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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?

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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 $C$.

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