Proof of The Borel Sigma-Algebra of a Product of Two Separable Metric Spaces is the Product Sigma-Algebra

Each result cited is universally quantified over the data in its own statement.

Write $\pi_{1}:Y\times Z\to Y$ and $\pi_{2}:Y\times Z\to Z$ for the coordinate projections, $\pi_{1}(y,z)=y$ and $\pi_{2}(y,z)=z$, and write $\mathcal{P}$ for the family of measurable rectangles $A\times B$ with $A\in\mathcal{B}(Y)$ and $B\in\mathcal{B}(Z)$. By Product Sigma-Algebra, $\mathcal{B}(Y)\otimes\mathcal{B}(Z)=\sigma(\mathcal{P})$, and by Generated Sigma-Algebra the family $\sigma(\mathcal{P})$ is a $\sigma$-algebra on $Y\times Z$ which contains $\mathcal{P}$ and is contained in every $\sigma$-algebra on $Y\times Z$ containing $\mathcal{P}$. Each of $\mathcal{B}(Y)$, $\mathcal{B}(Z)$ and $\mathcal{B}(Y\times Z)$ is, by Borel Sigma-Algebra of a Metric Space, the $\sigma$-algebra generated by the open subsets of the corresponding metric space, hence a $\sigma$-algebra on $Y$, $Z$, respectively $Y\times Z$, by claim 2 of Intersections of Sigma-Algebras and Minimality of the Generated Sigma-Algebra.

Step 1: $\mathcal{B}(Y)\otimes\mathcal{B}(Z)\subseteq\mathcal{B}(Y\times Z)$. We first show that $\pi_{1}$ is continuous on $Y\times Z$ in the sense of Continuous Map Between Metric Spaces. Let $p=(y_{1},z_{1})\in Y\times Z$ and let $\varepsilon$ be a positive real number; put $\delta=\varepsilon$. Let $q=(y_{2},z_{2})\in Y\times Z$ satisfy $d_{Y\times Z}(p,q)<\delta$. Since $d_{Y\times Z}$ is a metric by claim 1 of The Product Metric is a Metric, it is symmetric, so $d_{Y\times Z}(q,p)=d_{Y\times Z}(p,q)$; and claim 2 of The Product Metric is a Metric, applied to the points $q$ and $p$ in that order, gives

$$d_{Y}\bigl(\pi_{1}(q),\pi_{1}(p)\bigr)=d_{Y}(y_{2},y_{1})\le d_{Y\times Z}(q,p)=d_{Y\times Z}(p,q)<\varepsilon .$$

Thus $\pi_{1}$ is continuous at every point of $Y\times Z$, and in the same way, using the second inequality of claim 2 of The Product Metric is a Metric, so is $\pi_{2}$. By claim 3 of Borel Measurability and Bounded Integration on a Metric Space, $\pi_{1}$ is measurable with respect to $\mathcal{B}(Y\times Z)$ and $\mathcal{B}(Y)$, and $\pi_{2}$ is measurable with respect to $\mathcal{B}(Y\times Z)$ and $\mathcal{B}(Z)$, in the sense of Measurable Function and Real-Valued Measurable Function. Now let $A\in\mathcal{B}(Y)$ and $B\in\mathcal{B}(Z)$. Then

$$A\times B=\pi_{1}^{-1}(A)\cap\pi_{2}^{-1}(B),$$

and both sets on the right belong to $\mathcal{B}(Y\times Z)$, which is closed under finite intersections by Sigma-Algebra and Measurable Space. Hence $\mathcal{P}\subseteq\mathcal{B}(Y\times Z)$, and since $\mathcal{B}(Y\times Z)$ is a $\sigma$-algebra on $Y\times Z$, the minimality recorded in Generated Sigma-Algebra gives $\sigma(\mathcal{P})\subseteq\mathcal{B}(Y\times Z)$.

Step 2: the case of an empty factor. If $Y=\varnothing$ or $Z=\varnothing$, then $Y\times Z=\varnothing$, whose only subset is $\varnothing$. Every $\sigma$-algebra on $\varnothing$ contains the whole set $\varnothing$ by property 1 of Sigma-Algebra and Measurable Space, so both $\mathcal{B}(Y\times Z)$ and $\mathcal{B}(Y)\otimes\mathcal{B}(Z)$ are equal to $\{\varnothing\}$, and the claim holds. From now on assume that $Y$ and $Z$ are both nonempty.

Step 3: separable metric data. By Separable Metric Space there is a countable subset $D_{Y}\subseteq Y$ that is dense in $Y$ for the topology $\mathcal{T}_{d_{Y}}$ of open subsets of $(Y,d_{Y})$. We check that $D_{Y}$ is nonempty. Choose $y\in Y$. The set $Y$ is open in $(Y,d_{Y})$ by Open Subset of a Metric Space, since every open ball in $Y$ is a subset of $Y$; so $Y\in\mathcal{T}_{d_{Y}}$, $y\in Y$ and $Y\cap\varnothing=\varnothing$, whence $y$ does not belong to the closure of $\varnothing$ by Closure of a Subset of a Topological Space. Since $D_{Y}$ is dense, its closure is $Y$ by Dense Subset of a Topological Space, and this closure contains $y$. If $D_{Y}$ were empty, its closure would be the closure of $\varnothing$, which does not contain $y$, a contradiction; hence $D_{Y}\ne\varnothing$. Thus $\bigl((Y,d_{Y}),D_{Y}\bigr)$ is a separable metric datum in the sense of Borel Sets and Measurable Maps in a Separable Metric Space, and in the same way there is a countable dense subset $D_{Z}$ of $Z$ with $\bigl((Z,d_{Z}),D_{Z}\bigr)$ a separable metric datum.

Step 4: $\mathcal{B}(Y\times Z)\subseteq\mathcal{B}(Y)\otimes\mathcal{B}(Z)$. Put $\Omega=Y\times Z$ and $\mathcal{F}=\sigma(\mathcal{P})=\mathcal{B}(Y)\otimes\mathcal{B}(Z)$, so that $(\Omega,\mathcal{F})$ is a measurable space. The map $\pi_{1}$ is measurable with respect to $\mathcal{F}$ and $\mathcal{B}(Y)$: for $A\in\mathcal{B}(Y)$ we have $\pi_{1}^{-1}(A)=A\times Z$, and $Z\in\mathcal{B}(Z)$ by property 1 of Sigma-Algebra and Measurable Space, so $A\times Z\in\mathcal{P}\subseteq\mathcal{F}$. Likewise $\pi_{2}^{-1}(B)=Y\times B\in\mathcal{P}\subseteq\mathcal{F}$ for $B\in\mathcal{B}(Z)$, so $\pi_{2}$ is measurable with respect to $\mathcal{F}$ and $\mathcal{B}(Z)$. Apply claim 4 of Borel Sets and Measurable Maps in a Separable Metric Space to the measurable space $(\Omega,\mathcal{F})$, the separable metric data $\bigl((Y,d_{Y}),D_{Y}\bigr)$ and $\bigl((Z,d_{Z}),D_{Z}\bigr)$ of Step 3, and the maps $\pi_{1}$ and $\pi_{2}$ in the roles of the two measurable maps of that claim: the map $\omega\mapsto(\pi_{1}(\omega),\pi_{2}(\omega))$ from $\Omega$ to $Y\times Z$ is measurable with respect to $\mathcal{F}$ and the Borel $\sigma$-algebra of $(Y\times Z,d_{Y\times Z})$, which is $\mathcal{B}(Y\times Z)$. This map is the identity map of $Y\times Z$, since $(\pi_{1}(y,z),\pi_{2}(y,z))=(y,z)$. Hence every $E\in\mathcal{B}(Y\times Z)$ satisfies $E=\mathrm{id}^{-1}(E)\in\mathcal{F}$, that is, $\mathcal{B}(Y\times Z)\subseteq\mathcal{B}(Y)\otimes\mathcal{B}(Z)$.

Steps 1 and 4 together give $\mathcal{B}(Y\times Z)=\mathcal{B}(Y)\otimes\mathcal{B}(Z)$ when $Y$ and $Z$ are nonempty, and Step 2 covers the remaining case.