# product spaces separability implies same sigma algebra

Assume we have two metric spaces $S^{\prime},\tilde{S}$ with their respective metrics $\rho^{\prime},\tilde{\rho}$. Consider the product space $S= S^{\prime}\times \tilde{S}$, with metric $\rho = \sqrt{\rho^{\prime 2}+\tilde{\rho}^{2}}$. Assume Borel sigma algebras for the spaces are $\mathcal{S^{\prime}},\mathcal{\tilde{S}},\mathcal{S}$ respectively.

Now consider the sigma algebras formed by rectangles of the form $A^{\prime} \times \tilde{A}$, lets call it $S_{1}$. It is known that $S_{1} = \mathcal{S}$ if $S$ is separable. Can anyone give counterexample?, as in the case where $S$ is not separable and the two sigma algebras are not equal.

Giving $S$ the metric $\rho$ is the same as giving it the product topology, so the question is when the Borel $\sigma$-algebra generated by the product topology is different from the product $\sigma$-algebra.
Let's look at the definitions: In the Borel $\sigma$-algebra, you first generate the topology from open rectangles (sets of the form $A \times B$, where $A$ is open in $S'$, and $B$ is open in $\tilde{S}$), then generate the $\sigma$-algebra from the topology. In the product $\sigma$-algebra, you generate the $\sigma$-algebra directly from open rectangles. The difference is that uncountable unions are involved in the former case, but not the latter: The union of uncountably many open rectangles is open, but may not be in the product $\sigma$-algebra.
Here is a specific example: Take an uncountable set $X$ and give it a metric that defines the discrete topology. You can show that the Borel $\sigma$-algebra of $X \times X$ is the full power set $\mathcal{P} (X \times X)$, while the product $\sigma$-algebra is the set of countable unions of rectangles. In particular, the set $D = \{ (x,x): x \in X \}$ is in the former, but not the latter.
• let $D$ be the space of cadlag functions on $[0,1]$.Now let $T_0$ be a dense subset of $[0,1]$. Let $F_{T_0}$ be the class of sets $\pi^{-1}_{t_1,t_2,\ldots,t_k}$, where $t_i$ belong to $T_0$. We want to show that the Skorohod topology on $D$ coincides with that generated by $F_{T_0}$. Billingsley writes that we will show that any open set in $D$ of the form of $S_{d_0}(x,r)$ will lie in the sigma algebra generated by $F_{T_0}$ and result will follow since $D$ is separable.why do we need the fact that $D$ is separable? – user24367 Oct 25 '13 at 10:59