# Application of Blumenthal's Zero-One Law to Brownian Motion

Let $$W_t$$ be a Brownian motion. I wish to show that the stopping time $$\tau \equiv \inf\left\{t \ge 0 : W_t >0\right\} = 0$$ almost surely.

We have $$\{\tau = 0\} = \bigcap_{k=1}^\infty \quad\bigcup_{0 \leq t < \frac{1}{k}, t \in \mathbb{Q}} \{W_t > 0\} = \bigcap_{k=m}^\infty \quad \underbrace{\bigcup_{0 \leq t < \frac{1}{k}, t \in \mathbb{Q}} \{W_t > 0\}}_{\in \mathcal{F}_{1/m}^0 \forall m \in \mathbb{N}} \in \bigcap_{m=1}^\infty \mathcal{F}_{1/m}^0 = \mathcal{F}_0^+$$

Thus by Blumenthal's zero one law, we have $$P(\tau = 0) \in \{0, 1\}$$ so it suffices to show that $$P(\tau = 0) > 0$$ but I find this impossible. Please help if you can.

Suppose that $$\mathbb{P}(\tau=0)=0$$, then $$\mathbb{P}(\exists t_0>0\, \forall t \leq t_0\::\: W_t \leq 0)=1.$$ Since $$(-W_t)_{t \geq 0}$$ is also a Brownian motion, this implies $$\mathbb{P}(\exists t_0>0\, \forall t \leq t_0\::\: W_t \geq 0)=1.$$ Hence, $$\mathbb{P}(\exists t_0>0\, \forall t \leq t_0, t \in \mathbb{Q}\::\: W_t \geq 0)=1.$$ As $$\mathbb{P}(W_t=0)=0$$ for each $$t \geq 0$$, this gives $$\mathbb{P}(\exists t_0>0\, \forall t \leq t_0, t \in \mathbb{Q}\::\: W_t > 0)=1,$$
i.e. $$\mathbb{P}(\tau=0)=1$$, which clearly contradicts our assumption.
Hence, $$\mathbb{P}(\tau=0)>0$$, and by Blumenthal's 0-1-law we conclude that $$\mathbb{P}(\tau=0)=1$$.
• I like this solution so I'll mark it as correct but I'm curious if the following also works: $P(\tau = 0) = P(\cap_{n \in \mathbb{N}}\{\tau \leq 1/n\}) = \lim_{n \rightarrow \infty} P(\{\tau \leq 1/n\}) \ge \limsup P(W_{1/n} \ge 1/n)$ $= P(N(0,1) \ge 0) = 1/2$, where the second equality holds due to nested decreasing subsets. Thanks for your solution! Apr 6, 2020 at 7:10
Because $$\{W_t>0\}\subset\{\tau\le t\}$$, you have $$1/2\le\Bbb P[\tau\le t]$$, for each $$t>0$$. It follows that $$\Bbb P[\tau=0]\ge 1/2$$.