[proofplan]
We apply the bracket identity from part (ii) of the [Itô Isometry](/theorems/2092) twice in succession. First, $\langle H \cdot M, K \cdot N \rangle_t = \int_0^t H_s \, d\langle M, K \cdot N \rangle_s$ (treating $K \cdot N$ as the second martingale). Then the inner bracket $\langle M, K \cdot N \rangle$ is itself expressed as $\int_0^{(\cdot)} K_s \, d\langle M, N \rangle_s$ by applying part (ii) again (with the roles of integrand and martingale swapped). Substituting and using the associativity of Lebesgue–Stieltjes integration against the measure $d\langle M, N \rangle_s$ produces the result.
[/proofplan]
[step:Express $\langle H \cdot M, K \cdot N \rangle_t$ using the bracket identity for $H \cdot M$]
Both $H \cdot M$ and $K \cdot N$ belong to $\mathcal{M}^2_c$ by the [Itô Isometry](/theorems/2092) (part (i)), since $H \in L^2(M)$ and $K \in L^2(N)$. By part (ii) of the [Itô Isometry](/theorems/2092), applied with the continuous square-integrable martingale $H \cdot M$ and the test martingale $K \cdot N \in \mathcal{M}^2_c$,
\begin{align*}
\langle H \cdot M, K \cdot N \rangle_t = \int_0^t H_s \, d\langle M, K \cdot N \rangle_s.
\end{align*}
[/step]
[step:Compute the inner bracket $\langle M, K \cdot N \rangle_s$ using the bracket identity for $K \cdot N$]
By symmetry of covariation ([Properties of Covariation](/theorems/2086), part (ii)), $\langle M, K \cdot N \rangle = \langle K \cdot N, M \rangle$. Applying part (ii) of the [Itô Isometry](/theorems/2092) with integrand $K$, integrator $N$, and test martingale $M$,
\begin{align*}
\langle K \cdot N, M \rangle_s = \int_0^s K_u \, d\langle N, M \rangle_u = \int_0^s K_u \, d\langle M, N \rangle_u,
\end{align*}
where the second equality uses symmetry $\langle N, M \rangle = \langle M, N \rangle$.
[/step]
[step:Substitute and identify the iterated integral as $\int_0^t H_s K_s \, d\langle M, N \rangle_s$]
The measure $d\langle M, K \cdot N \rangle_s$ is the Lebesgue–Stieltjes measure of the function $s \mapsto \langle M, K \cdot N \rangle_s = \int_0^s K_u \, d\langle M, N \rangle_u$. By the definition of the Lebesgue–Stieltjes integral, this measure has density $K_s$ with respect to $d\langle M, N \rangle_s$: for any Borel set $B \subseteq [0, \infty)$,
\begin{align*}
\int_B d\langle M, K \cdot N \rangle_s = \int_B K_s \, d\langle M, N \rangle_s.
\end{align*}
Substituting into the expression from the first step,
\begin{align*}
\langle H \cdot M, K \cdot N \rangle_t = \int_0^t H_s \, d\langle M, K \cdot N \rangle_s = \int_0^t H_s K_s \, d\langle M, N \rangle_s.
\end{align*}
[guided]
The key idea is that the bracket identity in part (ii) of the [Itô Isometry](/theorems/2092) can be applied in two different ways: once to "peel off" the integrand $H$ from $H \cdot M$, and once to "peel off" the integrand $K$ from $K \cdot N$.
Starting from the bracket $\langle H \cdot M, K \cdot N \rangle_t$, part (ii) of the [Itô Isometry](/theorems/2092) gives
\begin{align*}
\langle H \cdot M, K \cdot N \rangle_t = \int_0^t H_s \, d\langle M, K \cdot N \rangle_s.
\end{align*}
Now the inner bracket $\langle M, K \cdot N \rangle$ involves a stochastic integral $K \cdot N$. By symmetry of covariation ([Properties of Covariation](/theorems/2086)), $\langle M, K \cdot N \rangle = \langle K \cdot N, M \rangle$. Applying part (ii) of the [Itô Isometry](/theorems/2092) again, now with integrand $K$, integrator $N$, and test martingale $M$:
\begin{align*}
\langle K \cdot N, M \rangle_s = \int_0^s K_u \, d\langle N, M \rangle_u = \int_0^s K_u \, d\langle M, N \rangle_u.
\end{align*}
What does this mean for the measure $d\langle M, K \cdot N \rangle_s$? The function $s \mapsto \langle M, K \cdot N \rangle_s$ is a Lebesgue–Stieltjes integral of $K$ against $d\langle M, N \rangle$. Therefore the Radon–Nikodym derivative of the signed measure $d\langle M, K \cdot N \rangle$ with respect to $d\langle M, N \rangle$ is $K_s$. This is the chain-rule property of Lebesgue–Stieltjes integrals: if $A(t) = \int_0^t f(s) \, d\mu(s)$, then $\int g \, dA = \int g \, f \, d\mu$. Applying this with $g = H$, $f = K$, $\mu = d\langle M, N \rangle$:
\begin{align*}
\langle H \cdot M, K \cdot N \rangle_t = \int_0^t H_s \, d\langle M, K \cdot N \rangle_s = \int_0^t H_s K_s \, d\langle M, N \rangle_s.
\end{align*}
[/guided]
[/step]
