[proofplan] For countable state spaces, point events generate all events. Sum the pointwise factorisation over arbitrary subsets, and use independence for the converse. [/proofplan] [step:Independence gives pointwise factorisation] If $X_1,\ldots,X_n$ are independent, apply the definition to the singleton sets $\{x_i\}\subset E_i$ to get \begin{align*} \mathbb P(X_1=x_1,\ldots,X_n=x_n) =\prod_{i=1}^n\mathbb P(X_i=x_i). \end{align*} [/step] [step:Pointwise factorisation gives independence] Assume the displayed identity holds for every point $(x_1,\ldots,x_n)$. Let $A_i\subset E_i$. Since the $E_i$ are countable and the events \begin{align*} \{X_1=x_1,\ldots,X_n=x_n\} \end{align*} are disjoint as $(x_1,\ldots,x_n)$ varies, \begin{align*} \mathbb P(X_1\in A_1,\ldots,X_n\in A_n) &=\sum_{x_1\in A_1}\cdots\sum_{x_n\in A_n} \mathbb P(X_1=x_1,\ldots,X_n=x_n) \\ &=\sum_{x_1\in A_1}\cdots\sum_{x_n\in A_n} \prod_{i=1}^n\mathbb P(X_i=x_i) \\ &=\prod_{i=1}^n\sum_{x_i\in A_i}\mathbb P(X_i=x_i) \\ &=\prod_{i=1}^n\mathbb P(X_i\in A_i). \end{align*} Thus $X_1,\ldots,X_n$ are independent. [/step]