Fundamentally my question is very simple: does the following argument make sense (it seems counterintuitive to me).
Given an arbitrary set $A$, for any $a \in A$ let $(X_a,B_a,\mu_a)$ be a probability space such that $X_a$ is a locally compact, $\sigma$-compact metric space with Borel $\sigma$-algebra $B_a$. By the existence of product measures theorem there exists a unique probability measure, $\mu_A$, on $(\Pi_{a \in A}X_a,\Pi_{a \in A}B_a)$ with the condition: $$\mu_A\left( \Pi_{a \in A}E_a\right) = \Pi_{a\in A} \mu_a(E_A)$$ wherever $E_a\in B_a$ for each $a\in A$ and one has $E_a=X_a$ for all but finitely many $a$.
In particular, let $A = \mathbb{N}$ and each $X_a = \{0,1,2,\dots,a\}$ with the discrete topology, let $B_a$ be the discrete Borel $\sigma$-algebra (i.e. the power set of $X_a$) and $\mu_a$ be the uniform probability measure.
Each $X_{a}$ is compact and therefore locally compact and $\sigma$-compact so there exists a unique probability measure $\mu_{A}$ on $(\Pi_{a = 1}^{\infty}, B_A)$.
The problem for me is that these discrete measures can extend to an infinite product space.