Getting almost certainty from uncountably many low-probability events

Question

Let $(\Omega,\Sigma,\mathbb{P})$ be a complete probability space, $B\subseteq X$ be a non-empty Borel subset of a polish space $X$, $A$ be an uncountable indexing set, and $\{X_{\alpha,n}\}_{a \in A, n \in \mathbb{N}}$ a set of $X$-valued random elements with the following properties:

• For every pair of distinct $\alpha,\alpha'\in A$ and every $\omega \in \Omega$ there does not exist $n,n'\in\mathbb{N}$ satisfying $$ X_{\alpha,n}(\omega) = X_{\alpha',n'}(\omega). $$
• For every $\alpha \in A$, and every non-empty open subset $O'\subseteq X$, $0<\mathbb{P}\left(\bigcup_{n \in \mathbb{N}}\{X_{\alpha,n} \in O'\}\right)$.

Then can we conclude that $1=\mathbb{P}\left(\bigcup_{n \in \mathbb{N},\alpha \in A}\{X_{\alpha,n} \in B\}\right)$?

Answer 1

Here is a "very regular" counterexample:

Let $X=\mathbb R$, $A:=(0,1)=:B$, and $X_{a,n}:=Z+a$ for all $a\in A$ and $n\in \mathbb R$, where $Z\sim N(0,1)$. Then all your conditions hold. However, $$P\Big(\bigcup_{n\in\mathbb N,a\in A}\{X_{a,n}\in B\}\Big)=P(|Z|<1)\ne1.$$

Answer 2

An attempt at a counterexample. Pick any $(\Omega,\Sigma,\mathbb{P})$. Pick $X = \{0,1\}^{\omega}$ with standard Borel structure, and $B = \{0^\omega\}$. Pick $A = \mathbb{R}$. Have each $X_{\alpha,n}$ be the constant function at $x_{\alpha,n}$ for all $\alpha,n$, and have $(x_{\alpha,n})_n$ enumerate a dense set in $X$ for each $\alpha$. It's easy to pick the sequences $(x_{\alpha,n})_n$ so they are all disjoint (note that each $X_{\alpha,n}$ is measurable trivially as it's constant, so it's just about cardinality), and you can pick them so that their values are not equal to $0^\omega$. Now $\mathbb{P}(\bigcup_{n \in \mathbb{N}} \{X_{\alpha, n} \in O'\}) = 1$ for all $\alpha$ and open $O'$ (if I'm interpreting it correctly). But no values are in $B$ so the other probability is zero.