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.

  • $\begingroup$ 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? $\endgroup$ – user24367 Oct 25 '13 at 10:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.