[proofplan] We extend the stochastic integral from simple processes to all of $L^2(M)$ by a density-and-completeness argument. The [Itô Isometry on Simple Processes](/theorems/2089) shows that $H \mapsto H \cdot M$ is an isometry from $\mathcal{E}$ into $\mathcal{M}^2_c$. Since $\mathcal{E}$ is dense in $L^2(M)$ by the [Density of Simple Processes](/theorems/2091) and $\mathcal{M}^2_c$ is complete by [M2 is a Hilbert Space](/theorems/2082), the isometry extends uniquely to all of $L^2(M)$. The bracket identity in part (ii) is established by passing to the limit from the simple-process case using the [Kunita–Watanabe Inequality](/theorems/2087), and uniqueness follows from the non-degeneracy of the bracket. Part (iii) is deduced from part (ii) by verifying that both sides satisfy the same bracket identity. [/proofplan] [step:Extend the isometry from $\mathcal{E}$ to $L^2(M)$ by density and completeness] By the [Itô Isometry on Simple Processes](/theorems/2089), the map \begin{align*} \mathcal{E} &\to \mathcal{M}^2_c, \quad H \mapsto H \cdot M \end{align*} is a linear isometry: $\|H \cdot M\|_{\mathcal{M}^2}^2 = \mathbb{E}\!\left[\int_0^\infty H_s^2 \, d\langle M \rangle_s\right] = \|H\|_{L^2(M)}^2$ for every $H \in \mathcal{E}$. By the [Density of Simple Processes](/theorems/2091), $\mathcal{E}$ is dense in $L^2(M)$. Since $\mathcal{M}^2_c$ is a Hilbert space by [$\mathcal{M}^2$ is a Hilbert Space](/theorems/2082), and in particular complete, the bounded linear extension theorem (BLT lemma) provides a unique isometric extension \begin{align*} L^2(M) &\to \mathcal{M}^2_c, \quad H \mapsto H \cdot M. \end{align*} Concretely, for $H \in L^2(M)$, choose any sequence $(H^n)_{n \geq 1}$ in $\mathcal{E}$ with $\|H^n - H\|_{L^2(M)} \to 0$. The isometry on $\mathcal{E}$ gives \begin{align*} \|H^n \cdot M - H^m \cdot M\|_{\mathcal{M}^2} = \|H^n - H^m\|_{L^2(M)} \to 0 \quad \text{as } n, m \to \infty, \end{align*} so $(H^n \cdot M)_{n \geq 1}$ is Cauchy in $\mathcal{M}^2_c$. Define $H \cdot M := \lim_{n \to \infty} H^n \cdot M$ in $\mathcal{M}^2_c$. The limit is independent of the approximating sequence, and the isometry identity $\|H \cdot M\|_{\mathcal{M}^2} = \|H\|_{L^2(M)}$ follows by continuity of the norm. [guided] The goal is to extend $H \mapsto H \cdot M$ from the "easy" domain $\mathcal{E}$ (bounded simple previsible processes) to the full space $L^2(M)$. The strategy is the standard Hilbert-space extension argument: an isometry from a dense subspace into a complete space extends uniquely. We need three ingredients: **Ingredient 1: Isometry on $\mathcal{E}$.** The [Itô Isometry on Simple Processes](/theorems/2089) gives $\|H \cdot M\|_{\mathcal{M}^2}^2 = \mathbb{E}\!\left[\int_0^\infty H_s^2 \, d\langle M \rangle_s\right] = \|H\|_{L^2(M)}^2$ for $H \in \mathcal{E}$. This is the crucial norm-preserving property. **Ingredient 2: Density of $\mathcal{E}$.** By the [Density of Simple Processes](/theorems/2091), $\mathcal{E}$ is dense in $L^2(M)$. So every $H \in L^2(M)$ can be approximated by a sequence $(H^n)_{n \geq 1}$ in $\mathcal{E}$ with $\|H^n - H\|_{L^2(M)} \to 0$. **Ingredient 3: Completeness of the target.** The space $\mathcal{M}^2_c$ is a Hilbert space by [$\mathcal{M}^2$ is a Hilbert Space](/theorems/2082), so Cauchy sequences converge. Given these, the extension proceeds as follows. For $H \in L^2(M)$, choose $(H^n)_{n \geq 1}$ in $\mathcal{E}$ with $\|H^n - H\|_{L^2(M)} \to 0$. Since the map is an isometry on $\mathcal{E}$, \begin{align*} \|H^n \cdot M - H^m \cdot M\|_{\mathcal{M}^2} = \|(H^n - H^m) \cdot M\|_{\mathcal{M}^2} = \|H^n - H^m\|_{L^2(M)} \to 0 \quad \text{as } n, m \to \infty, \end{align*} where the first equality uses linearity of the stochastic integral on $\mathcal{E}$. The sequence $(H^n \cdot M)_{n \geq 1}$ is therefore Cauchy in $\mathcal{M}^2_c$, hence converges to a limit, which we define as $H \cdot M$. Why is the limit independent of the approximating sequence? If $(K^n)_{n \geq 1}$ is another sequence in $\mathcal{E}$ with $\|K^n - H\|_{L^2(M)} \to 0$, then $\|H^n \cdot M - K^n \cdot M\|_{\mathcal{M}^2} = \|H^n - K^n\|_{L^2(M)} \leq \|H^n - H\|_{L^2(M)} + \|H - K^n\|_{L^2(M)} \to 0$, so both sequences converge to the same limit in $\mathcal{M}^2_c$. The isometry identity $\|H \cdot M\|_{\mathcal{M}^2} = \|H\|_{L^2(M)}$ for $H \in L^2(M)$ follows by continuity: $\|H \cdot M\|_{\mathcal{M}^2} = \lim_{n \to \infty} \|H^n \cdot M\|_{\mathcal{M}^2} = \lim_{n \to \infty} \|H^n\|_{L^2(M)} = \|H\|_{L^2(M)}$. [/guided] [/step] [step:Establish the bracket identity $\langle H \cdot M, N \rangle_t = \int_0^t H_s \, d\langle M, N \rangle_s$ for all $N \in \mathcal{M}^2_c$] Fix $N \in \mathcal{M}^2_c$ and let $(H^n)_{n \geq 1}$ in $\mathcal{E}$ converge to $H$ in $L^2(M)$. By the [Bracket of Stochastic Integral](/theorems/2090), for each $n$, \begin{align*} \langle H^n \cdot M, N \rangle_t = \int_0^t H^n_s \, d\langle M, N \rangle_s. \end{align*} We show that both sides converge appropriately as $n \to \infty$. **Left-hand side.** Since $H^n \cdot M \to H \cdot M$ in $\mathcal{M}^2_c$, the [Properties of Covariation](/theorems/2086) (part (v)) give \begin{align*} \mathbb{E}\!\left[\sup_{s \leq t} |\langle H^n \cdot M, N \rangle_s - \langle H \cdot M, N \rangle_s|\right] &= \mathbb{E}\!\left[\sup_{s \leq t} |\langle (H^n - H) \cdot M, N \rangle_s|\right]. \end{align*} By the [Kunita–Watanabe Inequality](/theorems/2087) applied with $K = 1$, \begin{align*} \sup_{s \leq t} \left|\langle (H^n - H) \cdot M, N \rangle_s\right| &\leq \int_0^t |1| \cdot |1| \, |d\langle (H^n - H) \cdot M, N \rangle_s| \\ &\leq \left(\int_0^t d\langle (H^n - H) \cdot M \rangle_s\right)^{1/2} \left(\int_0^t d\langle N \rangle_s\right)^{1/2} \\ &= \langle (H^n - H) \cdot M \rangle_t^{1/2} \cdot \langle N \rangle_t^{1/2}. \end{align*} Taking expectations and applying the Cauchy–Schwarz inequality in $L^2(\mathbb{P})$, \begin{align*} \mathbb{E}\!\left[\sup_{s \leq t} |\langle H^n \cdot M, N \rangle_s - \langle H \cdot M, N \rangle_s|\right] &\leq \left(\mathbb{E}\!\left[\langle (H^n - H) \cdot M \rangle_t\right]\right)^{1/2} \left(\mathbb{E}\!\left[\langle N \rangle_t\right]\right)^{1/2} \\ &\leq \|H^n - H\|_{L^2(M)} \cdot \|N\|_{\mathcal{M}^2} \to 0. \end{align*} **Right-hand side.** By the [Kunita–Watanabe Inequality](/theorems/2087), \begin{align*} \int_0^t |H^n_s - H_s| \, |d\langle M, N \rangle_s| &\leq \left(\int_0^t (H^n_s - H_s)^2 \, d\langle M \rangle_s\right)^{1/2} \left(\int_0^t d\langle N \rangle_s\right)^{1/2}. \end{align*} Taking expectations, \begin{align*} \mathbb{E}\!\left[\left|\int_0^t H^n_s \, d\langle M, N \rangle_s - \int_0^t H_s \, d\langle M, N \rangle_s\right|\right] &\leq \|H^n - H\|_{L^2(M)} \cdot \|N\|_{\mathcal{M}^2} \to 0. \end{align*} Since both sides converge in $L^1(\mathbb{P})$ and the identity holds at every $n$, the limiting identity $\langle H \cdot M, N \rangle_t = \int_0^t H_s \, d\langle M, N \rangle_s$ holds a.s. for each $t$. Since both sides are continuous processes, the identity holds simultaneously for all $t \geq 0$, a.s. [guided] We want to show that the bracket identity, known for simple integrands by the [Bracket of Stochastic Integral](/theorems/2090), survives the passage to the limit when $H^n \to H$ in $L^2(M)$. For each $n$, the identity $\langle H^n \cdot M, N \rangle_t = \int_0^t H^n_s \, d\langle M, N \rangle_s$ holds. We need to show that the left-hand side converges to $\langle H \cdot M, N \rangle_t$ and the right-hand side converges to $\int_0^t H_s \, d\langle M, N \rangle_s$. **Convergence of the left-hand side.** The covariation is bilinear ([Properties of Covariation](/theorems/2086), part (ii)), so $\langle H^n \cdot M, N \rangle - \langle H \cdot M, N \rangle = \langle (H^n - H) \cdot M, N \rangle$. We need this to tend to zero. The [Kunita–Watanabe Inequality](/theorems/2087) is the tool: it bounds the total variation of a covariation in terms of the quadratic variations. Applying it with both integrands equal to $1$, the total variation of $\langle (H^n - H) \cdot M, N \rangle$ on $[0, t]$ is bounded by $\langle (H^n - H) \cdot M \rangle_t^{1/2} \cdot \langle N \rangle_t^{1/2}$. Taking expectations and applying Cauchy–Schwarz in $L^2(\mathbb{P})$ gives a bound of $\|H^n - H\|_{L^2(M)} \cdot \|N\|_{\mathcal{M}^2}$, which tends to zero. **Convergence of the right-hand side.** The difference $\int_0^t (H^n_s - H_s) \, d\langle M, N \rangle_s$ is bounded in absolute value by $\int_0^t |H^n_s - H_s| \, |d\langle M, N \rangle_s|$. By the [Kunita–Watanabe Inequality](/theorems/2087), this is at most $\left(\int_0^t (H^n_s - H_s)^2 \, d\langle M \rangle_s\right)^{1/2} \cdot \langle N \rangle_t^{1/2}$. Taking expectations and applying Cauchy–Schwarz gives the same bound as above. Since both sides converge in $L^1(\mathbb{P})$ to the same limit, the bracket identity passes to the limit. The identity holds for each fixed $t$, a.s. Since both $\langle H \cdot M, N \rangle_t$ and $\int_0^t H_s \, d\langle M, N \rangle_s$ have continuous paths, the null sets can be absorbed: the identity holds for all $t$ simultaneously, a.s. [/guided] [/step] [step:Prove uniqueness of $H \cdot M$ via the bracket characterization] Suppose $X \in \mathcal{M}^2_c$ also satisfies $\langle X, N \rangle_t = \int_0^t H_s \, d\langle M, N \rangle_s$ for all $N \in \mathcal{M}^2_c$. Define $Y := X - H \cdot M \in \mathcal{M}^2_c$. By bilinearity of the bracket, \begin{align*} \langle Y, N \rangle_t = \langle X, N \rangle_t - \langle H \cdot M, N \rangle_t = 0 \end{align*} for all $N \in \mathcal{M}^2_c$ and all $t \geq 0$. Taking $N = Y$ gives $\langle Y, Y \rangle_t = \langle Y \rangle_t = 0$ for all $t$. By the [Quadratic Variation Vanishes Iff the Martingale Vanishes](/theorems/2084), $Y$ is indistinguishable from the zero process, so $X = H \cdot M$ in $\mathcal{M}^2_c$. [/step] [step:Verify the stopping identity $(H \cdot M)^T = (\mathbb{1}_{[0,T]} H) \cdot M = H \cdot M^T$] Let $T$ be a stopping time and $N \in \mathcal{M}^2_c$ arbitrary. We verify the bracket identities for each expression against $N$. By [Properties of Covariation](/theorems/2086) (part (iv)), \begin{align*} \langle (H \cdot M)^T, N \rangle_t = \langle H \cdot M, N \rangle_{t \wedge T}. \end{align*} By the bracket identity from part (ii), \begin{align*} \langle H \cdot M, N \rangle_{t \wedge T} = \int_0^{t \wedge T} H_s \, d\langle M, N \rangle_s = \int_0^t \mathbb{1}_{[0,T]}(s) H_s \, d\langle M, N \rangle_s. \end{align*} On the other hand, applying part (ii) to $(\mathbb{1}_{[0,T]} H) \cdot M$ gives \begin{align*} \langle (\mathbb{1}_{[0,T]} H) \cdot M, N \rangle_t = \int_0^t \mathbb{1}_{[0,T]}(s) H_s \, d\langle M, N \rangle_s. \end{align*} Since both $(H \cdot M)^T$ and $(\mathbb{1}_{[0,T]} H) \cdot M$ belong to $\mathcal{M}^2_c$ and have the same bracket against every $N \in \mathcal{M}^2_c$, the uniqueness result from the previous step gives $(H \cdot M)^T = (\mathbb{1}_{[0,T]} H) \cdot M$. For the second identity, applying part (ii) to $H \cdot M^T$ and using $\langle M^T, N \rangle_t = \langle M, N \rangle_{t \wedge T}$ ([Properties of Covariation](/theorems/2086), part (iv)): \begin{align*} \langle H \cdot M^T, N \rangle_t = \int_0^t H_s \, d\langle M^T, N \rangle_s = \int_0^t H_s \, d\langle M, N \rangle_{s \wedge T} = \int_0^{t \wedge T} H_s \, d\langle M, N \rangle_s, \end{align*} which equals $\langle (H \cdot M)^T, N \rangle_t$ as computed above. By uniqueness, $H \cdot M^T = (H \cdot M)^T$. [guided] Part (iii) states three processes are indistinguishable: $(H \cdot M)^T$, $(\mathbb{1}_{[0,T]} H) \cdot M$, and $H \cdot M^T$. The strategy for proving equalities of elements in $\mathcal{M}^2_c$ is to check that they satisfy the same bracket identity against all $N \in \mathcal{M}^2_c$ and invoke the uniqueness from part (ii). **First identity: $(H \cdot M)^T = (\mathbb{1}_{[0,T]} H) \cdot M$.** We compute the bracket of each side against an arbitrary $N \in \mathcal{M}^2_c$. For the left-hand side, using the stopped-bracket property from [Properties of Covariation](/theorems/2086) (part (iv)), $\langle (H \cdot M)^T, N \rangle_t = \langle H \cdot M, N \rangle_{t \wedge T}$. By part (ii), $\langle H \cdot M, N \rangle_{t \wedge T} = \int_0^{t \wedge T} H_s \, d\langle M, N \rangle_s$. Since $\int_0^{t \wedge T} = \int_0^t \mathbb{1}_{[0,T]}(s) \, \cdot$, this equals $\int_0^t \mathbb{1}_{[0,T]}(s) H_s \, d\langle M, N \rangle_s$. For the right-hand side, part (ii) applied directly gives $\langle (\mathbb{1}_{[0,T]} H) \cdot M, N \rangle_t = \int_0^t \mathbb{1}_{[0,T]}(s) H_s \, d\langle M, N \rangle_s$. The brackets coincide for all $N$, so by uniqueness $(H \cdot M)^T = (\mathbb{1}_{[0,T]} H) \cdot M$. **Second identity: $H \cdot M^T = (H \cdot M)^T$.** For the left-hand side, apply part (ii) with integrator $M^T$: \begin{align*} \langle H \cdot M^T, N \rangle_t = \int_0^t H_s \, d\langle M^T, N \rangle_s. \end{align*} By [Properties of Covariation](/theorems/2086) (part (iv)), $\langle M^T, N \rangle_t = \langle M, N \rangle_{t \wedge T}$. The measure $d\langle M^T, N \rangle_s$ is therefore the restriction of $d\langle M, N \rangle$ to $[0, T]$, so $\int_0^t H_s \, d\langle M^T, N \rangle_s = \int_0^{t \wedge T} H_s \, d\langle M, N \rangle_s$, which agrees with the bracket of $(H \cdot M)^T$ computed above. Uniqueness gives $H \cdot M^T = (H \cdot M)^T$. [/guided] [/step]