TITLE: Generated $\sigma$-algebras with cylinder set doesn't contain the space of continuous functions
QUESTION [5 upvotes]: Consider $\mathbb R^{[0,1]}$ the space of all functions from $[0,1]$  to $\mathbb R$  and the cylindrical sigma algebra $\mathcal B$  on it. The question is: how to prove that $C[0,1]\notin \mathcal B$.

REPLY [5 votes]: Using saz's lemma (sounds great btw) and supposing that the set of continuous functions $\mathcal{C}$ is a measurable set of $\mathcal{B}$, let's us be given $S_C$ the countable set that characterizes it in the lemma.  
Now let's pick up one $\omega\in \mathcal{C}$ and take $f=\omega+ 1_{u}$ where $u\not \in S_C$. 
As $\omega|_{S_C}=f|_{S_C}$ we must have $f\in\mathcal{C}$ but by construction $f$ isn't continuous. This shows that there's no such countable set $S_C$ in $[0,1]$ and so $\mathcal{C}$ cannot be a measurable set of $\mathcal{B}$. 
Sorry to saz that deserves all the credit.
Regards