Product space probability measures finding convergence determining class

Question

This is part of Theorem 2.8 in Billingsley's Convergence of probaility measures that has been bothering me a lot. To explain the setting, we assume that the product $T=S'\times S''$ is separable, with the regular Borel sigma fields. Let $P'$ be the marginal measure for $S'$ and $P''$ for $S''$. We take distance functions $\rho'(x',y')$ for the first component and $\rho''(x'',y'')$ for the second. And we take the distance between $(x',x'')$, $(y',y'')$ to be $\rho(x',y')\vee \rho(x'',y'')$ Then the open balls in $T$ have the form $$B_t((x',x''),r))=B_{\rho'}(x',r)\times B_{\rho''} (x'',r).$$

Let $\mathscr{A}$ be the $\pi$-system of measurable rectangles $A'\times A''$. Let $\mathscr{A}_P$ be the class of $A'\times A''$ in $\mathscr{A}$ such that $P'(\partial A')=P''(\partial A'')=0.$Then we can show that $\mathscr{A}_P$ is a $\pi-$system and easch set in $\mathscr{A}_P$ is a $P$-continuity set (i.e. $P(\partial A)=0$.

Now is the part of the text I don't understand. The statement argues that since the $B_{\rho'}(x',r)$ have disjoint boundaries for different values of $r$ (which I assume is because they each are $\subset \{\rho'(x',\cdot)=r\}$. But what I don't get is why $B_t((x',x''),r)$ which is (2.7) lies in $\mathscr{A}_P$ because each ball can be made arbitrarily small. But as I stated $\mathscr{A}_P$ is the class of sets of boundary zero. So how do we know that we would always find such a set lying inside a small ball? I would greatly appreciate some help.

Answer 1

Let's try to work our way through the proof. It is clear that we want to apply Theorem 2.3., and hence we need to check the following conditions.

1. $\mathcal{A}_P$ is a $\pi$-system
2. $T=S'\times S''$ is separable
3. For every $(x', x'')\in T$ and $\epsilon>0$, there exists $A\in \mathcal{A}_P$ such that $x\in A^{\circ}\subset A\subset B_t((x', x''), \epsilon).$

The condition (2) is already given to us and you have verified (1). So we are left to check (3). Here we show something stronger. We will be done if we could show that \begin{equation}\label{eon:Needed} B_t((x', x''), \epsilon)\in \mathcal{A}_P, \quad \forall (x', x'')\in S'\times S'', \text{and } \epsilon>0. \end{equation}

We don't quite get that, but something close and that would be enough. note that it is enough to show that~\eqref{eon:Needed} holds for some sequence $\epsilon_n \downarrow 0$. This is what is being argued.

To argue this it is enough to show that $P'(\partial B_{\rho'}(x', \epsilon_n))=0=P''(\partial B(x'', \epsilon_n))$ for some $\epsilon_n\downarrow 0$. Use the proof of Theorem 2.4. to get this. Since there are uncountably many values of $\epsilon$ and $\partial B_{\rho'}(x', \epsilon)$ are disjoint for different $\epsilon$ (and similarly for $B_{\rho''}(x'', \epsilon)$) the result follows.

Below I am trying to expand on the last paragraph!

Since $B_{\rho'}(x', \epsilon)$ have disjoint boundaries for different values of $\epsilon$ (and similarly for $B(x'', \epsilon)$) and since there are uncountably many values of $\epsilon$, there can be at most countably many $\epsilon$ for which $P'(\partial B_{\rho'}(x', \epsilon))\neq 0$. Let's define the set of `bad' epsilons. Set \begin{align} \mathcal{E}'&=\{\epsilon>0: P'(\partial B_{\rho'}(x', \epsilon))\neq 0\}\\ \mathcal{E}''&=\{\epsilon>0: P''(\partial B_{\rho''}(x'', \epsilon))\neq 0\}\;. \end{align}

Note that $\mathcal{E}',\mathcal{E}''$ are all countable and hence so is there union. In particular, there exists a sequence $\epsilon_n\downarrow 0$ such that $P'(\partial B_{\rho'}(x', \epsilon_n))=0=P''(\partial B_{\rho''}(x'', \epsilon))$.

In particular, $B_{t}((x', x''), \epsilon_n)=B_{\rho'}(x', \epsilon_n)\times B_{\rho''}(x'', \epsilon)\in \mathcal{A}_P$ and clearly satisfies (3).

I hope that it clarifies the proof.