Let $(X,\mathcal{A})$ and $(Y,\mathcal{B})$ be standard Borel spaces (for example, $\Bbb{R}$ with its Borel sigma-algebra).
Let $p$ be a probability measure on $(X,\mathcal{A})$ and $q$ be a probability measure on $(Y,\mathcal{B})$.
Let now $r$ be a finitely additive probability measure on $(X\times Y, \mathcal{A}\otimes\mathcal{B})$, such that for all $A\in\mathcal{A}$ and $B\in\mathcal{B}$,
$r(A\times Y)=p(A)$ and $r(X\times B)=q(B)$.
Can we conclude that $r$ is countably additive? If not, are there conditions under which we can?
Best Answer
No. Let $X = Y = [0, 1)$ with its usual Borel $\sigma$-algebra, $\lambda$ be the Lebesgue measure. Consider the decreasing sequence $A_n$ of subsets of $X \times Y$ given by $A_n = \cup_{i = 1}^{2^n} [\frac{i - 1}{2^n}, \frac{i}{2^n}) \times [\frac{i - 1}{2^n}, \frac{i}{2^n})$. We see that $\lambda \times \lambda(A_n) = 2^{-n}$. Define $\mu_n$ on $X \times Y$ by $\mu_n = 2^n1_{A_n}(\lambda \times \lambda)$. We observe that both marginals of $\mu_n$ are $\lambda$. Now, let $\mathcal{U}$ be a free ultrafilter on $\mathbb{N}$ and $\mu = \lim_{n \rightarrow \mathcal{U}} \mu_n$. Then $\mu$ is a finitely additive measure with both of its marginals $\lambda$. However, $\mu$ is not countably additive. Indeed, if it were, then we would have $\mu(\cap_m A_m) = \lim_{m \rightarrow \infty} \mu(A_m)$. Since $\lambda \times \lambda(A_m) = 2^{-m}$, we have $\lambda \times \lambda(\cap_m A_m) = 0$, so $\mu_n(\cap_m A_m) = 0$, whence $\mu(\cap_m A_m) = \lim_{n \rightarrow \mathcal{U}} \mu_n(\cap_m A_m) = 0$. On the other hand, whenever $n \geq m$, we have $\mu_n(A_m) = 2^n(\lambda \times \lambda)(A_n \cap A_m) = 2^n(\lambda \times \lambda)(A_n) = 1$, so $\mu(A_m) = \lim_{n \rightarrow \mathcal{U}} \mu_n(A_m) = 1$. But then $\lim_{m \rightarrow \infty} \mu(A_m) = 1 \neq 0 = \mu(\cap_m A_m)$, so $\mu$ is not countably additive.