[proofplan]
Each property follows from the corresponding property for the martingale part $H \cdot M$ (established by the [Itô Isometry](/theorems/2092)) and the finite variation part $H \cdot A$ (defined as a Lebesgue–Stieltjes integral), since the semimartingale integral is $H \cdot X := H \cdot M + H \cdot A$. Bilinearity, associativity, and stopping are inherited by linearity of both components. Structure preservation holds because $H \cdot M$ remains a continuous local martingale and $H \cdot A$ remains a finite variation process. For the simple process formula with unbounded $H_{i-1}$, the finite variation part is immediate from the Lebesgue–Stieltjes integral of a step function, and the martingale part is recovered by localizing with stopping times $T_n = \inf\{t : |H_t| \geq n\}$ and passing to the limit.
[/proofplan]
[step:Establish bilinearity of $(H, X) \mapsto H \cdot X$]
Write $X = X_0 + M + A$ where $M \in \mathcal{M}^2_{c,\text{loc}}$ and $A$ is a continuous adapted process of finite variation with $A_0 = 0$. The semimartingale integral is defined as
\begin{align*}
H \cdot X := H \cdot M + H \cdot A,
\end{align*}
where $H \cdot M$ is the stochastic integral and $H \cdot A$ is the Lebesgue–Stieltjes integral $\int_0^{(\cdot)} H_s \, dA_s$.
**Linearity in $H$.** For $\alpha, \beta \in \mathbb{R}$ and locally bounded previsible $H$, $K$:
\begin{align*}
(\alpha H + \beta K) \cdot X &= (\alpha H + \beta K) \cdot M + (\alpha H + \beta K) \cdot A \\
&= \alpha(H \cdot M) + \beta(K \cdot M) + \alpha(H \cdot A) + \beta(K \cdot A) \\
&= \alpha(H \cdot M + H \cdot A) + \beta(K \cdot M + K \cdot A) \\
&= \alpha(H \cdot X) + \beta(K \cdot X),
\end{align*}
where the second equality uses linearity of the stochastic integral in the integrand (from the [Itô Isometry](/theorems/2092)) and linearity of the Lebesgue–Stieltjes integral.
**Linearity in $X$.** Write $Y = Y_0 + N + B$. Then $\alpha X + \beta Y = (\alpha X_0 + \beta Y_0) + (\alpha M + \beta N) + (\alpha A + \beta B)$, and
\begin{align*}
H \cdot (\alpha X + \beta Y) &= H \cdot (\alpha M + \beta N) + H \cdot (\alpha A + \beta B) \\
&= \alpha(H \cdot M) + \beta(H \cdot N) + \alpha(H \cdot A) + \beta(H \cdot B) \\
&= \alpha(H \cdot X) + \beta(H \cdot Y).
\end{align*}
[/step]
[step:Prove associativity $H \cdot (K \cdot X) = (HK) \cdot X$]
Write $X = X_0 + M + A$. Then $K \cdot X = K \cdot M + K \cdot A$, where $K \cdot M$ is a continuous local martingale and $K \cdot A$ is a continuous finite variation process. Therefore $K \cdot X$ is a continuous semimartingale with decomposition $(K \cdot X)_0 + (K \cdot M) + (K \cdot A)$, where $(K \cdot X)_0 = 0$.
Applying the definition of the semimartingale integral:
\begin{align*}
H \cdot (K \cdot X) = H \cdot (K \cdot M) + H \cdot (K \cdot A).
\end{align*}
For the martingale component, the [Associativity of Stochastic Integration](/theorems/2094) gives $H \cdot (K \cdot M) = (HK) \cdot M$. For the finite variation component, associativity of the Lebesgue–Stieltjes integral gives $H \cdot (K \cdot A) = (HK) \cdot A$ pathwise: if $B(t) = \int_0^t K_s \, dA_s$ is a finite variation function, then $\int_0^t H_s \, dB_s = \int_0^t H_s K_s \, dA_s$ by the chain rule for Lebesgue–Stieltjes integrals. Therefore
\begin{align*}
H \cdot (K \cdot X) = (HK) \cdot M + (HK) \cdot A = (HK) \cdot X.
\end{align*}
[/step]
[step:Verify the stopping identity $(H \cdot X)^T = (\mathbb{1}_{[0,T]} H) \cdot X = H \cdot X^T$]
For the martingale component, part (iii) of the [Itô Isometry](/theorems/2092) gives $(H \cdot M)^T = (\mathbb{1}_{[0,T]} H) \cdot M = H \cdot M^T$.
For the finite variation component, the Lebesgue–Stieltjes integral satisfies $(H \cdot A)_t^T = \int_0^{t \wedge T} H_s \, dA_s = \int_0^t \mathbb{1}_{[0,T]}(s) H_s \, dA_s = (\mathbb{1}_{[0,T]} H \cdot A)_t$, and also $\int_0^{t \wedge T} H_s \, dA_s = \int_0^t H_s \, dA^T_s = (H \cdot A^T)_t$, since the measure $dA^T_s$ is the restriction of $dA_s$ to $[0, T]$.
Adding the components:
\begin{align*}
(H \cdot X)^T_t &= (H \cdot M)^T_t + (H \cdot A)^T_t = (\mathbb{1}_{[0,T]} H) \cdot M_t + (\mathbb{1}_{[0,T]} H) \cdot A_t = (\mathbb{1}_{[0,T]} H) \cdot X_t, \\
(H \cdot X)^T_t &= (H \cdot M^T)_t + (H \cdot A^T)_t = (H \cdot X^T)_t.
\end{align*}
[/step]
[step:Show structure preservation: $H \cdot X$ inherits the type of $X$]
**If $X$ is a continuous local martingale,** then $A = 0$ in the decomposition $X = X_0 + M + A$, and $H \cdot X = H \cdot M$. The stochastic integral $H \cdot M$ of a locally bounded previsible process against a continuous local martingale is a continuous local martingale. This follows from the [Itô Isometry](/theorems/2092) by localization: choose stopping times $T_n \nearrow \infty$ such that $\mathbb{1}_{[0,T_n]} H$ is bounded and $M^{T_n} \in \mathcal{M}^2_c$. Then $(H \cdot M)^{T_n} = (\mathbb{1}_{[0,T_n]} H) \cdot M^{T_n} \in \mathcal{M}^2_c$ by part (i) of the [Itô Isometry](/theorems/2092), so $(T_n)$ is a reducing sequence for $H \cdot M$.
**If $X$ is a continuous finite variation process,** then $M = 0$ and $H \cdot X = H \cdot A = \int_0^{(\cdot)} H_s \, dA_s$, which is a continuous finite variation process: its total variation on $[0, t]$ is $\int_0^t |H_s| \, |dA_s| < \infty$ since $H$ is locally bounded and $A$ is of finite variation on compact intervals.
[/step]
[step:Verify the simple process formula for unbounded $H_{i-1}$]
Let $H = \sum_{i=1}^n H_{i-1} \, \mathbb{1}_{(t_{i-1}, t_i]}$ with $H_{i-1} \in \mathcal{F}_{t_{i-1}}$ (not necessarily bounded). We must show
\begin{align*}
(H \cdot X)_t = \sum_{i=1}^n H_{i-1}(X_{t_i \wedge t} - X_{t_{i-1} \wedge t}).
\end{align*}
**Finite variation part.** For each fixed $\omega$, $H_{i-1}(\omega)$ is a real number, and $A(\cdot, \omega)$ is a finite variation function. The Lebesgue–Stieltjes integral of a simple function against a signed measure evaluates to
\begin{align*}
(H \cdot A)_t(\omega) = \sum_{i=1}^n H_{i-1}(\omega) \bigl(A_{t_i \wedge t}(\omega) - A_{t_{i-1} \wedge t}(\omega)\bigr).
\end{align*}
**Martingale part.** Define stopping times $T_m := \inf\{t \geq 0 : \max_{0 \leq i \leq n-1} |H_i| \geq m\}$. Since each $H_i$ is finite a.s., $T_m \nearrow \infty$ a.s. The process $\mathbb{1}_{[0,T_m]} H$ is bounded by $m$ and is a simple previsible process, so $\mathbb{1}_{[0,T_m]} H \in \mathcal{E}$. By the definition of the stochastic integral on $\mathcal{E}$,
\begin{align*}
(\mathbb{1}_{[0,T_m]} H \cdot M)_t = \sum_{i=1}^n H_{i-1} \, \mathbb{1}_{\{T_m > t_{i-1}\}} (M_{t_i \wedge t \wedge T_m} - M_{t_{i-1} \wedge t \wedge T_m}).
\end{align*}
By the stopping identity (part (iii)), $(H \cdot M)^{T_m} = (\mathbb{1}_{[0,T_m]} H) \cdot M$. On the event $\{T_m > t\}$ (which has $\mathbb{P}(\{T_m > t\}) \to 1$ as $m \to \infty$), $T_m \wedge t = t$ and the formula reduces to
\begin{align*}
(H \cdot M)_t = \sum_{i=1}^n H_{i-1}(M_{t_i \wedge t} - M_{t_{i-1} \wedge t}).
\end{align*}
Since $T_m \to \infty$ a.s., this identity holds a.s. Adding the finite variation part gives the stated formula.
[guided]
When $H_{i-1}$ is bounded, $H \in \mathcal{E}$ and the formula is the definition of the stochastic integral for simple processes. The difficulty arises when $H_{i-1}$ is $\mathcal{F}_{t_{i-1}}$-measurable but unbounded.
The finite variation part poses no problem: the Lebesgue–Stieltjes integral is defined pathwise, and for each fixed $\omega$ the integrand $H(\cdot, \omega)$ is a step function multiplied by the finite real numbers $H_{i-1}(\omega)$. The integral evaluates to the finite sum $\sum_i H_{i-1}(\omega)(A_{t_i \wedge t}(\omega) - A_{t_{i-1} \wedge t}(\omega))$ regardless of boundedness.
For the martingale part, we use a localization argument. Define
\begin{align*}
T_m := \inf\!\left\{t \geq 0 : \max_{0 \leq i \leq n-1} |H_i| \geq m\right\}.
\end{align*}
Since $H$ has finitely many values $H_0, \ldots, H_{n-1}$, each finite a.s., $T_m \nearrow \infty$ a.s. The truncated process $\mathbb{1}_{[0,T_m]} H$ is bounded by $m$ and remains a simple previsible process, so $\mathbb{1}_{[0,T_m]} H \in \mathcal{E}$.
By the definition of the stochastic integral for $\mathcal{E}$, $(\mathbb{1}_{[0,T_m]} H \cdot M)_t$ equals the sum $\sum_i H_{i-1} \, \mathbb{1}_{\{T_m > t_{i-1}\}}(M_{t_i \wedge t \wedge T_m} - M_{t_{i-1} \wedge t \wedge T_m})$. By the stopping identity (part (iii)), this equals $(H \cdot M)_{t \wedge T_m}$. On the event $\{T_m > t_n\}$ (where $t_n$ is the last partition point), $T_m$ does not intervene and the formula simplifies to $(H \cdot M)_t = \sum_i H_{i-1}(M_{t_i \wedge t} - M_{t_{i-1} \wedge t})$.
Since $\mathbb{P}(T_m > t_n) \to 1$ as $m \to \infty$, the formula holds a.s. Combining with the finite variation part gives the full result.
[/guided]
[/step]
