[proofplan] We compute the cross-bracket $\langle H \cdot M, N \rangle_t$ for a simple previsible process $H \in \mathcal{E}$ by decomposing $H \cdot M = \sum_i X^i$ into elementary pieces, using bilinearity of the bracket to reduce to $\sum_i \langle X^i, N \rangle$, computing each $\langle X^i, N \rangle$ via the scaling and stopping properties of the covariation, and summing to recognise the result as the Lebesgue--Stieltjes integral $\int_0^t H_s \, d\langle M, N \rangle_s$. [/proofplan]

[step:Decompose $H \cdot M$ into elementary pieces and apply bilinearity of the bracket] Let $H \in \mathcal{E}$ be a simple previsible process: \begin{align*} H_s = \sum_{i=1}^{k} H_{t_{i-1}} \, \mathbb{1}_{(t_{i-1}, t_i]}(s), \end{align*} where $0 = t_0 < t_1 < \cdots < t_k$ and each $H_{t_{i-1}}$ is a bounded $\mathcal{F}_{t_{i-1}}$-measurable random variable. As in the proof of the [Ito Isometry on Simple Processes](/theorems/2089), define \begin{align*} X^i_t := H_{t_{i-1}} (M_{t_i \wedge t} - M_{t_{i-1} \wedge t}), \end{align*} so that $(H \cdot M)_t = \sum_{i=1}^k X^i_t$. By part (ii) of the [Properties of Covariation](/theorems/2086), the bracket is bilinear, so \begin{align*} \langle H \cdot M, N \rangle_t = \left\langle \sum_{i=1}^k X^i, N \right\rangle_t = \sum_{i=1}^k \langle X^i, N \rangle_t. \end{align*} [/step]

[step:Compute $\langle X^i, N \rangle_t = H_{t_{i-1}} (\langle M, N \rangle_{t_i \wedge t} - \langle M, N \rangle_{t_{i-1} \wedge t})$]Fix $i \in \{1, \ldots, k\}$. The process $X^i = H_{t_{i-1}} \widetilde{M}^i$ where $\widetilde{M}^i_t = M_{t_i \wedge t} - M_{t_{i-1} \wedge t}$. By the characterisation of covariation (part (i) of the [Properties of Covariation](/theorems/2086)), $\langle X^i, N \rangle$ is the unique continuous finite variation process starting at $0$ such that $X^i_t N_t - \langle X^i, N \rangle_t$ is a continuous local martingale. We claim $\langle X^i, N \rangle_t = H_{t_{i-1}} \langle \widetilde{M}^i, N \rangle_t$. To verify this, consider \begin{align*} X^i_t N_t - H_{t_{i-1}} \langle \widetilde{M}^i, N \rangle_t = H_{t_{i-1}} \left[ \widetilde{M}^i_t N_t - \langle \widetilde{M}^i, N \rangle_t \right]. \end{align*} The bracketed process $\widetilde{M}^i N - \langle \widetilde{M}^i, N \rangle$ is a continuous local martingale by the characterisation of covariation. Since $H_{t_{i-1}}$ is bounded and $\mathcal{F}_{t_{i-1}}$-measurable, the product $H_{t_{i-1}}[\widetilde{M}^i N - \langle \widetilde{M}^i, N \rangle]$ is again a continuous local martingale (for $t \geq t_{i-1}$, $H_{t_{i-1}}$ is $\mathcal{F}_s$-measurable for $s \geq t_{i-1}$ and can be factored out of conditional expectations; for $t < t_{i-1}$, both $\widetilde{M}^i_t = 0$ and $\langle \widetilde{M}^i, N \rangle_t = 0$). The process $H_{t_{i-1}} \langle \widetilde{M}^i, N \rangle$ is continuous and of finite variation (since $\langle \widetilde{M}^i, N \rangle$ is, and $H_{t_{i-1}}$ is $\omega$-dependent but does not depend on $t$). By uniqueness, \begin{align*} \langle X^i, N \rangle_t = H_{t_{i-1}} \langle \widetilde{M}^i, N \rangle_t. \end{align*} Now compute $\langle \widetilde{M}^i, N \rangle_t$. Since $\widetilde{M}^i = M^{t_i} - M^{t_{i-1}}$, bilinearity of the bracket gives \begin{align*} \langle \widetilde{M}^i, N \rangle_t = \langle M^{t_i}, N \rangle_t - \langle M^{t_{i-1}}, N \rangle_t. \end{align*} By part (iv) of the [Properties of Covariation](/theorems/2086), $\langle M^{t_j}, N \rangle_t = \langle M, N \rangle_{t_j \wedge t}$ for each $j$. Therefore \begin{align*} \langle X^i, N \rangle_t = H_{t_{i-1}} \left(\langle M, N \rangle_{t_i \wedge t} - \langle M, N \rangle_{t_{i-1} \wedge t}\right). \end{align*}[/step]

[guided]The computation parallels the quadratic variation calculation in the [Ito Isometry on Simple Processes](/theorems/2089), but now we work with the cross-bracket $\langle \cdot, N \rangle$ instead of the self-bracket $\langle \cdot \rangle$. The key ingredients are the same: 1. Scaling by a predictable random variable: $\langle H_{t_{i-1}} L, N \rangle = H_{t_{i-1}} \langle L, N \rangle$. This holds because the defining property of the bracket (that $LN - \langle L, N \rangle$ is a local martingale) is preserved under multiplication by the bounded $\mathcal{F}_{t_{i-1}}$-measurable factor $H_{t_{i-1}}$. 2. Stopping property: $\langle M^{t_j}, N \rangle_t = \langle M, N \rangle_{t_j \wedge t}$. This is part (iv) of the [Properties of Covariation](/theorems/2086). 3. Bilinearity: $\langle M^{t_i} - M^{t_{i-1}}, N \rangle = \langle M^{t_i}, N \rangle - \langle M^{t_{i-1}}, N \rangle$. Combining these, $\langle X^i, N \rangle_t = H_{t_{i-1}}(\langle M, N \rangle_{t_i \wedge t} - \langle M, N \rangle_{t_{i-1} \wedge t})$, which is the increment of $\langle M, N \rangle$ over the interval $(t_{i-1} \wedge t, \, t_i \wedge t]$, scaled by $H_{t_{i-1}}$.[/guided]

[step:Sum and recognise the Lebesgue--Stieltjes integral] Summing over $i = 1, \ldots, k$: \begin{align*} \langle H \cdot M, N \rangle_t = \sum_{i=1}^k H_{t_{i-1}} \left(\langle M, N \rangle_{t_i \wedge t} - \langle M, N \rangle_{t_{i-1} \wedge t}\right). \end{align*} Since $H_s = \sum_{i=1}^k H_{t_{i-1}} \mathbb{1}_{(t_{i-1}, t_i]}(s)$ is a step function, the right-hand side is exactly the Lebesgue--Stieltjes integral of $H$ against the signed measure $d\langle M, N \rangle$ on $(0, t]$: \begin{align*} \sum_{i=1}^k H_{t_{i-1}} \left(\langle M, N \rangle_{t_i \wedge t} - \langle M, N \rangle_{t_{i-1} \wedge t}\right) = \int_0^t H_s \, d\langle M, N \rangle_s. \end{align*} This is the definition of the Lebesgue--Stieltjes integral for a step function: the integral of $H_{t_{i-1}} \mathbb{1}_{(t_{i-1}, t_i]}$ against the measure induced by the function $s \mapsto \langle M, N \rangle_s$ is $H_{t_{i-1}}$ times the increment of $\langle M, N \rangle$ over $(t_{i-1} \wedge t, \, t_i \wedge t]$. Therefore \begin{align*} \langle H \cdot M, N \rangle_t = \int_0^t H_s \, d\langle M, N \rangle_s. \end{align*} [/step]