Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've come across a statement that i cannot grasp.

When $T=[0,\infty)$, $E=\mathbb{R}$ and $\xi=B(\mathbb{R})$ then the collection of all continuous E-valued functions is not $\xi^T$-measurable.

Here we have the measurable space $(E,\xi)$ and $\xi^n=\sigma\{E_{1}\times...\times E_{n}|E_{1},...,E_{n}\in \xi\} $.

Could anyone prove why this is the case?

share|improve this question
add comment

1 Answer 1

The argument is similar to the one given here.

This follows from the general result that if $A\in\sigma(\mathcal{F})$, then there exists a countable family $\mathcal{C}\subseteq \mathcal{F}$ such that $A\in\sigma(\mathcal{C})$. One can show this by verifying that the family of sets generated by a countable sub-family of $\mathcal{F}$ forms a $\sigma$-algebra that contains $\mathcal{F}$.

Applied to our case, an event in the product $\sigma$-algebra must be generated by countably many coordinate projections. Let $\mathscr{C}$ be the set of continuous functions from $T$ to $E$. If there would be a countable set of coordinates $C\subseteq T$ such that $\mathscr{C}\in\sigma\{\pi_n:n\in C\}$ and $f$ is continuous, then a function $g$ that agrees with $f$ on $C$ will also be in $\mathscr{C}$. But if $E$ has at least two points, there will be a function that is discontinuous but agrees with $f$ on $C$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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