[proofplan]
We prove the three parts in order. First, we show $\langle M \rangle_\infty \in L^1$ by localizing with stopping times $S_n = \inf\{t : \langle M \rangle_t \geq n\}$, applying the martingale identity $\mathbb{E}[M^2_{t \wedge S_n}] = \mathbb{E}[M^2_0] + \mathbb{E}[\langle M \rangle_{t \wedge S_n}]$ at each localization level, and passing to the limit using Doob's maximal inequality and the Monotone Convergence Theorem. Second, we show $M^2 - \langle M \rangle$ is uniformly integrable by exhibiting an $L^1$ dominator. Third, the norm formula follows by evaluating the martingale identity at $t = \infty$.
[/proofplan]
[step:Define the localizing stopping times and verify they reduce the problem]
For each $n \geq 1$, define
\begin{align*}
S_n := \inf\{t \geq 0 : \langle M \rangle_t \geq n\}.
\end{align*}
Since $\langle M \rangle$ is continuous and increasing, $S_n$ is a stopping time for each $n$. The sequence $(S_n)$ is increasing, and $S_n \to \infty$ a.s. because $\langle M \rangle_t < \infty$ for all $t$ (since $M$ is a continuous local martingale). By definition of $S_n$, we have $\langle M \rangle_{t \wedge S_n} \leq n$ for all $t \geq 0$.
By the [Quadratic Variation of a Stopped Process](/theorems/2083) theorem, $\langle M^{S_n} \rangle_t = \langle M \rangle_{t \wedge S_n} \leq n$. Therefore $M^{S_n}$ is a continuous local martingale with bounded quadratic variation.
[guided]
Why do we localize by stopping $\langle M \rangle$ rather than $M$ itself? The issue is that we need to show $M^2_t - \langle M \rangle_t$ is a *true* martingale (not just a local martingale), and the direct approach requires integrability control on both $M^2$ and $\langle M \rangle$. By stopping at $S_n = \inf\{t : \langle M \rangle_t \geq n\}$, we guarantee $\langle M \rangle_{t \wedge S_n} \leq n$, which gives us the boundedness needed to promote the local martingale to a true martingale.
The stopping times $S_n$ are well-defined stopping times because $\langle M \rangle$ is continuous and adapted, so $\{S_n \leq t\} = \{\sup_{s \leq t} \langle M \rangle_s \geq n\} = \{\langle M \rangle_t \geq n\} \in \mathcal{F}_t$ (using the fact that $\langle M \rangle$ is increasing). The sequence increases to $\infty$ because $\langle M \rangle$ is continuous and finite at each time.
[/guided]
[/step]
[step:Show that $M^2_{t \wedge S_n} - \langle M \rangle_{t \wedge S_n}$ is a true martingale]
The process $M^2_t - \langle M \rangle_t$ is a continuous local martingale (by definition of quadratic variation). Stopping at $S_n$ gives the process $M^2_{t \wedge S_n} - \langle M \rangle_{t \wedge S_n}$, which is also a continuous local martingale.
We claim this stopped process is a genuine martingale. To verify this, we show it is dominated by an integrable random variable. Since $M \in \mathcal{M}^2_c$, [Doob's $L^2$ Maximal Inequality](/theorems/2112) gives
\begin{align*}
\mathbb{E}\!\left[\sup_{t \geq 0} M_t^2\right] \leq 4 \, \|M\|_{\mathcal{M}^2}^2 < \infty.
\end{align*}
Therefore
\begin{align*}
|M^2_{t \wedge S_n} - \langle M \rangle_{t \wedge S_n}| \leq \sup_{t \geq 0} M_t^2 + n,
\end{align*}
and the right-hand side is in $L^1$. By the [Dominated Local Martingale is a Martingale](/theorems/2079) theorem, $M^2_{t \wedge S_n} - \langle M \rangle_{t \wedge S_n}$ is a true martingale.
In particular, for each $t \geq 0$,
\begin{align*}
\mathbb{E}[M^2_{t \wedge S_n}] - \mathbb{E}[\langle M \rangle_{t \wedge S_n}] = \mathbb{E}[M^2_0].
\end{align*}
[guided]
The process $M^2 - \langle M \rangle$ is in general only a local martingale. To use the martingale identity $\mathbb{E}[N_t] = \mathbb{E}[N_0]$, we need to promote it to a true martingale. Stopping at $S_n$ controls the $\langle M \rangle$ term (bounded by $n$), but the $M^2$ term could still be unbounded.
This is where the hypothesis $M \in \mathcal{M}^2_c$ is crucial. Doob's $L^2$ maximal inequality gives $\mathbb{E}[\sup_t M_t^2] \leq 4\|M\|_{\mathcal{M}^2}^2 < \infty$. So we can dominate:
\begin{align*}
|M^2_{t \wedge S_n} - \langle M \rangle_{t \wedge S_n}| \leq M^2_{t \wedge S_n} + \langle M \rangle_{t \wedge S_n} \leq \sup_{t \geq 0} M_t^2 + n =: Z_n.
\end{align*}
Since $Z_n \in L^1$, the [Dominated Local Martingale is a Martingale](/theorems/2079) theorem applies, and the stopped process is a genuine martingale. The martingale identity then gives
\begin{align*}
\mathbb{E}[M^2_{t \wedge S_n}] = \mathbb{E}[M^2_0] + \mathbb{E}[\langle M \rangle_{t \wedge S_n}].
\end{align*}
[/guided]
[/step]
[step:Pass to the limit to establish $\langle M \rangle_\infty \in L^1$]
From the martingale identity, $\mathbb{E}[\langle M \rangle_{t \wedge S_n}] = \mathbb{E}[M^2_{t \wedge S_n}] - \mathbb{E}[M^2_0]$.
**Limit as $t \to \infty$ (fixed $n$).** As $t \to \infty$, $\langle M \rangle_{t \wedge S_n} \nearrow \langle M \rangle_{S_n}$ (monotone increase). By the Monotone Convergence Theorem,
\begin{align*}
\mathbb{E}[\langle M \rangle_{S_n}] = \lim_{t \to \infty} \mathbb{E}[\langle M \rangle_{t \wedge S_n}] = \lim_{t \to \infty} \mathbb{E}[M^2_{t \wedge S_n}] - \mathbb{E}[M^2_0].
\end{align*}
For the right-hand side, $M^2_{t \wedge S_n} \leq \sup_{t \geq 0} M_t^2 \in L^1$, so the Dominated Convergence Theorem gives $\lim_{t \to \infty} \mathbb{E}[M^2_{t \wedge S_n}] = \mathbb{E}[M^2_{S_n}]$. Therefore
\begin{align*}
\mathbb{E}[\langle M \rangle_{S_n}] = \mathbb{E}[M^2_{S_n}] - \mathbb{E}[M^2_0].
\end{align*}
**Limit as $n \to \infty$.** As $n \to \infty$, $S_n \to \infty$ a.s., so $\langle M \rangle_{S_n} \nearrow \langle M \rangle_\infty$ (monotone increase). By the Monotone Convergence Theorem,
\begin{align*}
\mathbb{E}[\langle M \rangle_\infty] = \lim_{n \to \infty} \mathbb{E}[\langle M \rangle_{S_n}] = \lim_{n \to \infty} \left(\mathbb{E}[M^2_{S_n}] - \mathbb{E}[M^2_0]\right).
\end{align*}
Since $M \in \mathcal{M}^2_c$, the martingale $M$ is uniformly integrable and $M_t \to M_\infty$ in $L^2$. As $S_n \to \infty$, $M_{S_n} \to M_\infty$ a.s. (by continuity of paths), and $M^2_{S_n} \leq \sup_t M_t^2 \in L^1$, so the Dominated Convergence Theorem gives $\mathbb{E}[M^2_{S_n}] \to \mathbb{E}[M^2_\infty]$. Therefore
\begin{align*}
\mathbb{E}[\langle M \rangle_\infty] = \mathbb{E}[M^2_\infty] - \mathbb{E}[M^2_0] \leq \mathbb{E}[M^2_\infty] < \infty,
\end{align*}
which shows $\langle M \rangle_\infty \in L^1(\Omega, \mathcal{F}, \mathbb{P})$.
[guided]
The argument proceeds by a double limit: first $t \to \infty$ (at fixed localization level $n$), then $n \to \infty$ (removing the localization). At each stage, we must justify the passage to the limit.
**First limit ($t \to \infty$, $n$ fixed).** The left-hand side uses monotone convergence: $\langle M \rangle_{t \wedge S_n}$ increases in $t$ to $\langle M \rangle_{S_n}$. The right-hand side uses dominated convergence: $M^2_{t \wedge S_n}$ converges a.s. to $M^2_{S_n}$ as $t \to \infty$ (since eventually $t \wedge S_n = S_n$), and is dominated by $\sup_t M_t^2 \in L^1$. This gives the identity $\mathbb{E}[\langle M \rangle_{S_n}] = \mathbb{E}[M^2_{S_n}] - \mathbb{E}[M^2_0]$.
**Second limit ($n \to \infty$).** Again, the left-hand side uses monotone convergence: $\langle M \rangle_{S_n} \nearrow \langle M \rangle_\infty$. For the right-hand side, $M_{S_n} \to M_\infty$ a.s. by continuity, and $M^2_{S_n} \leq \sup_t M_t^2 \in L^1$, so dominated convergence gives $\mathbb{E}[M^2_{S_n}] \to \mathbb{E}[M^2_\infty]$.
The final bound $\mathbb{E}[\langle M \rangle_\infty] = \mathbb{E}[M^2_\infty] - \mathbb{E}[M^2_0] \leq \mathbb{E}[M^2_\infty] \leq \|M\|_{\mathcal{M}^2}^2 < \infty$ shows $\langle M \rangle_\infty \in L^1$.
[/guided]
[/step]
[step:Show $M^2_t - \langle M \rangle_t$ is a uniformly integrable martingale]
Define $N_t := M^2_t - \langle M \rangle_t$. We already know $N$ is a continuous local martingale. We show it is uniformly integrable by exhibiting an $L^1$ dominator.
For all $t \geq 0$,
\begin{align*}
|N_t| = |M^2_t - \langle M \rangle_t| \leq M^2_t + \langle M \rangle_t \leq \sup_{s \geq 0} M_s^2 + \langle M \rangle_\infty =: Z.
\end{align*}
By Doob's $L^2$ maximal inequality, $\mathbb{E}[\sup_s M_s^2] \leq 4\|M\|_{\mathcal{M}^2}^2 < \infty$. By the previous step, $\mathbb{E}[\langle M \rangle_\infty] < \infty$. Therefore $Z \in L^1(\Omega)$.
Since $|N_t| \leq Z \in L^1$ for all $t$, the family $\{N_t : t \geq 0\}$ is uniformly integrable. A uniformly integrable local martingale is a true martingale (by the [Characterization of Martingales Among Local Martingales](/theorems/2078) theorem: uniform integrability of $\{N_T : T \leq t\}$ follows from domination by $Z$, which implies uniform integrability of $\chi_t$ for every $t$).
[guided]
We need to promote the local martingale $N_t = M^2_t - \langle M \rangle_t$ to a uniformly integrable martingale. The strategy is simple: find a single integrable random variable $Z$ that dominates $|N_t|$ for all $t$.
The bound is
\begin{align*}
|N_t| = |M^2_t - \langle M \rangle_t| \leq M^2_t + \langle M \rangle_t \leq \sup_{s \geq 0} M_s^2 + \langle M \rangle_\infty.
\end{align*}
Both terms on the right are non-negative, do not depend on $t$, and are integrable:
- $\mathbb{E}[\sup_s M_s^2] \leq 4\|M\|_{\mathcal{M}^2}^2 < \infty$ by Doob's maximal inequality.
- $\mathbb{E}[\langle M \rangle_\infty] < \infty$ as shown in the previous step.
Therefore $Z := \sup_s M_s^2 + \langle M \rangle_\infty \in L^1$, and $|N_t| \leq Z$ for all $t$.
Domination by an $L^1$ random variable implies uniform integrability: for any $\lambda > 0$,
\begin{align*}
\sup_{t \geq 0} \mathbb{E}[|N_t| \mathbb{1}_{\{|N_t| > \lambda\}}] \leq \mathbb{E}[Z \mathbb{1}_{\{Z > \lambda\}}] \to 0 \quad \text{as } \lambda \to \infty,
\end{align*}
by integrability of $Z$. A uniformly integrable local martingale is a genuine (uniformly integrable) martingale by the [Characterization of Martingales Among Local Martingales](/theorems/2078) theorem.
[/guided]
[/step]
[step:Derive the norm formula $\|M - M_0\|_{\mathcal{M}^2}^2 = \mathbb{E}[\langle M \rangle_\infty]$]
Since $N_t = M^2_t - \langle M \rangle_t$ is a uniformly integrable martingale, we have $\mathbb{E}[N_t] = \mathbb{E}[N_0]$ for all $t \geq 0$. Evaluating at $t = 0$: $\mathbb{E}[N_0] = \mathbb{E}[M^2_0 - \langle M \rangle_0] = \mathbb{E}[M^2_0]$. Therefore
\begin{align*}
\mathbb{E}[M^2_t] - \mathbb{E}[\langle M \rangle_t] = \mathbb{E}[M^2_0] \quad \text{for all } t \geq 0.
\end{align*}
Since $N$ is uniformly integrable, $N_t \to N_\infty := M^2_\infty - \langle M \rangle_\infty$ in $L^1$ as $t \to \infty$, and the martingale identity extends to $t = \infty$:
\begin{align*}
\mathbb{E}[M^2_\infty] - \mathbb{E}[\langle M \rangle_\infty] = \mathbb{E}[M^2_0].
\end{align*}
Rearranging,
\begin{align*}
\mathbb{E}[\langle M \rangle_\infty] = \mathbb{E}[M^2_\infty] - \mathbb{E}[M^2_0] = \mathbb{E}[(M_\infty - M_0)^2] + 2\mathbb{E}[M_\infty M_0] - 2\mathbb{E}[M^2_0].
\end{align*}
Since $M$ is a martingale, $\mathbb{E}[M_\infty \mid \mathcal{F}_0] = M_0$, so $\mathbb{E}[M_\infty M_0] = \mathbb{E}[\mathbb{E}[M_\infty \mid \mathcal{F}_0] M_0] = \mathbb{E}[M^2_0]$. Therefore
\begin{align*}
\mathbb{E}[\langle M \rangle_\infty] = \mathbb{E}[(M_\infty - M_0)^2] = \|M - M_0\|_{\mathcal{M}^2}^2,
\end{align*}
where the last equality uses $\|M - M_0\|_{\mathcal{M}^2}^2 = \mathbb{E}[(M_\infty - M_0)^2] = \mathbb{E}[M^2_\infty] - \mathbb{E}[M^2_0]$ (since the $\mathcal{M}^2$-norm of $M - M_0$ is the $L^2$-norm of the terminal value $(M - M_0)_\infty = M_\infty - M_0$).
[guided]
The norm formula connects two objects: the $\mathcal{M}^2$-norm of $M - M_0$ (an $L^2$-quantity of the terminal value) and the expected total quadratic variation $\mathbb{E}[\langle M \rangle_\infty]$ (a path-level quantity).
Since $N_t = M^2_t - \langle M \rangle_t$ is a uniformly integrable martingale, the constant-expectation property $\mathbb{E}[N_t] = \mathbb{E}[N_0] = \mathbb{E}[M^2_0]$ holds for all $t$, including $t = \infty$:
\begin{align*}
\mathbb{E}[M^2_\infty - \langle M \rangle_\infty] = \mathbb{E}[M^2_0].
\end{align*}
Rearranging: $\mathbb{E}[\langle M \rangle_\infty] = \mathbb{E}[M^2_\infty] - \mathbb{E}[M^2_0]$.
To connect this with $\|M - M_0\|_{\mathcal{M}^2}^2 = \mathbb{E}[(M_\infty - M_0)^2]$, expand the square:
\begin{align*}
\mathbb{E}[(M_\infty - M_0)^2] = \mathbb{E}[M^2_\infty] - 2\mathbb{E}[M_\infty M_0] + \mathbb{E}[M^2_0].
\end{align*}
The cross term simplifies using the martingale property: $\mathbb{E}[M_\infty M_0] = \mathbb{E}[\mathbb{E}[M_\infty \mid \mathcal{F}_0] \cdot M_0] = \mathbb{E}[M_0 \cdot M_0] = \mathbb{E}[M^2_0]$, where we used the tower property and the fact that $M_0$ is $\mathcal{F}_0$-measurable. Substituting:
\begin{align*}
\mathbb{E}[(M_\infty - M_0)^2] = \mathbb{E}[M^2_\infty] - 2\mathbb{E}[M^2_0] + \mathbb{E}[M^2_0] = \mathbb{E}[M^2_\infty] - \mathbb{E}[M^2_0] = \mathbb{E}[\langle M \rangle_\infty].
\end{align*}
This identity is the stochastic calculus analogue of the Pythagorean theorem: the expected total squared displacement of the martingale equals the expected total quadratic variation. It provides a concrete formula for computing the $\mathcal{M}^2$-norm via the quadratic variation, which is often easier to compute in practice.
[/guided]
[/step]
