= Solution
<Independence of random variables> factors the <joint probability mass function> as $\mathbb P(X=x,Y=y)=\mathbb P(X=x)\mathbb P(Y=y)$. Therefore
$$
\begin{aligned}
\mathbb E[f(X)g(Y)]
&=\sum_x\sum_y f(x)g(y)\mathbb P(X=x,Y=y)\\
&=\sum_x\sum_y f(x)\mathbb P(X=x)g(y)\mathbb P(Y=y)\\
&=\left(\sum_x f(x)\mathbb P(X=x)\right)\left(\sum_y g(y)\mathbb P(Y=y)\right).
\end{aligned}
$$
With finite <expected values>, these rearrangements are justified by absolute convergence: the analogous sum with absolute values equals $\mathbb E[|f(X)|]\mathbb E[|g(Y)|]<\infty$. Thus
$$
\boxed{\mathbb E[f(X)g(Y)]=\mathbb E[f(X)]\mathbb E[g(Y)].}
$$
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