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Consider the space $\mathbb R^{[0,1]}$ of all functions from $[0,1]$ to $\mathbb R$ and the cylindrical sigma algebra $\mathcal B$ on it. I know how to prove that $C[0,1]\not \in \mathcal B$. My question is the following: does there exist a subset of $C[0,1]$ that is in $\mathcal B$? Thank you.

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By cylindrical $\sigma$-algebra, do you mean the $\sigma$-algebra generated by the sets of the form $\{x\in \Bbb R^{[0,1]},x(t_j)\in B_j,j\in J\}$ where $J\subset [0,1]$ is finite and $B_j$ are Borel subsets of $\Bbb R$? – Davide Giraudo Sep 23 '12 at 13:18
@DavideGiraudo: You and I seem to be in lockstep this morning :) Do you want to take this one? – Nate Eldredge Sep 23 '12 at 13:19
Yes, presicely this. – Tarasenya Sep 23 '12 at 13:20
@NateEldredge I just saw your comment, and I've taken it. Hoping it's correct... – Davide Giraudo Sep 23 '12 at 13:46
up vote 2 down vote accepted

We have $$\mathcal B=\{\{x\colon [0,1]\to \Bbb R,x(t_j)\in B_j,\mbox{ for all }j\in J\}, J\subset [0,1]\mbox{ at most countable},B_j\in\mathcal B(\Bbb R),\\t_j\in [0,1]\}.$$ Indeed, this class contains the cylindrical sets and is a $\sigma$-algebra. If a $\sigma$-algebra contains the cylindrical sets, it will contain $\mathcal B$ as a countable intersection of such sets.

Now, we can see that no $S\subset C[0,1]$ (except the emptyset) is in $\mathcal B$. Indeed, $\mathcal B$ only gives conditions on the values of the map over a countable set. If $S\in\mathcal B$, let $\{t_j,j\in J\}\subset [0,1]$ as in the definition. Then taking $x(t_j)=a_j$, where $a_j\in B_j$, and $x(t)=\alpha\neq a_0$ when $t\notin \{t_j,j\in J\}$, this map is in $\mathcal B$ but not continuous (a neighborhood of $t_0$ will contain points which are not $t_j$).

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wow! and the same method seems to work when we want to prove that no subset of bounded or measurable functions, for instance, is in $\mathcal B$. Great! – Tarasenya Sep 23 '12 at 15:09
I was wondering: in which context do you use cylindrical $\sigma$-algebra? – Davide Giraudo Sep 23 '12 at 15:16
I do not understand your question. It was just an attempt to refresh some notions in my head. – Tarasenya Sep 23 '12 at 20:21
I thought for example it could be use full in probability theory, or things like that. But you answered my question: it was an exercise. – Davide Giraudo Sep 24 '12 at 7:05

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