[proofplan] We construct $H \cdot M$ by localization: choose stopping times $S_n \nearrow \infty$ so that $M^{S_n} \in \mathcal{M}^2_c$ and $H \in L^2(M^{S_n})$, which places us in the scope of the Ito isometry construction. The key technical step is verifying consistency — that the integrals at different stopping levels agree on their common domain — so that the limit $H \cdot M$ is well-defined as a continuous local martingale. Once existence is established, we prove the bracket formula (i) by a second localization that also controls $N$, then deduce uniqueness from the bracket characterization. Properties (ii)–(iv) are inherited from the $L^2$ theory by passing through the same localization. [/proofplan] [step:Define localizing stopping times $S_n$ so that $M^{S_n} \in \mathcal{M}^2_c$ and $H \in L^2(M^{S_n})$] We may assume $M_0 = 0$ without loss of generality (the general case reduces to this by replacing $M$ with $M - M_0$, which does not affect the quadratic variation or the stochastic integral). Define the stopping times \begin{align*} S_n = \inf\!\left\{ t \geq 0 : \int_0^t (1 + H_s^2)\, d\langle M \rangle_s \geq n \right\}. \end{align*} The process $t \mapsto \int_0^t (1 + H_s^2)\, d\langle M \rangle_s$ is continuous (since $\langle M \rangle$ is continuous) and adapted (since $H$ is previsible and $\langle M \rangle$ is adapted), so each $S_n$ is a stopping time by the debut theorem applied to the closed set $[n, \infty)$. Since $H \in L^2_{\mathrm{loc}}(M)$, the integral $\int_0^t H_s^2\, d\langle M \rangle_s$ is finite almost surely for each $t \geq 0$. Therefore $\int_0^t (1 + H_s^2)\, d\langle M \rangle_s < \infty$ almost surely for each $t$, which implies $S_n \nearrow \infty$ almost surely. We verify that $M^{S_n} \in \mathcal{M}^2_c$. By [Quadratic Variation of a Stopped Process](/theorems/2083), \begin{align*} \langle M^{S_n} \rangle_t = \langle M \rangle_{t \wedge S_n} \leq \int_0^{S_n} (1 + H_s^2)\, d\langle M \rangle_s \leq n. \end{align*} Since $\langle M^{S_n} \rangle_\infty \leq n < \infty$, the [Quadratic Variation Norm Formula](/theorems/2085) gives $\mathbb{E}[\sup_{t \geq 0} (M^{S_n}_t)^2] \leq 4n$ (by [Doob's $L^2$ Maximal Inequality](/theorems/2112)), so $M^{S_n} \in \mathcal{M}^2_c$. We also verify that $H \in L^2(M^{S_n})$: \begin{align*} \int_0^\infty H_s^2\, d\langle M^{S_n} \rangle_s = \int_0^{S_n} H_s^2\, d\langle M \rangle_s \leq \int_0^{S_n} (1 + H_s^2)\, d\langle M \rangle_s \leq n < \infty. \end{align*} Therefore $H \in L^2(M^{S_n})$, and the stochastic integral $H \cdot M^{S_n}$ is defined by the [Ito Isometry](/theorems/2092) construction for $L^2$ martingales. [guided] The choice of the integrand $1 + H_s^2$ in the definition of $S_n$ is deliberate. Using $H_s^2$ alone would ensure $H \in L^2(M^{S_n})$, but including the extra $1$ also ensures $\langle M \rangle_{t \wedge S_n} \leq n$, which forces $M^{S_n}$ into $\mathcal{M}^2_c$. We need both properties simultaneously: $M^{S_n}$ must be a square-integrable martingale (so that the $L^2$ stochastic integral is defined on it) and $H$ must be in $L^2(M^{S_n})$ (so that $H$ is an admissible integrand for that integral). Why does $\langle M^{S_n} \rangle_\infty \leq n$ imply $M^{S_n} \in \mathcal{M}^2_c$? By the [Quadratic Variation Norm Formula](/theorems/2085), $\mathbb{E}[(M^{S_n}_\infty)^2] = \mathbb{E}[\langle M^{S_n} \rangle_\infty] \leq n$. The process $M^{S_n}$ is a continuous local martingale with bounded quadratic variation, so it is bounded in $L^2$ and therefore a true $L^2$ martingale. Combined with [Doob's $L^2$ Maximal Inequality](/theorems/2112), $\mathbb{E}[\sup_t (M^{S_n}_t)^2] \leq 4\mathbb{E}[(M^{S_n}_\infty)^2] \leq 4n < \infty$. [/guided] [/step] [step:Verify consistency: $(H \cdot M^{S_n})^{S_m} = H \cdot M^{S_m}$ for $m \leq n$] Fix $m \leq n$. Then $S_m \leq S_n$ (since the integrand $1 + H_s^2 \geq 0$ makes $t \mapsto \int_0^t (1 + H_s^2)\, d\langle M \rangle_s$ non-decreasing, so the first hitting time of $m$ occurs no later than the first hitting time of $n$). We have $M^{S_m} = (M^{S_n})^{S_m}$, so by [Associativity of Stochastic Integration](/theorems/2094) applied to the indicator $\mathbb{1}_{[0, S_m]}$: \begin{align*} (H \cdot M^{S_n})^{S_m} = (\mathbb{1}_{[0, S_m]} H) \cdot M^{S_n} = H \cdot (M^{S_n})^{S_m} = H \cdot M^{S_m}. \end{align*} This consistency relation holds for all $m \leq n$. [guided] The consistency check is the crux of the localization argument. We are building a sequence of processes $H \cdot M^{S_1}, H \cdot M^{S_2}, \ldots$, and we need to know that the $n$-th approximation, when stopped at time $S_m$ (for $m \leq n$), gives exactly the $m$-th approximation. Without this, the limit process would depend on the choice of localizing sequence. Why does $(H \cdot M^{S_n})^{S_m} = H \cdot M^{S_m}$ hold? By [Associativity of Stochastic Integration](/theorems/2094), stopping $H \cdot M^{S_n}$ at $S_m$ is the same as integrating $\mathbb{1}_{[0, S_m]} H$ against $M^{S_n}$. But $\mathbb{1}_{[0, S_m]} H$ integrated against $M^{S_n}$ equals $H$ integrated against $\mathbb{1}_{[0, S_m]} \cdot M^{S_n} = (M^{S_n})^{S_m} = M^{S_m}$. The last equality uses $S_m \leq S_n$, so stopping $M^{S_n}$ at $S_m$ simply stops $M$ at $S_m$. [/guided] [/step] [step:Define $H \cdot M$ as the unique process satisfying $(H \cdot M)^{S_n} = H \cdot M^{S_n}$] By the consistency relation, the family $\{H \cdot M^{S_n}\}_{n \geq 1}$ is coherent: for each $\omega$ and each $n$, the path $t \mapsto (H \cdot M^{S_n})_t(\omega)$ agrees with $(H \cdot M^{S_m})_t(\omega)$ for all $t \leq S_m(\omega)$ and all $m \leq n$. Since $S_n(\omega) \nearrow \infty$ for almost every $\omega$, we define \begin{align*} (H \cdot M)_t(\omega) = (H \cdot M^{S_n})_t(\omega) \quad \text{for any } n \text{ with } S_n(\omega) > t. \end{align*} This is well-defined almost surely: for each $t$, the set $\{n : S_n > t\}$ is eventually all of $\mathbb{N}$ (since $S_n \nearrow \infty$), and the consistency relation ensures the values agree for different choices of $n$. The process $H \cdot M$ is continuous (since each $H \cdot M^{S_n}$ is continuous and the paths are defined by local agreement). It is adapted (since each $H \cdot M^{S_n}$ is adapted). By construction, $(H \cdot M)^{S_n} = H \cdot M^{S_n}$, which is a martingale (in $\mathcal{M}^2_c$). Since $S_n \nearrow \infty$, the sequence $(S_n)$ is a reducing sequence for $H \cdot M$, making $H \cdot M$ a continuous local martingale. Moreover, $(H \cdot M)_0 = (H \cdot M^{S_1})_0 = 0$. [/step] [step:Prove the bracket formula $\langle H \cdot M, N \rangle = H \cdot \langle M, N \rangle$ for every continuous local martingale $N$] Let $N$ be a continuous local martingale. We introduce additional stopping times to bring $N$ into $\mathcal{M}^2_c$. Define \begin{align*} S'_n = \inf\{t \geq 0 : |N_t| \geq n\}. \end{align*} Since $N$ is continuous, $S'_n \nearrow \infty$ almost surely. Set $T_n = S_n \wedge S'_n$, so that $T_n \nearrow \infty$ almost surely. The stopped process $N^{S'_n}$ is a continuous local martingale bounded by $n$, hence $N^{S'_n} \in \mathcal{M}^2_c$ (by the same argument as before: $\sup_t |N^{S'_n}_t| \leq n$ implies $\mathbb{E}[\sup_t (N^{S'_n}_t)^2] \leq n^2 < \infty$). At the stopped level $T_n$, both $H \cdot M^{S_n}$ and $N^{S'_n}$ lie in $\mathcal{M}^2_c$. The [Bracket of Two Stochastic Integrals](/theorems/2093) in the $L^2$ theory gives \begin{align*} \langle H \cdot M^{S_n}, N^{S'_n} \rangle = H \cdot \langle M^{S_n}, N^{S'_n} \rangle. \end{align*} Stopping both sides at $T_n$, and using $\langle M^{S_n}, N^{S'_n} \rangle_t = \langle M, N \rangle_{t \wedge S_n \wedge S'_n} = \langle M, N \rangle_{t \wedge T_n}$: \begin{align*} \langle H \cdot M, N \rangle^{T_n} = \langle H \cdot M^{S_n}, N^{S'_n} \rangle^{T_n} = (H \cdot \langle M^{S_n}, N^{S'_n} \rangle)^{T_n} = (H \cdot \langle M, N \rangle)^{T_n}. \end{align*} Since this holds for every $n$ and $T_n \nearrow \infty$, we conclude \begin{align*} \langle H \cdot M, N \rangle = H \cdot \langle M, N \rangle. \end{align*} [guided] We need to localize $N$ as well because the bracket formula $\langle H \cdot M^{S_n}, N \rangle = H \cdot \langle M^{S_n}, N \rangle$ from the $L^2$ theory requires both arguments to lie in $\mathcal{M}^2_c$. The first argument $H \cdot M^{S_n}$ is already in $\mathcal{M}^2_c$ by the construction above, but $N$ may not be square-integrable. Stopping $N$ at $S'_n = \inf\{t : |N_t| \geq n\}$ ensures $N^{S'_n}$ is bounded, hence in $\mathcal{M}^2_c$. The identity $\langle H \cdot M, N \rangle^{T_n} = (H \cdot \langle M, N \rangle)^{T_n}$ uses two facts. First, the covariation is bilinear and respects stopping: $\langle H \cdot M, N \rangle^{T_n} = \langle (H \cdot M)^{T_n}, N^{T_n} \rangle$. Second, $(H \cdot M)^{T_n} = (H \cdot M)^{S_n \wedge S'_n} = ((H \cdot M)^{S_n})^{S'_n} = (H \cdot M^{S_n})^{S'_n}$, and similarly for the bracket. Since the identity holds at each stopped level and $T_n \nearrow \infty$, the identity holds globally as an equality between continuous finite-variation processes. [/guided] [/step] [step:Prove uniqueness from the bracket characterization] Suppose $\tilde{M}$ is another continuous local martingale with $\tilde{M}_0 = 0$ satisfying $\langle \tilde{M}, N \rangle = H \cdot \langle M, N \rangle$ for every continuous local martingale $N$. Define $\Delta = H \cdot M - \tilde{M}$. Then $\Delta$ is a continuous local martingale with $\Delta_0 = 0$, and for every continuous local martingale $N$: \begin{align*} \langle \Delta, N \rangle = \langle H \cdot M, N \rangle - \langle \tilde{M}, N \rangle = H \cdot \langle M, N \rangle - H \cdot \langle M, N \rangle = 0. \end{align*} In particular, taking $N = \Delta$, we obtain $\langle \Delta \rangle = \langle \Delta, \Delta \rangle = 0$. By the [Quadratic Variation Vanishes Iff the Martingale Vanishes](/theorems/2084) theorem, $\Delta_t = 0$ for all $t \geq 0$ almost surely. Therefore $H \cdot M = \tilde{M}$. [/step] [step:Derive property (ii): stopping and indicator commutativity] Let $T$ be a stopping time. We must show $(\mathbb{1}_{[0,T]} H) \cdot M = (H \cdot M)^T = H \cdot M^T$. For the first equality, note that $(H \cdot M)^T$ is defined by stopping $H \cdot M$ at $T$. At the localized level, $((H \cdot M)^{S_n})^T = (H \cdot M^{S_n})^T$. By [Associativity of Stochastic Integration](/theorems/2094), $(H \cdot M^{S_n})^T = (\mathbb{1}_{[0,T]} H) \cdot M^{S_n}$. This means $(\mathbb{1}_{[0,T]} H) \cdot M^{S_n}$ and $(H \cdot M)^T$ agree when stopped at $S_n$, for every $n$. Since $S_n \nearrow \infty$, the two processes are indistinguishable: $(\mathbb{1}_{[0,T]} H) \cdot M = (H \cdot M)^T$. For the second equality, the process $H \cdot M^T$ is defined by integrating $H$ against the stopped martingale $M^T$. Using the localizing sequence $(S_n)$: $(M^T)^{S_n} = M^{T \wedge S_n}$ and $H \cdot M^{T \wedge S_n} = (\mathbb{1}_{[0,T]} H) \cdot M^{S_n}$ (by associativity in the $L^2$ theory). Therefore $H \cdot M^T$ and $(\mathbb{1}_{[0,T]} H) \cdot M$ agree at every stopped level, hence are indistinguishable. Combining: $(H \cdot M)^T = (\mathbb{1}_{[0,T]} H) \cdot M = H \cdot M^T$. [/step] [step:Derive property (iii): associativity of the stochastic integral] Let $K$ be previsible. We prove that $K \in L^2_{\mathrm{loc}}(H \cdot M)$ if and only if $KH \in L^2_{\mathrm{loc}}(M)$, and that $K \cdot (H \cdot M) = (KH) \cdot M$ when either condition holds. By the bracket formula (i) applied with $N = H \cdot M$: \begin{align*} \langle H \cdot M \rangle = \langle H \cdot M, H \cdot M \rangle = H \cdot \langle M, H \cdot M \rangle. \end{align*} By the [Bracket of Two Stochastic Integrals](/theorems/2093) at the localized level, $\langle M, H \cdot M \rangle = H \cdot \langle M \rangle$ (applying the bracket formula (i) with $N = M$). Therefore \begin{align*} \langle H \cdot M \rangle = H \cdot (H \cdot \langle M \rangle) = H^2 \cdot \langle M \rangle, \end{align*} where the last equality is [Associativity of Stochastic Integration](/theorems/2094) for Lebesgue-Stieltjes integrals. This means $d\langle H \cdot M \rangle_s = H_s^2\, d\langle M \rangle_s$. The condition $K \in L^2_{\mathrm{loc}}(H \cdot M)$ means $\int_0^t K_s^2\, d\langle H \cdot M \rangle_s < \infty$ almost surely for every $t$, which equals $\int_0^t K_s^2 H_s^2\, d\langle M \rangle_s < \infty$ for every $t$, which is exactly the condition $KH \in L^2_{\mathrm{loc}}(M)$. When these equivalent conditions hold, we verify $K \cdot (H \cdot M) = (KH) \cdot M$ via the bracket characterization (uniqueness). For any continuous local martingale $N$: \begin{align*} \langle K \cdot (H \cdot M), N \rangle = K \cdot \langle H \cdot M, N \rangle = K \cdot (H \cdot \langle M, N \rangle) = (KH) \cdot \langle M, N \rangle = \langle (KH) \cdot M, N \rangle, \end{align*} where we applied the bracket formula (i) twice (once for $K$ against $H \cdot M$, once for $KH$ against $M$) and associativity of Lebesgue-Stieltjes integration in the middle step. Both $K \cdot (H \cdot M)$ and $(KH) \cdot M$ are continuous local martingales starting at $0$ with the same bracket against every continuous local martingale $N$, so by uniqueness they are indistinguishable. [/step] [step:Derive property (iv): coincidence with the $L^2$ definition when $M \in \mathcal{M}^2_c$ and $H \in L^2(M)$] Suppose $M \in \mathcal{M}^2_c$ and $H \in L^2(M)$, so that the stochastic integral $H \cdot M$ is already defined by the [Ito Isometry](/theorems/2092) construction. Denote this $L^2$ integral by $(H \cdot M)^{L^2}$. Since $M \in \mathcal{M}^2_c$ and $H \in L^2(M)$, we have $H \in L^2_{\mathrm{loc}}(M)$ (the $L^2$ condition is stronger than $L^2_{\mathrm{loc}}$). The localization construction above produces a continuous local martingale $H \cdot M$ with bracket formula (i). But $(H \cdot M)^{L^2}$ also satisfies the bracket formula: by the [Bracket of Two Stochastic Integrals](/theorems/2093) in the $L^2$ theory, $\langle (H \cdot M)^{L^2}, N \rangle = H \cdot \langle M, N \rangle$ for every $N \in \mathcal{M}^2_c$. Both $(H \cdot M)^{L^2}$ and $H \cdot M$ are continuous local martingales starting at $0$ with the same bracket against every continuous local martingale, so the uniqueness established above gives $H \cdot M = (H \cdot M)^{L^2}$. [/step]