# A nonmeasurable set on $\mathbb{R}^{\left[0,\infty\right)}$.

Let $\left\{ X_{t}\right\} _{t\geq o}$ the canonical version of Brownian motion, i.e., if we consider $\Omega:=\mathbb{R}^{\left[0,\infty\right)}$ the set of the real valued functions on $\left[0,\infty\right)$ and for each $t\geq0$, $X_{t}\left(\omega\right)=\omega\left(t\right)$ and $\mathcal{F}:=\sigma\left\{ \omega\in\Omega|X_{t_{1}}\left(\omega\right)\in B_{1},\ldots,X_{t_{k}}\left(\omega\right)\in B_{k},B_{i}\in\mathcal{B}\left(\mathbb{R}\right)\right\}$ then the process $\left\{ X_{t}\right\} _{t\geq o}$ is called the canonical version of Brownian motion. Proof that

$$A=\left\{ \omega\in\Omega|\sup_{a\leq t\leq b}X_{t}\left(\omega\right)<c\right\} \notin\mathcal{F}$$,

for some $c\in\mathbb{R}$.

I cheked Introduction to Stochastic Integration-Kuo Hui-Hsiung, but here there is a particular case. However, I think I have to find a nonmeasurable set on $\mathcal{F}$ and then proceed in the similar way, but I have been unable to make any headway in that respect.

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Hint: Elements of $\mathcal F$ depend on $X_t$ for $t$ in at most a countable subset of $[0,\infty)$. Your set $A$ does not. – Did Sep 3 '12 at 18:33
Eugenio: Did you reach an answer, based on my hint? – Did Sep 9 '12 at 15:12

Hint: Elements of $F$ depend on $X_t$ for $t$ in at most a countable subset of $[0,\infty)$. Your set $A$ does not.