[proofplan]
The forward direction ($M = 0 \implies \langle M \rangle = 0$) follows immediately from the Riemann-sum characterization of quadratic variation: every increment of $M$ is zero, so every Riemann sum vanishes. The reverse direction ($\langle M \rangle = 0 \implies M = 0$) uses the fact that $M^2_t - \langle M \rangle_t = M^2_t$ is a continuous non-negative local martingale starting at zero. By the [Non-negative Local Martingale is a Supermartingale](/theorems/2077) theorem, $M^2$ is a supermartingale, which forces $M^2_t = 0$ a.s. for each $t$, and continuity upgrades this to indistinguishability.
[/proofplan]
[step:Forward direction: $M = 0$ implies $\langle M \rangle = 0$]
Suppose $M$ is indistinguishable from zero. Then $M^2_t = 0$ for all $t \geq 0$ a.s. Since $M^2_t - \langle M \rangle_t$ is a continuous local martingale and $M^2_t = 0$ a.s. for all $t$, the process $-\langle M \rangle_t$ is a continuous local martingale. But $\langle M \rangle$ is a continuous, increasing process starting at zero, so $-\langle M \rangle$ is a continuous, decreasing process starting at zero. A continuous local martingale that is also of finite variation and starts at zero must be identically zero by the [Continuous Finite Variation Local Martingale is Identically Zero](/theorems/2081) theorem. Therefore $\langle M \rangle_t = 0$ for all $t \geq 0$ a.s.
[/step]
[step:Reverse direction: $\langle M \rangle = 0$ implies $M^2$ is a non-negative continuous local martingale starting at zero]
Suppose $\langle M \rangle_t = 0$ for all $t \geq 0$ a.s. Since $M$ is a continuous local martingale with $M_0 = 0$, the process $M^2_t - \langle M \rangle_t$ is a continuous local martingale. Substituting $\langle M \rangle_t = 0$, the process $M^2_t$ is itself a continuous local martingale.
The process $M^2$ satisfies:
- $M^2_t \geq 0$ for all $t \geq 0$ (since it is a square),
- $M^2_0 = M_0^2 = 0$ (by hypothesis),
- $M^2$ is a continuous local martingale (as just established).
[/step]
[step:Apply the supermartingale property to conclude $M^2 = 0$]
Since $M^2$ is a non-negative local martingale, by the [Non-negative Local Martingale is a Supermartingale](/theorems/2077) theorem, $M^2$ is a supermartingale. The supermartingale property gives, for all $t \geq 0$,
\begin{align*}
0 \leq \mathbb{E}[M^2_t] \leq \mathbb{E}[M^2_0] = 0.
\end{align*}
Therefore $\mathbb{E}[M^2_t] = 0$ for all $t \geq 0$, which implies $M^2_t = 0$ a.s. for each $t$.
Since $M$ has continuous paths, $\{M_t = 0\}$ holds simultaneously for all $t \geq 0$ on a single event of full probability: define $\Omega_0 = \bigcap_{q \in \mathbb{Q}_+} \{M_q = 0\}$. Since $\mathbb{Q}_+$ is countable, $\mathbb{P}(\Omega_0) = 1$. For $\omega \in \Omega_0$ and any $t \geq 0$, by continuity of $s \mapsto M_s(\omega)$ and the fact that $M_q(\omega) = 0$ for all $q \in \mathbb{Q}_+$, taking a sequence of rationals $q_n \to t$ gives $M_t(\omega) = \lim_{n \to \infty} M_{q_n}(\omega) = 0$. Therefore $M$ is indistinguishable from zero.
[guided]
The key insight is that the hypothesis $\langle M \rangle = 0$ turns the squared process $M^2$ into a local martingale: normally $M^2_t - \langle M \rangle_t$ is the local martingale, but when $\langle M \rangle = 0$, the subtracted correction vanishes and $M^2$ itself is the local martingale.
Now $M^2$ is a *non-negative* local martingale, and the [Non-negative Local Martingale is a Supermartingale](/theorems/2077) theorem applies: it says that any non-negative local martingale is automatically a supermartingale, without any additional integrability assumptions. The supermartingale inequality $\mathbb{E}[M^2_t] \leq \mathbb{E}[M^2_0] = 0$ then forces $M^2_t = 0$ a.s. for each fixed $t$.
But we need $M_t(\omega) = 0$ for *all* $t$ simultaneously, not just for each fixed $t$ individually (the difference is between "for all $t$, $\mathbb{P}(M_t = 0) = 1$" and "$\mathbb{P}(M_t = 0 \text{ for all } t) = 1$"). This is where continuity of paths is used. We know $M_q = 0$ a.s. for each rational $q \geq 0$. Since the rationals are countable, a union bound gives $\mathbb{P}(M_q = 0 \text{ for all } q \in \mathbb{Q}_+) = 1$. On this full-probability event, continuity of paths forces $M_t = 0$ for all $t \geq 0$: for any $t$, pick rationals $q_n \to t$ and use $M_t = \lim_{n} M_{q_n} = 0$.
This upgrades the conclusion from "for each $t$, $M_t = 0$ a.s." to "$M$ is indistinguishable from zero."
[/guided]
[/step]
