TITLE: Is $\mathcal {F}$ a sigma algebra?
QUESTION [5 upvotes]: $(C[0,1],d)$ be metric space with usual 'sup-norm' metric.

Let $(C[0,1],\mathcal {B})$ be a measurable space where $\mathcal {B}$ is Borel sigma algebra on $C[0,1]$.Let $\mathcal{F}_t=\sigma(W_s:s \in [0,t])$ where $W_s$ denotes the evaluation map.Let $\mathcal {F}= \bigcup_{ t \in [0,1)} \mathcal{F}_t$. Is $\mathcal {F}$ a $\sigma-$ algebra ?

I am having trouble in showing that $\mathcal {F}$ is closed under countable union.I really dont have any intuition whether $\mathcal {F}$ is a $\sigma-$ algebra or not? Any idea to prove or disprove this?

REPLY [3 votes]: $\mathcal F$ is an algebra but not a $\sigma$-algebra. Let $\mathcal G:=\sigma(\mathcal F)$. Clearly $\mathcal G\supset\mathcal F$. Consider now the set $B:=\{\omega\in C[0,1]: \sup_{n\in\Bbb N}W_{1-1/n}(\omega)\le W_1(\omega)\}$.Then $B\in \mathcal G$, but $B\notin\mathcal F$. To see the latter assertion, note that if $B$ were an element of $\mathcal F$, then there would be a $t_0\in(0,1)$ with $B\in\mathcal F_{t_0}$. But it's easy enough to exhibit two elements $\omega$ and $\omega'$ of $C[0,1]$ with $\omega(s)=\omega'(s)$ for all $s\in[0,t_0]$ but $\sup_{n\in\Bbb N}W_{1-1/n}(\omega)>0\ge\sup_{n\in\Bbb N}W_{1-1/n}(\omega')$, so that $\omega\notin B$, $\omega'\in B$.