[proofplan] We verify three properties of the sequence $(S_n)$ defined by $S_n = \inf\{t \geq 0 : |X_t| = n\}$: (i) each $S_n$ is a stopping time, (ii) $S_n \to \infty$ a.s., and (iii) $X^{S_n}$ is a bounded martingale for each $n$. Property (i) follows from the fact that the hitting time of a closed set by a continuous adapted process is a stopping time w.r.t. the right-continuous filtration. Property (ii) follows from the continuity of $X$ on compact intervals, which forces $\sup_{s \leq t} |X_s| < \infty$ a.s. for each $t$, and hence $S_n \geq t$ for large $n$. Property (iii) uses the bound $|X^{S_n}_t| \leq n$ together with the [Dominated Local Martingale is a Martingale](/theorems/2079) result: a bounded local martingale is a genuine martingale. [/proofplan] [step:Show each $S_n$ is a stopping time with respect to the filtration] For each $n \geq 1$, define \begin{align*} S_n := \inf\{t \geq 0 : |X_t| = n\}. \end{align*} Since $X$ is a continuous adapted process and the set $\{x \in \mathbb{R} : |x| = n\}$ is closed, $S_n$ is the first hitting time of a closed set by a continuous adapted process. For a right-continuous filtration $(\mathcal{F}_t)_{t \geq 0}$, the hitting time of any closed set by a continuous adapted process is a stopping time. To verify directly: \begin{align*} \{S_n < t\} = \bigcup_{s \in [0, t) \cap \mathbb{Q}} \{|X_s| \geq n\} \in \mathcal{F}_t \end{align*} for each $t > 0$. Here we used continuity of $X$: if $|X_s| = n$ for some $s < t$, then there exist rational times $q < t$ with $|X_q|$ arbitrarily close to $n$, and by continuity $|X_q| \geq n$ for some rational $q$ (in fact $|X_s| = n$ is achieved, but $|X_q| \geq n$ suffices since $|X_s| = n$ and $X$ is continuous, so $|X|$ must reach $n$ at or before any close rational). Since the filtration is right-continuous, \begin{align*} \{S_n \leq t\} = \bigcap_{k=1}^\infty \{S_n < t + 1/k\} \in \mathcal{F}_t, \end{align*} confirming $S_n$ is a stopping time. [guided] We need to show that $S_n = \inf\{t \geq 0 : |X_t| = n\}$ is a stopping time with respect to the right-continuous filtration $(\mathcal{F}_t)_{t \geq 0}$. For general adapted processes, the hitting time of a set can fail to be a stopping time. However, continuity of $X$ simplifies the situation considerably. For a continuous process, the event $\{S_n < t\}$ — "the process $|X|$ has reached level $n$ strictly before time $t$" — can be detected by observing $X$ at rational times: \begin{align*} \{S_n < t\} = \bigcup_{s \in [0, t) \cap \mathbb{Q}} \{|X_s| \geq n\}. \end{align*} Why does this equality hold? The inclusion $\supseteq$ is immediate: if $|X_q| \geq n$ for some rational $q < t$, then $S_n \leq q < t$. For the reverse inclusion $\subseteq$: if $S_n < t$, then there exists $s < t$ with $|X_s| = n$. Since $X$ is continuous and $|X_0| = 0 < n$, the intermediate value theorem applied to $|X|$ on $[0, s]$ shows that $|X|$ must pass through the value $n$. By continuity of $|X|$ at the time $S_n$, for any $\varepsilon > 0$ we have $|X_{S_n + \varepsilon}|$ close to $n$, so there exist rational times $q \in [0, t)$ with $|X_q| \geq n$. Each set $\{|X_s| \geq n\}$ is $\mathcal{F}_s$-measurable (since $X$ is adapted) and hence $\mathcal{F}_t$-measurable (since $s < t$ implies $\mathcal{F}_s \subseteq \mathcal{F}_t$). The countable union is therefore $\mathcal{F}_t$-measurable, so $\{S_n < t\} \in \mathcal{F}_t$. To get $\{S_n \leq t\} \in \mathcal{F}_t$ (the stopping time condition), we use right-continuity of the filtration: \begin{align*} \{S_n \leq t\} = \bigcap_{k=1}^\infty \{S_n < t + 1/k\} \in \bigcap_{k=1}^\infty \mathcal{F}_{t + 1/k} = \mathcal{F}_{t+} = \mathcal{F}_t, \end{align*} where the last equality is the right-continuity assumption $\mathcal{F}_t = \mathcal{F}_{t+}$. [/guided] [/step] [step:Prove $S_n \to \infty$ a.s. using compactness and continuity] The sequence $(S_n)_{n \geq 1}$ is increasing: since $|X_t| = n$ implies $|X_t| = n+1$ only if $n = n+1$ (which is impossible), we have $S_n \leq S_{n+1}$ (the level $n+1$ cannot be hit before level $n$ because $X$ is continuous with $X_0 = 0$, so $|X|$ must pass through $n$ before reaching $n+1$). Fix $t \geq 0$. Since $X$ is continuous and $[0, t]$ is compact, the function $s \mapsto X_s(\omega)$ attains its maximum on $[0, t]$ for each $\omega$. Therefore \begin{align*} \sup_{0 \leq s \leq t} |X_s(\omega)| < \infty \quad \text{for every } \omega \in \Omega. \end{align*} For any $n > \sup_{0 \leq s \leq t} |X_s(\omega)|$, we have $|X_s(\omega)| < n$ for all $s \in [0, t]$, so $S_n(\omega) > t$. This shows $S_n(\omega) \to \infty$ for every $\omega$, hence a.s. [guided] We need $S_n \to \infty$ a.s. First, note that the sequence is increasing. Why? Since $X$ is continuous with $X_0 = 0$, and $|X_{S_n}| = n$ (when $S_n < \infty$), the process $|X|$ must traverse the value $n$ before reaching $n + 1$. More precisely, if $S_{n+1} < S_n$, then $|X_{S_{n+1}}| = n + 1 > n$, but $|X_0| = 0 < n$, and by continuity $|X|$ would have to pass through $n$ at some time $s \leq S_{n+1} < S_n$, contradicting the definition of $S_n$ as the infimum. So $S_n \leq S_{n+1}$. Now fix $\omega \in \Omega$ and $t \geq 0$. The path $s \mapsto X_s(\omega)$ is continuous on the compact interval $[0, t]$, so it is bounded: there exists $M(\omega, t) < \infty$ such that $|X_s(\omega)| \leq M(\omega, t)$ for all $s \in [0, t]$. For any $n > M(\omega, t)$, the process $|X(\omega)|$ never reaches $n$ on $[0, t]$, so $S_n(\omega) > t$. Since this holds for every $t$, taking $t \to \infty$ gives $\lim_{n \to \infty} S_n(\omega) = \infty$. This argument works for every $\omega$ (not just a.s.), because continuity on compact intervals is a pointwise-in-$\omega$ property of the sample paths. [/guided] [/step] [step:Prove $X^{S_n}$ is a bounded martingale via the dominated local martingale theorem] Fix $n \geq 1$. The stopped process $X^{S_n}$ is defined by $X^{S_n}_t = X_{t \wedge S_n}$. We first establish the bound $|X^{S_n}_t| \leq n$ for all $t \geq 0$. If $t < S_n$, then $|X_t| < n$ by definition of $S_n$ (since $S_n$ is the first time $|X|$ reaches $n$, and it has not done so by time $t$). If $t \geq S_n$, then $X^{S_n}_t = X_{S_n}$, and $|X_{S_n}| = n$ by continuity of $X$ (the continuous process $|X|$ hits the level $n$ exactly at time $S_n$, since it starts at $|X_0| = 0 < n$). In both cases, $|X^{S_n}_t| \leq n$. Next, $X^{S_n}$ is a local martingale. Let $(T_m)_{m \geq 1}$ be a reducing sequence for $X$. Then $X^{T_m}$ is a martingale for each $m$, and by the optional stopping theorem applied to the bounded stopping time $S_n \wedge T_m$, the process $X^{T_m \wedge S_n}$ is also a martingale. Since $T_m \wedge S_n \to S_n$ as $m \to \infty$, the sequence $(T_m \wedge S_n)$ serves as a reducing sequence for $X^{S_n}$. Since $X^{S_n}$ is a local martingale with $|X^{S_n}_t| \leq n$ for all $t$, the [Dominated Local Martingale is a Martingale](/theorems/2079) theorem (with $Z = n$) shows that $X^{S_n}$ is a genuine martingale. Moreover, it is bounded by $n$. [guided] Fix $n \geq 1$. We need to show two things: (a) $X^{S_n}$ is a local martingale, and (b) $X^{S_n}$ is bounded, so that it is a genuine martingale. **Boundedness.** For $t \geq 0$, we have $X^{S_n}_t = X_{t \wedge S_n}$. There are two cases: - If $t < S_n(\omega)$: by definition of $S_n$, $|X_s(\omega)| < n$ for all $s < S_n(\omega)$. (Why? If $|X_s(\omega)| = n$ for some $s < S_n(\omega)$, this would contradict the definition of $S_n$ as the infimum of such times.) So $|X_t(\omega)| < n$, hence $|X^{S_n}_t(\omega)| < n \leq n$. - If $t \geq S_n(\omega)$: then $X^{S_n}_t(\omega) = X_{S_n(\omega)}(\omega)$. We claim $|X_{S_n}| = n$. Since $X$ is continuous and $|X_0| = 0 < n$, the set $\{t \geq 0 : |X_t| = n\}$ is closed (as the preimage of the closed set $\{n\}$ under the continuous function $t \mapsto |X_t|$). Therefore $S_n = \inf\{t : |X_t| = n\}$ belongs to this set (the infimum of a non-empty closed subset of $[0, \infty]$), so $|X_{S_n}| = n$. Hence $|X^{S_n}_t| = n$. In both cases $|X^{S_n}_t| \leq n$. **Local martingale property.** Let $(T_m)$ be a reducing sequence for $X$. The process $X^{T_m}$ is a martingale. Since $S_n$ is a stopping time, the stopped process $(X^{T_m})^{S_n} = X^{T_m \wedge S_n}$ is also a martingale (stopping a martingale at a stopping time yields a martingale). The sequence $(T_m \wedge S_n)_{m \geq 1}$ is increasing and satisfies $T_m \wedge S_n \to S_n$ as $m \to \infty$ (since $T_m \to \infty$). For $m$ large enough that $T_m \geq S_n$ a.s., we have $T_m \wedge S_n = S_n$, but even before that, $(T_m \wedge S_n)$ serves as a reducing sequence for $X^{S_n}$ (since $X^{T_m \wedge S_n} = (X^{S_n})^{T_m}$ is a martingale). So $X^{S_n}$ is a local martingale bounded by $n$. The [Dominated Local Martingale is a Martingale](/theorems/2079) theorem applies with $Z = n$ (a constant, which is in $L^1$): since $|X^{S_n}_t| \leq n = Z$ for all $t$, the process $X^{S_n}$ is a genuine martingale. This completes the proof: $(S_n)$ is an increasing sequence of stopping times with $S_n \to \infty$ a.s., and $X^{S_n}$ is a (bounded) martingale for each $n$. In particular, $(S_n)$ is a reducing sequence for $X$. [/guided] [/step]