I would like to prove the following theorem : Let $B$ be a brownian motion and $\mathcal{F_t}$ its natural filtration. Then for all $A\in\mathcal{F}_{0^{+}}$ we have $\mathbb{P}(A)\in\left\{0,1\right\}$
Here is my attempt, which has clearly a problem.
Consider $0<\delta$ and $A\in\mathcal{F}_{0^{+}}$. There exists $0<\epsilon<\delta$ such that $A\in\mathcal{F}_{\epsilon}$. Now for all $\epsilon<t<\delta$ we have $\mathcal{F}_{\epsilon}\perp \sigma(\cup_{t}\sigma(B_t - B_{\epsilon}))$. Then I infer that $\sigma(\cup_{t}\sigma(B_t - B_{\epsilon})) = \mathcal{F}_{\delta}$ but $\mathcal{F}_{\epsilon}\subset\mathcal{F}_{\delta}$. Thus $\mathbb{P}(A\cap A) = \mathbb{P}(A)\mathbb{P}(A)$ and we conclude that $\mathbb{P}(A)\in\left\{0,1\right\}$
I would like to have your opinion on this proof and what can I improve, clearly what I infer concerning $\sigma(\cup_{t}\sigma(B_t - B_{\epsilon})) = \mathcal{F}_{\delta}$ is false or at least not under a good form.
Thank you a lot