[proofplan] We show $\mathbb{P}(\tau = 0) = 1$ by verifying that $\{\tau = 0\} \in \mathcal{F}_0^+$ and $\mathbb{P}(\tau = 0) \geq 1/2$, then applying [Blumenthal's Zero-One Law](/theorems/1178). The same argument gives $\inf\{t > 0 : B_t < 0\} = 0$ a.s. Path [continuity](/page/Continuity) and the [Intermediate Value Theorem](/theorems/629) then force $\mathbb{P}(\sigma = 0) = 1$. [/proofplan] [step:Show $\{\tau = 0\} \in \mathcal{F}_0^+$ and apply [Blumenthal's zero-one law](/theorems/1178)] Write the event as \begin{align*} \{\tau = 0\} = \bigcap_{k \geq 1} \bigl\{\exists\, t \in (0, 1/k) : B_t > 0\bigr\}. \end{align*} For each $k \geq 1$, the event $\{\exists\, t \in (0, 1/k) : B_t > 0\}$ belongs to $\sigma(B_u : u \leq 1/k) = \mathcal{F}_{1/k}^B$. Since $\mathcal{F}_0^+ = \bigcap_{n \geq 1} \mathcal{F}_{1/n}^B$, the intersection over all $k$ belongs to $\mathcal{F}_{1/n}^B$ for every $n \geq 1$, hence $\{\tau = 0\} \in \mathcal{F}_0^+$. By [Blumenthal's Zero-One Law](/theorems/1178), $\mathbb{P}(\tau = 0) \in \{0, 1\}$. [/step] [step:Show $\mathbb{P}(\tau = 0) \geq 1/2$ to conclude $\mathbb{P}(\tau = 0) = 1$] For any $t > 0$, the event $\{\tau \leq t\} \supset \{B_t > 0\}$, so \begin{align*} \mathbb{P}(\tau \leq t) \geq \mathbb{P}(B_t > 0) = \frac{1}{2}, \end{align*} since $B_t \sim \mathcal{N}(0, t)$ is a symmetric [distribution](/page/Distribution). Since $\{\tau = 0\} = \bigcap_{n \geq 1} \{\tau \leq 1/n\}$ and this is a decreasing intersection, [continuity of probability](/theorems/1107) gives $\mathbb{P}(\tau = 0) = \lim_{n \to \infty} \mathbb{P}(\tau \leq 1/n) \geq 1/2$. Combined with the zero-one law, $\mathbb{P}(\tau = 0) = 1$. By symmetry (replacing $B$ with $-B$, which is also a standard Brownian motion by [Invariance Properties](/theorems/1175)(i)), $\inf\{t > 0 : B_t < 0\} = 0$ almost surely. [/step] [step:Deduce $\mathbb{P}(\sigma = 0) = 1$ via the [intermediate value theorem](/theorems/629)] Since $B$ has continuous paths, $B_0 = 0$, and $B$ takes both positive and negative values in every interval $(0, \varepsilon)$ (by the previous step), the [Intermediate Value Theorem](/theorems/629) applied to the continuous [function](/page/Function) $t \mapsto B_t$ on each interval $[0, \varepsilon]$ guarantees the existence of $t \in (0, \varepsilon)$ with $B_t = 0$. Therefore $\sigma = \inf\{t > 0 : B_t = 0\} = 0$ almost surely. [/step]