[proofplan] We prove associativity in two stages. First, we show the $L^2$ norm equivalence: $\|K\|_{L^2(H \cdot M)} = \|KH\|_{L^2(M)}$, which follows from the bracket identity $\langle H \cdot M \rangle_t = \int_0^t H_s^2 \, d\langle M \rangle_s$ (a consequence of part (ii) of the [Itô Isometry](/theorems/2092)). This establishes that $K \in L^2(H \cdot M)$ if and only if $KH \in L^2(M)$. Second, we verify that both $K \cdot (H \cdot M)$ and $(KH) \cdot M$ satisfy the same bracket identity against every $N \in \mathcal{M}^2_c$ by iterating the bracket formula from the [Itô Isometry](/theorems/2092) (part (ii)), and uniqueness identifies them. [/proofplan]

[step:Compute $\langle H \cdot M \rangle_t$ and establish the norm equivalence $\|K\|_{L^2(H \cdot M)} = \|KH\|_{L^2(M)}$]By part (ii) of the [Itô Isometry](/theorems/2092), taking $N = H \cdot M$, \begin{align*} \langle H \cdot M, H \cdot M \rangle_t = \int_0^t H_s \, d\langle M, H \cdot M \rangle_s. \end{align*} Applying part (ii) again to the inner bracket (with integrand $H$, integrator $M$, and test martingale $M$): \begin{align*} \langle M, H \cdot M \rangle_t = \langle H \cdot M, M \rangle_t = \int_0^t H_s \, d\langle M, M \rangle_s = \int_0^t H_s \, d\langle M \rangle_s, \end{align*} where symmetry of covariation ([Properties of Covariation](/theorems/2086)) is used for the first equality. Substituting and applying the chain rule for Lebesgue–Stieltjes integrals: \begin{align*} \langle H \cdot M \rangle_t = \int_0^t H_s \, d\!\left(\int_0^s H_u \, d\langle M \rangle_u\right) = \int_0^t H_s^2 \, d\langle M \rangle_s. \end{align*} Using this, the $L^2(H \cdot M)$ norm of $K$ is \begin{align*} \|K\|_{L^2(H \cdot M)}^2 = \mathbb{E}\!\left[\int_0^\infty K_s^2 \, d\langle H \cdot M \rangle_s\right] = \mathbb{E}\!\left[\int_0^\infty K_s^2 H_s^2 \, d\langle M \rangle_s\right] = \|KH\|_{L^2(M)}^2. \end{align*} In particular, $K \in L^2(H \cdot M)$ if and only if $KH \in L^2(M)$.[/step]

[guided]We need to show $\|K\|_{L^2(H \cdot M)} = \|KH\|_{L^2(M)}$. The $L^2(H \cdot M)$ norm is defined as $\mathbb{E}\!\left[\int_0^\infty K_s^2 \, d\langle H \cdot M \rangle_s\right]^{1/2}$, so we need to express $d\langle H \cdot M \rangle_s$ in terms of $d\langle M \rangle_s$. The quadratic variation of $H \cdot M$ is $\langle H \cdot M \rangle = \langle H \cdot M, H \cdot M \rangle$. Applying the [Bracket of Two Stochastic Integrals](/theorems/2093) with both integrands equal to $H$ and both integrators equal to $M$: \begin{align*} \langle H \cdot M \rangle_t = \langle H \cdot M, H \cdot M \rangle_t = \int_0^t H_s \cdot H_s \, d\langle M, M \rangle_s = \int_0^t H_s^2 \, d\langle M \rangle_s. \end{align*} This is the chain rule for quadratic variation under stochastic integration: integrating against $H$ scales the quadratic variation measure by $H^2$. Substituting into the norm: \begin{align*} \|K\|_{L^2(H \cdot M)}^2 = \mathbb{E}\!\left[\int_0^\infty K_s^2 \, d\langle H \cdot M \rangle_s\right] = \mathbb{E}\!\left[\int_0^\infty K_s^2 H_s^2 \, d\langle M \rangle_s\right] = \|KH\|_{L^2(M)}^2. \end{align*} The chain rule for Lebesgue–Stieltjes integrals justifies replacing $d\langle H \cdot M \rangle_s = H_s^2 \, d\langle M \rangle_s$ inside the integral: if $A(t) = \int_0^t f(s) \, d\mu(s)$ with $f \geq 0$, then $\int g \, dA = \int g \cdot f \, d\mu$. This norm identity shows that $K \in L^2(H \cdot M)$ iff $KH \in L^2(M)$, since both conditions are equivalent to $\mathbb{E}\!\left[\int_0^\infty K_s^2 H_s^2 \, d\langle M \rangle_s\right] < \infty$.[/guided]

[step:Verify both sides satisfy the same bracket identity against all $N \in \mathcal{M}^2_c$]Assume $KH \in L^2(M)$ (equivalently, $K \in L^2(H \cdot M)$). Both $K \cdot (H \cdot M)$ and $(KH) \cdot M$ belong to $\mathcal{M}^2_c$ by the [Itô Isometry](/theorems/2092) (part (i)). We compute their brackets against an arbitrary $N \in \mathcal{M}^2_c$. **Bracket of $K \cdot (H \cdot M)$.** By part (ii) of the [Itô Isometry](/theorems/2092), applied with integrand $K$ and integrator $H \cdot M$, \begin{align*} \langle K \cdot (H \cdot M), N \rangle_t = \int_0^t K_s \, d\langle H \cdot M, N \rangle_s. \end{align*} By part (ii) again, $\langle H \cdot M, N \rangle_s = \int_0^s H_u \, d\langle M, N \rangle_u$. The measure $d\langle H \cdot M, N \rangle_s$ therefore has density $H_s$ with respect to $d\langle M, N \rangle_s$, so the chain rule for Lebesgue–Stieltjes integrals gives \begin{align*} \langle K \cdot (H \cdot M), N \rangle_t = \int_0^t K_s H_s \, d\langle M, N \rangle_s. \end{align*} **Bracket of $(KH) \cdot M$.** By part (ii) of the [Itô Isometry](/theorems/2092), applied directly with integrand $KH$ and integrator $M$, \begin{align*} \langle (KH) \cdot M, N \rangle_t = \int_0^t K_s H_s \, d\langle M, N \rangle_s. \end{align*} The two brackets coincide for all $N \in \mathcal{M}^2_c$. By the uniqueness statement in part (ii) of the [Itô Isometry](/theorems/2092), $K \cdot (H \cdot M) = (KH) \cdot M$ in $\mathcal{M}^2_c$.[/step]

[guided]The strategy for proving equalities between stochastic integrals in $\mathcal{M}^2_c$ is always the same: show both sides have the same bracket against all test martingales $N \in \mathcal{M}^2_c$, then invoke the uniqueness from part (ii) of the [Itô Isometry](/theorems/2092). For $K \cdot (H \cdot M)$, we apply the bracket identity twice. The first application peels off the outer integrand $K$: \begin{align*} \langle K \cdot (H \cdot M), N \rangle_t = \int_0^t K_s \, d\langle H \cdot M, N \rangle_s. \end{align*} The second application peels off the inner integrand $H$ from the bracket $\langle H \cdot M, N \rangle$: \begin{align*} \langle H \cdot M, N \rangle_s = \int_0^s H_u \, d\langle M, N \rangle_u. \end{align*} Combining via the chain rule: $d\langle H \cdot M, N \rangle_s = H_s \, d\langle M, N \rangle_s$, so $\int_0^t K_s \, d\langle H \cdot M, N \rangle_s = \int_0^t K_s H_s \, d\langle M, N \rangle_s$. For $(KH) \cdot M$, the bracket identity applies in one shot: $\langle (KH) \cdot M, N \rangle_t = \int_0^t K_s H_s \, d\langle M, N \rangle_s$. Both expressions equal $\int_0^t K_s H_s \, d\langle M, N \rangle_s$ for all $N$. By the uniqueness in part (ii) of the [Itô Isometry](/theorems/2092), the two processes $K \cdot (H \cdot M)$ and $(KH) \cdot M$ are indistinguishable. Note that the whole argument is the "algebraic" counterpart of the chain rule: integrating $K$ against an integral of $H$ against $d\langle M, N \rangle$ is the same as integrating $KH$ directly. The stochastic integral inherits this property from the Lebesgue–Stieltjes integral because the bracket identity reduces everything to pathwise Lebesgue–Stieltjes integrals.[/guided]