Proof of Borel Sets and Measurable Maps in a Separable Metric Space

Let $\mathbb{N}$ denote the natural numbers, and write $\sigma(\mathcal{E})$ for the $\sigma$-algebra generated by $\mathcal{E}$. By the symmetry axiom of a metric we have $B_d(q,s)=\{y\in X:d(y,q)<s\}$ for every $q\in X$ and every real $s>0$, and we use this identification without further comment.

Claim 1.

Countability. The set $\mathbb{Q}_{>0}$ is a subset of $\mathbb{Q}$, which is countable by claim 2 of The Integers and the Rational Numbers are Countable; hence $\mathbb{Q}_{>0}$ is countable by claim 3 of Basic Properties of Countable Sets. Since $D$ is countable, the set $D\times\mathbb{Q}_{>0}$ is countable by claim 1 of Products and Powers of Countable Sets. The map sending $(q,s)\in D\times\mathbb{Q}_{>0}$ to $B_d(q,s)$ is, by the description of $\mathcal{E}$ in the statement, a surjection onto $\mathcal{E}$; so $\mathcal{E}$ is countable by claim 4 of Basic Properties of Countable Sets.

Basis property. Let $U\in\mathcal{T}_d$ and let $V$ be the union of those members of $\mathcal{E}$ that are contained in $U$, the union of an empty family being empty. Every such member is a subset of $U$, so $V\subseteq U$.

For the reverse inclusion let $x\in U$. Since $U$ is open in $(X,d)$ there is a real number $r>0$ with $B_d(x,r)\subseteq U$. By claim 8 of Elementary Order Arithmetic in an Ordered Field there is a real number $h$ with $0<h$ and $h+h=r$. By claim 1 of The Rational Numbers are Dense in the Real Numbers, applied to the pair $0<h$, there is $s\in\mathbb{Q}$ with $0<s<h$; thus $s\in\mathbb{Q}_{>0}$. Since $D$ is dense in $X$ for $\mathcal{T}_d$, the closure of $D$ in $X$ is $X$ and so contains $x$; by Characterization of the Closure in a Metric Space by Open Balls, condition 1 there implies condition 3, so there is $q\in D$ with $d(x,q)<s$. In particular $x\in B_d(q,s)$, and $B_d(q,s)$ is a member of $\mathcal{E}$.

We check that $B_d(q,s)\subseteq U$. Let $w\in B_d(q,s)$, so that $d(q,w)<s$. Adding the inequalities $d(x,q)<s$ and $d(q,w)<s$ by claim 3 of Elementary Order Arithmetic in an Ordered Field gives $d(x,q)+d(q,w)<s+s$. The triangle inequality for $d$ gives $d(x,w)\le d(x,q)+d(q,w)$, so claim 2 of Elementary Order Arithmetic in an Ordered Field yields $d(x,w)<s+s$. Adding the inequality $s<h$ to itself, again by claim 3 there, gives $s+s<h+h=r$; since $s+s<r$ implies $s+s\le r$, a second application of claim 2 gives $d(x,w)<r$. Hence $w\in B_d(x,r)\subseteq U$.

So $B_d(q,s)$ is a member of $\mathcal{E}$ contained in $U$, whence $x\in B_d(q,s)\subseteq V$. As $x\in U$ was arbitrary, $U\subseteq V$, and therefore $U=V$.

Claim 2. By Open Ball in a Metric Space is Open every member of $\mathcal{E}$ is open in $(X,d)$, so $\mathcal{E}\subseteq\mathcal{T}_d$. By claim 2 of Intersections of Sigma-Algebras and Minimality of the Generated Sigma-Algebra, $\mathcal{B}(X,d)$ is a $\sigma$-algebra on $X$ containing $\mathcal{T}_d$, hence containing $\mathcal{E}$; by the minimality assertion of that same claim, $\sigma(\mathcal{E})\subseteq\mathcal{B}(X,d)$.

For the reverse inclusion, let $U\in\mathcal{T}_d$ and let $\mathcal{E}_U$ be the family of those members of $\mathcal{E}$ that are contained in $U$, so that $U$ is the union of $\mathcal{E}_U$ by claim 1. Being a subfamily of the countable family $\mathcal{E}$, $\mathcal{E}_U$ is countable by claim 3 of Basic Properties of Countable Sets. By claim 2 of Intersections of Sigma-Algebras and Minimality of the Generated Sigma-Algebra the family $\sigma(\mathcal{E})$ is a $\sigma$-algebra on $X$ containing $\mathcal{E}$.

If $\mathcal{E}_U$ has no members then $U$ is empty; since $X\in\sigma(\mathcal{E})$ and $\sigma(\mathcal{E})$ is closed under complements relative to $X$ by the $\sigma$-algebra axioms, the set $X\setminus X$, which is empty, lies in $\sigma(\mathcal{E})$, so $U\in\sigma(\mathcal{E})$. Otherwise $\mathcal{E}_U$ is nonempty and countable, so by the definition of a countable set there is a sequence $(E_m)_{m\in\mathbb{N}}$ whose set of terms is $\mathcal{E}_U$; then $U$ is the union of the sets $E_m$ over $m\in\mathbb{N}$. Each $E_m$ belongs to $\mathcal{E}$ and hence to $\sigma(\mathcal{E})$, so $U\in\sigma(\mathcal{E})$ by the third $\sigma$-algebra axiom.

Thus $\mathcal{T}_d\subseteq\sigma(\mathcal{E})$. Since $\mathcal{B}(X,d)$ is by definition the $\sigma$-algebra generated by $\mathcal{T}_d$, minimality gives $\mathcal{B}(X,d)\subseteq\sigma(\mathcal{E})$. The two inclusions give $\sigma(\mathcal{E})=\mathcal{B}(X,d)$.

Claim 3. Suppose first that $Y$ is measurable with respect to $\mathcal{F}$ and $\mathcal{B}(X,d)$. Let $q\in D$ and $s\in\mathbb{Q}_{>0}$. As in claim 2, $B_d(q,s)$ is open in $(X,d)$ and hence belongs to $\mathcal{B}(X,d)$, and

By measurability this set belongs to $\mathcal{F}$.

Conversely, suppose that $\{\omega\in\Omega:d(Y(\omega),q)<s\}\in\mathcal{F}$ for every $q\in D$ and every $s\in\mathbb{Q}_{>0}$. Let $\mathcal{G}$ be the family of those subsets $B\subseteq X$ with $Y^{-1}(B)\in\mathcal{F}$. Then $\mathcal{G}$ is a $\sigma$-algebra on $X$: first, $Y^{-1}(X)=\Omega\in\mathcal{F}$, so $X\in\mathcal{G}$; second, if $B\in\mathcal{G}$ then $Y^{-1}(X\setminus B)=\Omega\setminus Y^{-1}(B)$, which lies in $\mathcal{F}$; third, if $(B_m)_{m\in\mathbb{N}}$ is a sequence in $\mathcal{G}$ then the preimage of the union of the $B_m$ is the union of the sets $Y^{-1}(B_m)$, which lies in $\mathcal{F}$. By the displayed hypothesis, $Y^{-1}(B_d(q,s))\in\mathcal{F}$ for all $q\in D$ and $s\in\mathbb{Q}_{>0}$, that is, $\mathcal{E}\subseteq\mathcal{G}$. Minimality (claim 2 of Intersections of Sigma-Algebras and Minimality of the Generated Sigma-Algebra) gives $\sigma(\mathcal{E})\subseteq\mathcal{G}$, and claim 2 above gives $\mathcal{B}(X,d)\subseteq\mathcal{G}$. Thus $Y^{-1}(B)\in\mathcal{F}$ for every $B\in\mathcal{B}(X,d)$, that is, $Y$ is measurable.

For the final assertion, let $d_{\mathbb{R}}$ be the absolute-value metric on $\mathbb{R}$, given by $d_{\mathbb{R}}(a,b)=|a-b|$, and assume that for every $q\in D$ the map $g_q:\Omega\to\mathbb{R}$, $g_q(\omega)=d(Y(\omega),q)$, is measurable with respect to $\mathcal{F}$ and $\mathcal{B}(\mathbb{R})$. Fix $q\in D$ and $s\in\mathbb{Q}_{>0}$ and set $H=\{t\in\mathbb{R}:t<s\}$.

We check that $H$ is open in $(\mathbb{R},d_{\mathbb{R}})$. Let $t_0\in H$, so $t_0<s$. By claim 1 of Elementary Order Arithmetic in an Ordered Field, adding $-t_0$ to both sides gives $t_0-t_0<s-t_0$, and $t_0-t_0=0$ by claim 3 of Additive Cancellation and Elementary Additive Identities in a Field, so the real number $r_0=s-t_0$ satisfies $0<r_0$. Let $t\in B_{d_{\mathbb{R}}}(t_0,r_0)$, that is $|t_0-t|<r_0$. By claim 3 of Properties of the Absolute Value in an Ordered Field we have $t-t_0\le|t-t_0|$, and $|t-t_0|=|t_0-t|$ by claim 2 there, so $t-t_0\le|t_0-t|$; since $|t_0-t|<r_0$, claim 2 of Elementary Order Arithmetic in an Ordered Field gives $t-t_0<r_0=s-t_0$. Adding $t_0$ to both sides, again by claim 1 of that lemma, and simplifying with claim 3 of Additive Cancellation and Elementary Additive Identities in a Field, gives $t<s$, so $t\in H$. Hence $B_{d_{\mathbb{R}}}(t_0,r_0)\subseteq H$, and $H$ is open in $(\mathbb{R},d_{\mathbb{R}})$.

Therefore $H$ belongs to the Borel $\sigma$-algebra of the metric space $(\mathbb{R},d_{\mathbb{R}})$, which equals $\mathcal{B}(\mathbb{R})$ by claim 2 of Borel Measurability and Bounded Integration on a Metric Space. Consequently

Since $q\in D$ and $s\in\mathbb{Q}_{>0}$ were arbitrary, the criterion proved above applies and $Y$ is measurable with respect to $\mathcal{F}$ and $\mathcal{B}(X,d)$.

Claim 4. By claim 3 of Coordinatewise Convergence, Sequential Compactness and Density in a Product Metric Space the set $D\times D'$ is countable and dense in $X\times X'$ for the topology of subsets open in $(X\times X',d_{X\times X'})$, and it is nonempty because $D$ and $D'$ are nonempty. Hence $\bigl((X\times X',d_{X\times X'}),D\times D'\bigr)$ is a separable metric datum, which is the first assertion of claim 4. Since claims 1, 2 and 3 are asserted for every separable metric datum, they apply to it, and we use them for it below.

Let $(q,q')\in D\times D'$ and $s\in\mathbb{Q}_{>0}$. By the definition of the product metric,

By the definition of the maximum of two elements of a totally ordered set, this maximum is one of the two numbers and each of the two numbers is at most the maximum. Hence the maximum is less than $s$ if and only if both numbers are less than $s$: if both are less than $s$ then so is the maximum, being one of them; and if the maximum is less than $s$ then each number is less than $s$ by claim 2 of Elementary Order Arithmetic in an Ordered Field. Therefore

Applying claim 3 to $Y$ with the dense set $D$, and to $Y'$ with the dense set $D'$, in the direction from measurability to the displayed condition, both sets on the right belong to $\mathcal{F}$; hence so does their intersection, a $\sigma$-algebra being closed under intersections of two of its members as recorded with the $\sigma$-algebra axioms.

Since $(q,q')\in D\times D'$ and $s\in\mathbb{Q}_{>0}$ were arbitrary, claim 3 applied to $(X\times X',d_{X\times X'})$ with the dense set $D\times D'$, in the direction from the condition to measurability, shows that $Z$ is measurable with respect to $\mathcal{F}$ and $\mathcal{B}(X\times X',d_{X\times X'})$.