[proofplan]
We show that $\langle M^T \rangle_t = \langle M \rangle_{t \wedge T}$ by verifying that the process $t \mapsto \langle M \rangle_{t \wedge T}$ satisfies the characterizing properties of the quadratic variation of $M^T$. Specifically, we show it is continuous, adapted, increasing, starts at zero, and that $(M^T)^2_t - \langle M \rangle_{t \wedge T}$ is a continuous local martingale. The result then follows from the uniqueness of quadratic variation.
[/proofplan]
[step:Verify that $t \mapsto \langle M \rangle_{t \wedge T}$ is a continuous, adapted, increasing process starting at zero]
Define $A_t := \langle M \rangle_{t \wedge T}$.
**Starting value.** $A_0 = \langle M \rangle_{0 \wedge T} = \langle M \rangle_0 = 0$, since the quadratic variation of $M$ starts at zero.
**Monotonicity.** For $s \leq t$, we have $s \wedge T \leq t \wedge T$, and $\langle M \rangle$ is increasing, so $A_s = \langle M \rangle_{s \wedge T} \leq \langle M \rangle_{t \wedge T} = A_t$.
**Continuity.** Since $\langle M \rangle$ is continuous and $t \mapsto t \wedge T(\omega)$ is continuous for each $\omega$, the composition $t \mapsto \langle M \rangle_{t \wedge T(\omega)}(\omega)$ is continuous.
**Adaptedness.** For each $t \geq 0$, the random variable $t \wedge T$ is a stopping time bounded by $t$. Since $\langle M \rangle$ is adapted and continuous, the random variable $\langle M \rangle_{t \wedge T}$ is $\mathcal{F}_{t \wedge T}$-measurable. Since $t \wedge T \leq t$, we have $\mathcal{F}_{t \wedge T} \subseteq \mathcal{F}_t$, so $A_t$ is $\mathcal{F}_t$-measurable.
[/step]
[step:Show that $(M^T)^2_t - \langle M \rangle_{t \wedge T}$ is a continuous local martingale]
By hypothesis, $M$ is a continuous local martingale, so the process $M^2_t - \langle M \rangle_t$ is a continuous local martingale. Let $(S_n)_{n \geq 1}$ be a reducing sequence for $M^2 - \langle M \rangle$.
Stopping $M^2_t - \langle M \rangle_t$ at $T$ yields the process
\begin{align*}
(M^2_t - \langle M \rangle_t)\big|_{t \to t \wedge T} = M^2_{t \wedge T} - \langle M \rangle_{t \wedge T} = (M^T_t)^2 - A_t.
\end{align*}
The stopped process of a local martingale is again a local martingale: the sequence $(S_n \wedge T)$ serves as a reducing sequence for $(M^T)^2 - A$, since stopping a martingale at a stopping time produces a martingale. Continuity of $(M^T)^2 - A$ follows from continuity of $M^T$ (which inherits continuity from $M$) and continuity of $A$ (established above).
Therefore $(M^T)^2_t - \langle M \rangle_{t \wedge T}$ is a continuous local martingale.
[guided]
The characterizing property of the quadratic variation $\langle M \rangle$ is that $M^2_t - \langle M \rangle_t$ is a continuous local martingale. We need to show that $\langle M \rangle_{t \wedge T}$ plays the same role for the stopped process $M^T$.
Write out the stopped process explicitly: $(M^T_t)^2 = M^2_{t \wedge T}$. We need $M^2_{t \wedge T} - \langle M \rangle_{t \wedge T}$ to be a continuous local martingale. But this is exactly the process $N_{t \wedge T}$, where $N_t := M^2_t - \langle M \rangle_t$.
Why is the stopped process $N^T$ a continuous local martingale? Since $N$ is a continuous local martingale with reducing sequence $(S_n)$, the doubly-stopped process $N^{S_n \wedge T}$ is a (bounded) martingale for each $n$. The sequence $(S_n \wedge T)$ increases to $T$ (since $S_n \to \infty$), and each $S_n \wedge T$ is a stopping time. Therefore $(S_n \wedge T)$ is a reducing sequence for $N^T$, confirming that $N^T = (M^T)^2 - A$ is a continuous local martingale.
Continuity of $N^T$ is inherited from continuity of $N$ and continuity of $t \mapsto t \wedge T(\omega)$.
[/guided]
[/step]
[step:Conclude by uniqueness of quadratic variation]
We have shown that $A_t = \langle M \rangle_{t \wedge T}$ is a continuous, adapted, increasing process starting at zero, and that $(M^T)^2_t - A_t$ is a continuous local martingale. By the [Continuous Finite Variation Local Martingale is Identically Zero](/theorems/2081) theorem, the quadratic variation of a continuous local martingale is unique: if $A$ and $A'$ are both continuous, adapted, increasing processes starting at zero such that $(M^T)^2 - A$ and $(M^T)^2 - A'$ are continuous local martingales, then $A - A' = ((M^T)^2 - A') - ((M^T)^2 - A)$ is a continuous local martingale of finite variation starting at zero, hence identically zero.
Since $\langle M^T \rangle$ is defined as the unique process with these properties, we conclude
\begin{align*}
\langle M^T \rangle_t = \langle M \rangle_{t \wedge T} \quad \text{a.s. for all } t \geq 0.
\end{align*}
[guided]
The uniqueness argument is the final ingredient. We know two candidates for the quadratic variation of $M^T$:
1. $\langle M^T \rangle_t$, which exists by the general theory of quadratic variation for continuous local martingales.
2. $\langle M \rangle_{t \wedge T}$, which we just verified satisfies the characterizing properties.
If both work, then their difference $\langle M^T \rangle_t - \langle M \rangle_{t \wedge T}$ is a continuous process of finite variation (difference of two increasing processes) starting at zero. Moreover,
\begin{align*}
(M^T)^2_t - \langle M^T \rangle_t \quad \text{and} \quad (M^T)^2_t - \langle M \rangle_{t \wedge T}
\end{align*}
are both continuous local martingales, so their difference $\langle M \rangle_{t \wedge T} - \langle M^T \rangle_t$ is a continuous local martingale of finite variation starting at zero. By the [Continuous Finite Variation Local Martingale is Identically Zero](/theorems/2081) theorem, this difference is indistinguishable from zero. Therefore $\langle M^T \rangle_t = \langle M \rangle_{t \wedge T}$ a.s. for all $t \geq 0$.
[/guided]
[/step]
