Proof of Factorized Joint Probability Mass Function Implies Independence

Conventions on countable sums. All sums below have nonnegative terms. As in Poisson Distribution, a sum over a countable index set is the supremum of its finite partial sums; when the index set is enumerated by a bijection with $\mathbb{N}$ (or is finite), this supremum equals the limit of the partial sums along the enumeration, because the partial sums are nondecreasing and every finite subset of the index set is contained in an initial segment of the enumeration. In particular $\sum_{c\in\mathbb{N}_0}g_i(c)=\sum_{c=0}^{\infty}g_i(c)=1$.

Step 1 (rectangle partition). For $c=(c_1,\dots,c_r)\in\mathbb{N}_0^r$ write $R(c)=\bigcap_{i=1}^{r}\{X_i=c_i\}\in\mathcal{F}$. Since every $X_i(\omega)$ lies in $\mathbb{N}_0$, each $\omega\in\Omega$ belongs to $R(c)$ for exactly one $c$, namely $c=(X_1(\omega),\dots,X_r(\omega))$. Hence the sets $R(c)$, $c\in\mathbb{N}_0^r$, are pairwise disjoint with union $\Omega$, and $\mathbb{N}_0^r$ is countable (a finite Cartesian product of countable sets).

Step 2 (a product identity for countable sums). Let $C_1,\dots,C_r\subseteq\mathbb{N}_0$ be nonempty. We claim

It suffices to treat $r=2$ and induct: grouping the tuple as $((c_1,\dots,c_{r-1}),c_r)$ reduces ($\dagger$) to the two-index-set case with term $a_{c'}b_{c_r}$, $a_{c'}=\prod_{i<r}g_i(c_i)$, $b=g_r$. So let $(a_{u})_{u\in U}$, $(b_{v})_{v\in V}$ be nonnegative families over countable sets with sums $\alpha=\sum_U a_u\le1$ and $\beta=\sum_V b_v\le1$ (all sums here are at most $1$, being suprema of sums of probabilities or products of such, so all suprema are finite). Every finite $T\subseteq U\times V$ is contained in some $T_U\times T_V$ with $T_U\subseteq U$, $T_V\subseteq V$ finite, and then $\sum_{T}a_ub_v\le\sum_{T_U\times T_V}a_ub_v=(\sum_{T_U}a_u)(\sum_{T_V}b_v)\le\alpha\beta$ by distributivity and monotonicity; hence $\sum_{U\times V}a_ub_v\le\alpha\beta$. Conversely, for finite $T_U,T_V$, $(\sum_{T_U}a_u)(\sum_{T_V}b_v)=\sum_{T_U\times T_V}a_ub_v\le\sum_{U\times V}a_ub_v$; taking the supremum over $T_U$ and then over $T_V$ gives $\alpha\beta\le\sum_{U\times V}a_ub_v$ (for fixed $T_V$ the map $s\mapsto s\sum_{T_V}b_v$ is nondecreasing in $s\ge0$, and then similarly in the second factor). This proves ($\dagger$).

Step 3 (probabilities of products of events). Let $S\subseteq\{1,\dots,r\}$ be nonempty and let $B_i$ be a Borel set for each $i\in S$. Put $C_i=B_i\cap\mathbb{N}_0$ for $i\in S$ and $C_i=\mathbb{N}_0$ for $i\notin S$. By Step 1,

a countable disjoint union: an $\omega$ lies in the left side if and only if its value tuple $c(\omega)$ satisfies $c_i(\omega)\in B_i$ for $i\in S$, i.e., $c(\omega)\in C_1\times\cdots\times C_r$ (if some $C_i=\emptyset$, both sides are empty and the identity below reads $0=0$ by convention that an empty sum is $0$). By countable additivity of the measure $P$ (applied along any enumeration of the countable index set; the series sum equals the unordered sum by the convention above), the hypothesis, and ($\dagger$),

where the factors with $i\notin S$ equal $\sum_{\mathbb{N}_0}g_i=1$ and were dropped.

Conclusion. Taking $S=\{i\}$ in ($\ddagger$) proves Part 1, and taking $B=\{c\}$ gives $P(X_i=c)=g_i(c)$. For Part 2, let $B_1,\dots,B_r$ be arbitrary Borel sets. For every nonempty $S\subseteq\{1,\dots,r\}$, ($\ddagger$) together with Part 1 gives

which is exactly the product formula required by Independence of Events and of Random Variables for the events $\{X_1\in B_1\},\dots,\{X_r\in B_r\}$. Hence $X_1,\dots,X_r$ are independent. $\blacksquare$