[proofplan]
We reduce to a bounded martingale via stopping, then show the second moment vanishes. Let $V_t$ denote the total variation process and $S_n = \inf\{t : V_t \geq n\}$. The stopped process $X^{S_n}$ is a bounded martingale (since $|X^{S_n}_t| \leq V_{t \wedge S_n} \leq n$). By the orthogonal increments property of $L^2$ martingales, $\mathbb{E}[(X^{S_n}_t)^2]$ equals a sum of squared increments over any partition. We bound each squared increment by the maximum oscillation times the corresponding absolute increment, yielding $\mathbb{E}[(X^{S_n}_t)^2] \leq n \cdot \mathbb{E}[\max_i |X^{S_n}_{t_i} - X^{S_n}_{t_{i-1}}|]$. By uniform continuity of $X$ on compact intervals, the maximum oscillation tends to zero as the mesh refines; dominated convergence then gives $\mathbb{E}[(X^{S_n}_t)^2] = 0$, hence $X^{S_n}_t = 0$ a.s. Sending $n \to \infty$ completes the proof.
[/proofplan]
[step:Construct stopping times $S_n$ and show $X^{S_n}$ is a bounded martingale]
Let $(V_t)_{t \geq 0}$ denote the total variation process of $X$:
\begin{align*}
V_t(\omega) := \sup \left\{ \sum_{i=1}^k |X_{t_i}(\omega) - X_{t_{i-1}}(\omega)| : 0 = t_0 < t_1 < \cdots < t_k = t, \; k \geq 1 \right\}.
\end{align*}
Since $X$ is a finite variation process, $V_t(\omega) < \infty$ for all $(\omega, t)$. Define
\begin{align*}
S_n := \inf\{t \geq 0 : V_t \geq n\}.
\end{align*}
Each $S_n$ is a stopping time (as the hitting time of the level $n$ by the continuous increasing adapted process $V$). Since $V_t < \infty$ for all $t$ a.s., we have $S_n \to \infty$ a.s. The process $V$ is continuous because $X$ is continuous and has finite variation (a continuous BV function has continuous total variation function).
The stopped process $X^{S_n}$ satisfies $|X^{S_n}_t| = |X_{t \wedge S_n}|$. Since $X_0 = 0$ and $X$ has total variation $V$, the triangle inequality gives
\begin{align*}
|X_{t \wedge S_n}| = |X_{t \wedge S_n} - X_0| \leq V_{t \wedge S_n} \leq n.
\end{align*}
Since $X$ is a continuous local martingale, $X^{S_n}$ is also a continuous local martingale. As $|X^{S_n}_t| \leq n$ for all $t$, the [Dominated Local Martingale is a Martingale](/theorems/2079) theorem shows that $X^{S_n}$ is a genuine martingale, bounded by $n$.
[guided]
We need to stop $X$ to get a bounded martingale. The natural stopping involves the total variation.
Define the total variation process
\begin{align*}
V_t(\omega) := \sup \left\{ \sum_{i=1}^k |X_{t_i}(\omega) - X_{t_{i-1}}(\omega)| : 0 = t_0 < t_1 < \cdots < t_k = t, \; k \geq 1 \right\}.
\end{align*}
Since $X$ is a finite variation process, $V_t(\omega) < \infty$ for all $(\omega, t)$. The process $V$ is increasing, adapted, and continuous. Why is $V$ continuous? Because $X$ is continuous: if $X$ has a continuous path of bounded variation, the total variation function $t \mapsto V_t$ is also continuous. (If $V$ had a jump at some time $t_0$, then $\lim_{t \uparrow t_0} V_t < V_{t_0}$, which would mean $X$ has a jump at $t_0$, contradicting continuity.)
Set $S_n = \inf\{t \geq 0 : V_t \geq n\}$. Since $V$ is continuous and increasing:
- $S_n$ is a stopping time (the hitting time of $[n, \infty)$ by a continuous adapted process).
- $S_n \to \infty$ a.s. (since $V_t < \infty$ for each $t$ a.s., so for any fixed $t$, $V_t < n$ for large $n$, hence $S_n > t$).
Now we bound $X^{S_n}$. For any $t \geq 0$,
\begin{align*}
|X_{t \wedge S_n}| = |X_{t \wedge S_n} - X_0| \leq \int_0^{t \wedge S_n} |dX_s| = V_{t \wedge S_n} \leq n,
\end{align*}
where the inequality uses $X_0 = 0$, the first equality is the definition of total variation (the total variation of $X$ on $[0, t \wedge S_n]$ bounds the net displacement), and $V_{t \wedge S_n} \leq n$ holds by definition of $S_n$ (if $t \wedge S_n < S_n$, then $V_{t \wedge S_n} < n$; if $t \wedge S_n = S_n$, then $V_{S_n} = n$ by continuity of $V$).
Since $X^{S_n}$ is a local martingale (as $X$ is a local martingale and stopping preserves the local martingale property) and $|X^{S_n}_t| \leq n$, the [Dominated Local Martingale is a Martingale](/theorems/2079) theorem gives that $X^{S_n}$ is a bounded martingale.
[/guided]
[/step]
[step:Express $\mathbb{E}[(X^{S_n}_t)^2]$ as a sum of squared increments using the martingale property]
Fix $n \geq 1$ and $t \geq 0$. Since $X^{S_n}$ is a bounded martingale (hence in $L^2$), its increments over disjoint intervals are orthogonal in $L^2(\mathbb{P})$: for $u < v \leq w < z$,
\begin{align*}
\mathbb{E}\!\left[(X^{S_n}_v - X^{S_n}_u)(X^{S_n}_z - X^{S_n}_w)\right] = \mathbb{E}\!\left[\mathbb{E}\!\left[(X^{S_n}_z - X^{S_n}_w)(X^{S_n}_v - X^{S_n}_u) \;\Big|\; \mathcal{F}_w\right]\right] = \mathbb{E}\!\left[(X^{S_n}_v - X^{S_n}_u) \cdot 0\right] = 0,
\end{align*}
since $\mathbb{E}[X^{S_n}_z - X^{S_n}_w \mid \mathcal{F}_w] = 0$ by the martingale property, and $X^{S_n}_v - X^{S_n}_u$ is $\mathcal{F}_w$-measurable.
For any partition $\pi: 0 = t_0 < t_1 < \cdots < t_k = t$, expanding the telescoping sum $X^{S_n}_t = \sum_{i=1}^k (X^{S_n}_{t_i} - X^{S_n}_{t_{i-1}})$ and using orthogonality of increments (with $X^{S_n}_0 = X_0 = 0$):
\begin{align*}
\mathbb{E}\!\left[(X^{S_n}_t)^2\right] = \sum_{i=1}^k \mathbb{E}\!\left[(X^{S_n}_{t_i} - X^{S_n}_{t_{i-1}})^2\right].
\end{align*}
[guided]
Fix $n \geq 1$ and $t \geq 0$. The bounded martingale $X^{S_n}$ is in $L^2(\mathbb{P})$ (since $|X^{S_n}_t| \leq n$ implies $\mathbb{E}[(X^{S_n}_t)^2] \leq n^2$).
We use the fundamental orthogonality property of martingale increments. If $M$ is a martingale with $M \in L^2$, then for $s < t$, the increment $M_t - M_s$ is orthogonal in $L^2$ to any $\mathcal{F}_s$-measurable random variable $Y$ with $\mathbb{E}[Y^2] < \infty$:
\begin{align*}
\mathbb{E}[(M_t - M_s) \cdot Y] = \mathbb{E}\!\left[\mathbb{E}[(M_t - M_s) \mid \mathcal{F}_s] \cdot Y\right] = \mathbb{E}[0 \cdot Y] = 0.
\end{align*}
In particular, for disjoint intervals $[t_{i-1}, t_i]$ and $[t_{j-1}, t_j]$ with $t_i \leq t_{j-1}$, the increments $M_{t_i} - M_{t_{i-1}}$ and $M_{t_j} - M_{t_{j-1}}$ are orthogonal (since the first is $\mathcal{F}_{t_i} \subseteq \mathcal{F}_{t_{j-1}}$-measurable).
Now take any partition $\pi: 0 = t_0 < t_1 < \cdots < t_k = t$. Since $X^{S_n}_0 = X_0 = 0$, we have $X^{S_n}_t = \sum_{i=1}^k (X^{S_n}_{t_i} - X^{S_n}_{t_{i-1}})$. Squaring and taking expectations, the cross terms vanish by orthogonality:
\begin{align*}
\mathbb{E}\!\left[(X^{S_n}_t)^2\right] &= \mathbb{E}\!\left[\left(\sum_{i=1}^k (X^{S_n}_{t_i} - X^{S_n}_{t_{i-1}})\right)^2\right] \\
&= \sum_{i=1}^k \mathbb{E}\!\left[(X^{S_n}_{t_i} - X^{S_n}_{t_{i-1}})^2\right] + 2\sum_{i < j} \underbrace{\mathbb{E}\!\left[(X^{S_n}_{t_i} - X^{S_n}_{t_{i-1}})(X^{S_n}_{t_j} - X^{S_n}_{t_{j-1}})\right]}_{= 0} \\
&= \sum_{i=1}^k \mathbb{E}\!\left[(X^{S_n}_{t_i} - X^{S_n}_{t_{i-1}})^2\right].
\end{align*}
This identity holds for every partition $\pi$ of $[0, t]$.
[/guided]
[/step]
[step:Bound the sum of squared increments by the maximum oscillation times the total variation]
Continuing from the previous step, bound each squared increment by the maximum oscillation times the absolute increment:
\begin{align*}
\sum_{i=1}^k (X^{S_n}_{t_i} - X^{S_n}_{t_{i-1}})^2 &\leq \max_{1 \leq i \leq k} |X^{S_n}_{t_i} - X^{S_n}_{t_{i-1}}| \cdot \sum_{i=1}^k |X^{S_n}_{t_i} - X^{S_n}_{t_{i-1}}|.
\end{align*}
The second factor is bounded by the total variation of $X^{S_n}$ on $[0, t]$:
\begin{align*}
\sum_{i=1}^k |X^{S_n}_{t_i} - X^{S_n}_{t_{i-1}}| \leq V_{t \wedge S_n} \leq n.
\end{align*}
Taking expectations:
\begin{align*}
\mathbb{E}\!\left[(X^{S_n}_t)^2\right] \leq \mathbb{E}\!\left[\max_{1 \leq i \leq k} |X^{S_n}_{t_i} - X^{S_n}_{t_{i-1}}| \cdot V_{t \wedge S_n}\right] \leq n \cdot \mathbb{E}\!\left[\max_{1 \leq i \leq k} |X^{S_n}_{t_i} - X^{S_n}_{t_{i-1}}|\right].
\end{align*}
[/step]
[step:Send the mesh to zero using uniform continuity and dominated convergence]
Since $X$ is continuous and $[0, t]$ is compact, the path $s \mapsto X_s(\omega)$ is uniformly continuous on $[0, t]$ for each $\omega$. Therefore $s \mapsto X^{S_n}_s(\omega) = X_{s \wedge S_n(\omega)}(\omega)$ is also uniformly continuous on $[0, t]$ (as the composition of a uniformly continuous function with $s \mapsto s \wedge S_n(\omega)$, which is Lipschitz with constant $1$).
Let $(\pi_m)_{m \geq 1}$ be a sequence of partitions of $[0, t]$ with mesh $|\pi_m| \to 0$. By uniform continuity,
\begin{align*}
\max_{1 \leq i \leq k_m} |X^{S_n}_{t_i^{(m)}} - X^{S_n}_{t_{i-1}^{(m)}}| \to 0 \quad \text{a.s. as } m \to \infty.
\end{align*}
Since $|X^{S_n}_{t_i} - X^{S_n}_{t_{i-1}}| \leq 2n$ for all $i$ (because $|X^{S_n}| \leq n$), the maximum is bounded by $2n$ deterministically. By the dominated convergence theorem (with the constant dominator $2n$),
\begin{align*}
\mathbb{E}\!\left[\max_{1 \leq i \leq k_m} |X^{S_n}_{t_i^{(m)}} - X^{S_n}_{t_{i-1}^{(m)}}|\right] \to 0 \quad \text{as } m \to \infty.
\end{align*}
Since the bound $\mathbb{E}[(X^{S_n}_t)^2] \leq n \cdot \mathbb{E}[\max_i |X^{S_n}_{t_i^{(m)}} - X^{S_n}_{t_{i-1}^{(m)}}|]$ holds for every partition $\pi_m$ (the left-hand side does not depend on $m$), sending $m \to \infty$ gives
\begin{align*}
\mathbb{E}\!\left[(X^{S_n}_t)^2\right] = 0.
\end{align*}
Therefore $X^{S_n}_t = 0$ a.s. for each fixed $t \geq 0$.
[guided]
We have established, for any partition $\pi: 0 = t_0 < t_1 < \cdots < t_k = t$,
\begin{align*}
\mathbb{E}\!\left[(X^{S_n}_t)^2\right] \leq n \cdot \mathbb{E}\!\left[\max_{1 \leq i \leq k} |X^{S_n}_{t_i} - X^{S_n}_{t_{i-1}}|\right].
\end{align*}
The left-hand side is a fixed number (independent of the partition), while the right-hand side depends on the partition. The idea is to choose partitions with increasingly fine mesh and show the right-hand side tends to $0$.
Why does $\max_i |X^{S_n}_{t_i} - X^{S_n}_{t_{i-1}}| \to 0$ as the mesh refines? This is precisely the definition of uniform continuity. The path $s \mapsto X^{S_n}_s(\omega) = X_{s \wedge S_n(\omega)}(\omega)$ is continuous on the compact interval $[0, t]$ (since $X$ is continuous and $s \mapsto s \wedge S_n(\omega)$ is continuous). By the Heine-Cantor theorem, a continuous function on a compact set is uniformly continuous. Therefore, for each $\omega$, the modulus of continuity $\omega_X(\delta) = \sup_{|s_1 - s_2| \leq \delta} |X^{S_n}_{s_1} - X^{S_n}_{s_2}|$ satisfies $\omega_X(\delta) \to 0$ as $\delta \to 0$. For a partition with mesh $|\pi| \leq \delta$, we have $\max_i |X^{S_n}_{t_i} - X^{S_n}_{t_{i-1}}| \leq \omega_X(\delta) \to 0$.
To pass the limit through the expectation, we need a dominator. Since $|X^{S_n}| \leq n$, each increment satisfies $|X^{S_n}_{t_i} - X^{S_n}_{t_{i-1}}| \leq 2n$, so $\max_i |X^{S_n}_{t_i} - X^{S_n}_{t_{i-1}}| \leq 2n$. The constant $2n$ is an integrable dominator. By the dominated convergence theorem,
\begin{align*}
\lim_{m \to \infty} \mathbb{E}\!\left[\max_{1 \leq i \leq k_m} |X^{S_n}_{t_i^{(m)}} - X^{S_n}_{t_{i-1}^{(m)}}|\right] = \mathbb{E}\!\left[\lim_{m \to \infty} \max_{1 \leq i \leq k_m} |X^{S_n}_{t_i^{(m)}} - X^{S_n}_{t_{i-1}^{(m)}}|\right] = \mathbb{E}[0] = 0.
\end{align*}
Since $0 \leq \mathbb{E}[(X^{S_n}_t)^2] \leq n \cdot \mathbb{E}[\max_i |X^{S_n}_{t_i^{(m)}} - X^{S_n}_{t_{i-1}^{(m)}}|]$ for every $m$, taking $m \to \infty$ gives $\mathbb{E}[(X^{S_n}_t)^2] = 0$. Since the variance of $X^{S_n}_t$ is zero and $\mathbb{E}[X^{S_n}_t] = \mathbb{E}[X_0] = 0$ (by the martingale property), this means $X^{S_n}_t = 0$ a.s.
The key insight is that a continuous finite variation martingale has "no quadratic variation" — the sum of squared increments vanishes as the mesh refines, because each increment is controlled by the modulus of continuity, while the total absolute variation remains bounded. This is what distinguishes finite variation processes from martingales like Brownian motion, where the sum of squared increments converges to a positive limit.
[/guided]
[/step]
[step:Send $n \to \infty$ to conclude $X_t = 0$ for all $t$]
For each $n$, we have shown $X^{S_n}_t = X_{t \wedge S_n} = 0$ a.s. for every $t \geq 0$. Since $S_n \to \infty$ a.s., for each fixed $t$ and a.e. $\omega$, there exists $N(\omega)$ such that $S_n(\omega) > t$ for all $n \geq N(\omega)$. For such $n$, $t \wedge S_n(\omega) = t$, so
\begin{align*}
X_t(\omega) = X_{t \wedge S_n(\omega)}(\omega) = 0.
\end{align*}
This shows $X_t = 0$ a.s. for each fixed $t \geq 0$. Since $X$ is continuous, the exceptional null set can be chosen independently of $t$: let $\Omega_0 = \{\omega : X_t(\omega) = 0 \text{ for all } t \in \mathbb{Q}_{\geq 0}\}$. Since $\mathbb{Q}_{\geq 0}$ is countable, $\mathbb{P}(\Omega_0) = 1$. For $\omega \in \Omega_0$ and any $t \geq 0$, choose rationals $q_k \to t$ and use continuity: $X_t(\omega) = \lim_{k \to \infty} X_{q_k}(\omega) = 0$. Therefore $X_t = 0$ for all $t \geq 0$ on an event of full probability.
[guided]
We have established: for each $n \geq 1$ and each $t \geq 0$, $X_{t \wedge S_n} = 0$ a.s. Since $S_n \to \infty$ a.s., for a.e. $\omega$ and each fixed $t$, eventually $S_n(\omega) > t$, so $X_t(\omega) = X_{t \wedge S_n(\omega)}(\omega) = 0$.
But this only gives $X_t = 0$ a.s. for each fixed $t$ — the exceptional null set may depend on $t$. To get $X_t(\omega) = 0$ for all $t$ simultaneously (outside a single null set), we use continuity of paths.
Define
\begin{align*}
\Omega_0 := \bigcap_{q \in \mathbb{Q}_{\geq 0}} \{X_q = 0\}.
\end{align*}
This is a countable intersection of events, each of full probability, so $\mathbb{P}(\Omega_0) = 1$. For $\omega \in \Omega_0$, $X_q(\omega) = 0$ for every rational $q \geq 0$. Now take any $t \geq 0$ and choose a sequence of rationals $q_k \to t$. By continuity of $s \mapsto X_s(\omega)$:
\begin{align*}
X_t(\omega) = \lim_{k \to \infty} X_{q_k}(\omega) = \lim_{k \to \infty} 0 = 0.
\end{align*}
Therefore $X_t(\omega) = 0$ for all $t \geq 0$ and all $\omega \in \Omega_0$, with $\mathbb{P}(\Omega_0) = 1$. This is indistinguishability: $X$ and the zero process agree on a full-probability event for all times simultaneously.
[/guided]
[/step]
