i}\right]\right)\cdot\ldots\cdot P_{n}\left(\text{pr}_{n}\left[A_{i}\right]\right)\right)=\sum_{i\in I}P\left(A_{i}\right)$
? ($\text{pr}_{j}\left[\cdot\right]$ denotes the $j$-th projection
mapping)
If the $\Omega_i$'s are finite I can prove my theorem, since together with
the disjointness of the $A_{j}$'s, this implies that $I$ has to
be finite, so everything is fine. But for countable infinite $\Omega_i$'s,
the sum from $\left(\star\right)$ can only be evaluated
using the Cauchy product for series, and this number wouldn't be the
same as $\sum_{i\in I}\left(P_{1}\left(\text{pr}_{1}\left[A_{i}\right]\right)\cdot\ldots\cdot P_{n}\left(\text{pr}_{n}\left[A_{i}\right]\right)\right)$,
I think.
Can product spaces maybe be defined in a meaningful way only under the conditions from above ?
 A: What you did right: You have correctly defined the product space $\Omega=\Omega_1\times\cdots\times \Omega_n$, and assigned probability values to individual points 
$\omega=(\omega_1,\dots,\omega_n)$ by the product 
$$P(\omega)=P_1(\omega_1)\cdots P_n(\omega_n).$$ This assigns a non-negative value to every point $\omega\in\Omega$ such that $\sum_{\omega\in\Omega}P(\omega)=1$. 
Where you went wrong: Your equation $(\star)$ seems to indicate that you want to assign a probability to an arbitrary subset $A$ of $\Omega$ using projections, as in (2). This is incorrect; you should simply define $$P(A)=\sum_{\omega\in A}P(\omega).\tag1$$ The formula $$P\left( A\right)=P_{1}\left(\text{pr}_{1} A\right)\cdots  P_{n}\left(\text{pr}_{n} A\right)\tag 2$$ works only for product sets, but this is a completely separate issue from countable additivity. 
What to do next: You want to show that $P$ is countably additive. At this point, you can completely forget about the product structure of $\Omega$. In general, if $\Omega$ is any countable space, and $P$ assigns non-negative values  to the points of $\Omega$, where $\sum_{\omega\in \Omega}P(\omega)=1$, then (1) defines a countably additive measure on all subsets of $\Omega$. The proof of this will involve manipulating countably many infinite sums, but I think you can do it.  
