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.

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.

share|improve this question
    
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
add comment

1 Answer

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 $t_j$).

share|improve this answer
1  
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
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.