[proofplan] Closedness of $\mathcal{Z}$ is immediate from path [continuity](/page/Continuity). The main content is that $\mathcal{Z}$ has no isolated points: we show this first for zeros of the form $\tau_q$ (using the strong Markov property and [Immediate Return to Zero](/theorems/1179)) and then extend to arbitrary zeros by approximation from the left with rational times. For the measure-zero claim, we use the occupation times formula. [/proofplan] [step:Verify $\mathcal{Z}$ is closed] The zero set $\mathcal{Z} = \{t \geq 0 : B_t = 0\} = B^{-1}(\{0\})$ is the preimage of the [closed set](/page/Closed%20Set) $\{0\}$ under the continuous map $t \mapsto B_t(\omega)$. Since preimages of closed [sets](/page/Set) under continuous maps are closed, $\mathcal{Z}$ is closed almost surely. [/step] [step:Show no zero of the form $\tau_q$ is isolated, using the strong Markov property] For each $q \in \mathbb{Q} \cap [0, \infty)$, define the stopping time $\tau_q = \inf\{t \geq q : B_t = 0\}$. By [Recurrence-Transience](/theorems/1185)(i), $\tau_q < \infty$ almost surely for each $q$, and $B_{\tau_q} = 0$ by path continuity. By the [Strong Markov Property](/theorems/1180) at $\tau_q$, the process $(B_{\tau_q + t})_{t \geq 0} = (B_{\tau_q + t} - B_{\tau_q})_{t \geq 0}$ is a standard Brownian motion (started at $0$). By [Immediate Return to Zero](/theorems/1179), this process satisfies $\inf\{t > 0 : B_{\tau_q + t} = 0\} = 0$ almost surely. Therefore in every interval $(\tau_q, \tau_q + \varepsilon)$ there exist zeros of $B$, so $\tau_q$ is not isolated from the right. This holds simultaneously for all $q \in \mathbb{Q} \cap [0, \infty)$ almost surely, since we take a countable intersection of almost sure events. [guided] The strong Markov property is applied here at the random time $\tau_q$, which is indeed a stopping time for the right-continuous filtration $(\mathcal{F}_t^+)$: the event $\{\tau_q \leq t\} = \{\inf_{s \in [q, t]} |B_s| = 0\}$ belongs to $\mathcal{F}_t^+$ because $B$ has continuous paths and the infimum over a closed interval can be computed from the path up to time $t$. The conclusion gives non-isolation from the right. What about non-isolation from the left? This is handled in the next step by a different argument — one cannot directly apply the strong Markov property "backwards in time." [/guided] [/step] [step:Show that every zero is a [limit](/page/Limit) of zeros from the left] Let $t > 0$ with $B_t(\omega) = 0$. We show $t$ is not isolated from the left. Choose rational numbers $q_n \uparrow t$ with $q_n < t$. For each $n$, $\tau_{q_n} = \inf\{s \geq q_n : B_s = 0\}$ satisfies $q_n \leq \tau_{q_n} \leq t$ (since $t \geq q_n$ and $B_t = 0$, the infimum is at most $t$). **Case 1:** If $\tau_{q_n} = t$ for all sufficiently large $n$, then by the previous step, $t = \tau_{q_n}$ is not isolated from the right. For non-isolation from the left: since $\tau_{q_n} = t$ and $q_n < t$, there are no zeros in $[q_n, t)$, so we need a different argument. But note that $t$ cannot be isolated from the left unless there exists an interval $(t - \varepsilon, t)$ with no zeros. If such an interval existed, then $B$ would have constant sign on $(t - \varepsilon, t)$ (by the [Intermediate Value Theorem](/theorems/629), since $B$ is continuous and any sign change would produce a zero). But then by [Immediate Return to Zero](/theorems/1179) applied at $t$ (using the strong Markov property), $B$ has zeros in $(t, t + \eta)$ for every $\eta > 0$, and by continuity, $B$ must change sign near $t$, forcing zeros in $(t - \varepsilon, t)$ as well — a contradiction. **Case 2:** If $\tau_{q_n} < t$ for infinitely many $n$, then we have zeros $\tau_{q_n} \in [q_n, t)$ with $\tau_{q_n} \to t$ (since $q_n \to t$ forces $\tau_{q_n} \to t$). So $t$ is a limit of zeros from the left. In both cases, $t$ is not isolated. Therefore $\mathcal{Z}$ has no isolated points, making it a perfect set. [guided] The subtlety here is that the strong Markov property gives non-isolation from the right (future zeros accumulate at any given zero) but not directly from the left. The left-hand approximation requires the indirect argument of squeezing $\tau_{q_n}$ between $q_n$ and $t$. Why is $\mathcal{Z}$ perfect rather than dense? Because Brownian motion spends intervals of time away from zero — the excursion intervals. Between any two zeros, $B$ makes an excursion of some positive length. The complement $[0, \infty) \setminus \mathcal{Z}$ is open (since $\mathcal{Z}$ is closed) and is a countable union of open intervals (the excursion intervals). So $\mathcal{Z}$ is a closed, nowhere-dense set with no isolated points — a Cantor-like set (though not self-similar in the strict sense). [/guided] [/step] [step:Show $\mathcal{Z}$ has Lebesgue measure zero] By Fubini's theorem, \begin{align*} \mathbb{E}\!\left[\mathcal{L}^1(\mathcal{Z} \cap [0, T])\right] = \mathbb{E}\!\left[\int_0^T \mathbb{1}_{\{B_t = 0\}} \, d\mathcal{L}^1(t)\right] = \int_0^T \mathbb{P}(B_t = 0) \, d\mathcal{L}^1(t). \end{align*} For $t > 0$, $B_t \sim \mathcal{N}(0, t)$ has a density with respect to $\mathcal{L}^1$, so $\mathbb{P}(B_t = 0) = 0$. For $t = 0$, the single point $\{0\}$ has $\mathcal{L}^1$-measure zero. Therefore $\int_0^T \mathbb{P}(B_t = 0) \, d\mathcal{L}^1(t) = 0$, giving $\mathbb{E}[\mathcal{L}^1(\mathcal{Z} \cap [0, T])] = 0$. Since a non-negative random variable with zero expectation is zero almost surely, $\mathcal{L}^1(\mathcal{Z} \cap [0, T]) = 0$ a.s. for each $T > 0$. Taking $T \to \infty$ (via countable union $\mathcal{Z} = \bigcup_{n=1}^\infty (\mathcal{Z} \cap [0,n])$) gives $\mathcal{L}^1(\mathcal{Z}) = 0$ almost surely. [/step]