Borel Sets and Measurable Maps in a Separable Metric Space

Let $\mathbb{Q}$ be the set of rational numbers and let $\mathbb{Q}_{>0}$ be the set of those $s\in\mathbb{Q}$ with $0<s$. Let $(\Omega,\mathcal{F})$ be a measurable space.

Call a pair $\bigl((X,d),D\bigr)$ a separable metric datum when $(X,d)$ is a metric space and $D$ is a nonempty countable subset of $X$ that is dense in $X$ for the collection $\mathcal{T}_d$ of subsets of $X$ that are open in $(X,d)$, which is a topology on $X$ by Metric Open Sets Form a Topology; such a $D$ witnesses that $(X,d)$ is separable. For such a pair write $\mathcal{B}(X,d)$ for the Borel $\sigma$-algebra of $(X,d)$, and write $\mathcal{E}$ for the family of subsets of $X$ whose members are exactly the open balls $B_d(q,s)$ with $q\in D$ and $s\in\mathbb{Q}_{>0}$.

Claims 1, 2 and 3 are asserted for every separable metric datum $\bigl((X,d),D\bigr)$, with $\mathcal{T}_d$, $\mathcal{B}(X,d)$ and $\mathcal{E}$ as just described.

1. (Countable ball basis.) $\mathcal{E}$ is countable, and every $U\in\mathcal{T}_d$ is the union of those members of $\mathcal{E}$ that are contained in $U$.

2. (Generation.) The $\sigma$-algebra generated by $\mathcal{E}$ is $\mathcal{B}(X,d)$.

3. (Criterion for measurability.) Let $Y:\Omega\to X$. Then $Y$ is measurable with respect to $\mathcal{F}$ and $\mathcal{B}(X,d)$ if and only if

In particular, if for every $q\in D$ the real-valued map $\omega\mapsto d(Y(\omega),q)$ is measurable with respect to $\mathcal{F}$ and the Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R})$ of the real line, then $Y$ is measurable with respect to $\mathcal{F}$ and $\mathcal{B}(X,d)$.

4. (Pairs.) Let $\bigl((X,d),D\bigr)$ and $\bigl((X',d'),D'\bigr)$ be separable metric data, and let $X\times X'$ carry the product metric $d_{X\times X'}$, a metric by claim 1 of The Product Metric is a Metric. Then $\bigl((X\times X',d_{X\times X'}),D\times D'\bigr)$ is again a separable metric datum. Moreover, if $Y:\Omega\to X$ is measurable with respect to $\mathcal{F}$ and $\mathcal{B}(X,d)$, and $Y':\Omega\to X'$ is measurable with respect to $\mathcal{F}$ and $\mathcal{B}(X',d')$, then the map $Z:\Omega\to X\times X'$ given by $Z(\omega)=(Y(\omega),Y'(\omega))$ is measurable with respect to $\mathcal{F}$ and $\mathcal{B}(X\times X',d_{X\times X'})$.