I am trying to understand in which metric spaces the metric is jointly measurable. 


There exist a metric space $(X,d)$ for which the Borel $\sigma$-algebra, does not coincide with the product Borel $\sigma$-algebra, that is $\mathcal{B}(X)\otimes \mathcal{B}(X) \subsetneq B(X\times X)$. Every construction, I have encountered, of such a (non-separable) metric space one considers a metric space of cardinality strictly greater than the continuum $\mathfrak{c}$. For in such a metric space the diagonal $I=\{(x,x):x\in X\}\in B(X\times X)$, is not an element of $\mathcal{B}(X)\otimes \mathcal{B}(X)$ (c.f. Nedoma's pathology). This also proves that $d$ is not jointly measurable, since $d^{-1}(\{0\})=I\not\in\mathcal{B}(X)\otimes \mathcal{B}(X)$.


For metric spaces $(X,d)$ with $\text{card}(X)\leq \aleph\_0$, we obviously have that $d$ is jointly measurable, but the area of interest is : What happends when the cardinality is $\aleph\_0 < \text{card}(X) \leq \mathfrak{c}$? 


More specifically, I have been wondering about the following questions: Let $(X,d)$ be a metric space,


1. Is the metric $d$ jointly-measurable, if $\aleph\_0 < \text{card}(X) \leq \mathfrak{c}$?
2. If not, is there equivalence between joint measurability of $d$ and separability of $X$?
3. If not, do you know property of $(X,d)$, weaker than separability, that ensures joint-measurability of $d$?



Joint measurability of metric
One can probably not go too far from separability, there must always exist a sub-$\sigma$-algebra that looks like the Borel $\sigma$-algebra of a separable metric space. 


If $\mathcal{A}$ is a family of subsets of a set $X$ and $A\in\sigma(A)$, then there exists a countable family $\mathcal{C}\subseteq\mathcal{A}$ such that $A\in\sigma(\mathcal{C})$. In particular, if the diagonal $D\subseteq X\times X$ is in the $\sigma$-algebra generated by measurable rectangles, then the countable family of sets $\mathcal{C}$ must have the property that for each $x,y\in X$ with $x\neq y$, there is $C\in\mathcal{C}$ such that $x\in C$ and $y\notin C$ or $x\notin C$ and $y\in C$. 


Let $\langle C\_n\rangle$ be a sequence that lists all elements of $\mathcal{C}$. Then the Marczewski-funtion $g:X\to\mathbb{R}$ given by
$$g(x)=\sum\_n 1/3^n 1\_{C\_n}(x)$$ 
has the property that $\sigma(\mathcal{C})$ consists exactly of inverse images of Borel sets in $\mathbb{R}$. Since $\mathcal{C}$ separates points, it is injective. Then $d:X\times X\to\mathbb{R}$ given by $d(x,y)=|g(x)-g(y)|$ turns $X$ into a separable metric space with Borel $\sigma$-algebra $\sigma(\mathcal{C})$.


However, separability is not necessary. Endow $\omega\_1$ with the discrete metric, so it becomes a non-separable metric space with Borel $\sigma$-algebra $2^{\omega\_1}$. In this case, the metric is still jointly measurable, for one can show that $2^{\omega\_1}\otimes2^{\omega\_1}=2^{\omega\_1\times\omega\_1}$. The proof can be found in: 


V. Rao, [On discrete Borel spaces and projective sets](https://projecteuclid.org/download/pdf_1/euclid.bams/1183530570) Bull. Amer. Math. Soc. Volume 75, Number 3 (1969), 614-617.


