[proofplan] We decompose the semimartingale $X = X_0 + M + A$ and handle the martingale and finite variation components separately. For the finite variation part, the classical dominated convergence theorem for Lebesgue–Stieltjes integrals gives pathwise (hence a.s.) convergence of $\int_0^t H^n_s \, dA_s$ to $\int_0^t H_s \, dA_s$. For the martingale part, we localize using stopping times $T_m = \inf\{t : \int_0^t K_s^2 \, d\langle M \rangle_s \geq m\}$ to reduce to the $L^2$ setting, apply the [Itô Isometry](/theorems/2092) to control $\mathbb{E}[(H^n \cdot M - H \cdot M)_{t \wedge T_m}^2]$ by $\mathbb{E}[\int_0^{t \wedge T_m} (H^n_s - H_s)^2 \, d\langle M \rangle_s]$, and invoke the classical dominated convergence theorem (with dominator $4K_s^2$ bounded by $4m$ after stopping) to send this to zero. Since $T_m \to \infty$ a.s., convergence in probability on $[0, t \wedge T_m]$ for every $m$ upgrades to convergence in probability on $[0, t]$. [/proofplan] [step:Handle the finite variation component via classical dominated convergence] Write $X = X_0 + M + A$. The finite variation part of the integral is the Lebesgue–Stieltjes integral \begin{align*} (H^n \cdot A)_t = \int_0^t H^n_s \, dA_s. \end{align*} For each fixed $\omega \in \Omega$, the function $s \mapsto A_s(\omega)$ is a continuous function of finite variation on $[0, t]$, so $dA_s(\omega)$ defines a signed Borel measure $\mu_\omega$ on $[0, t]$ with total mass $\int_0^t |dA_s(\omega)| < \infty$ by hypothesis (iii). The integrand satisfies: - $H^n_s(\omega) \to H_s(\omega)$ for all $s \in [0, t]$ by hypothesis (i), - $|H^n_s(\omega)| \leq K_s(\omega)$ for all $s, n$ by hypothesis (ii), - $\int_0^t K_s(\omega) \, |dA_s(\omega)| < \infty$ by hypothesis (iii). By the classical dominated convergence theorem for signed measures (applied pathwise to the measure $\mu_\omega$ with dominator $K_s(\omega)$), \begin{align*} \int_0^t H^n_s \, dA_s \to \int_0^t H_s \, dA_s \quad \text{a.s. as } n \to \infty. \end{align*} In particular, $(H^n \cdot A)_t \to (H \cdot A)_t$ in probability. [/step] [step:Localize the martingale component to reduce to the $L^2$ setting] Define stopping times \begin{align*} T_m := \inf\!\left\{t \geq 0 : \int_0^t K_s^2 \, d\langle M \rangle_s \geq m\right\}, \quad m \geq 1. \end{align*} By hypothesis (iii), $\int_0^t K_s^2 \, d\langle M \rangle_s < \infty$ a.s., so $T_m \nearrow \infty$ a.s. (in fact $T_m > t$ eventually for each $t$). For each $m$, the stopped integral satisfies \begin{align*} \int_0^{t \wedge T_m} (H^n_s)^2 \, d\langle M \rangle_s \leq \int_0^{t \wedge T_m} K_s^2 \, d\langle M \rangle_s \leq m < \infty \quad \text{a.s.} \end{align*} by hypothesis (ii). In particular, $\mathbb{1}_{[0, T_m]} H^n$ and $\mathbb{1}_{[0, T_m]} H$ both belong to $L^2(M)$, so the stopped stochastic integrals $(H^n \cdot M)_{t \wedge T_m}$ and $(H \cdot M)_{t \wedge T_m}$ are well-defined elements of $\mathcal{M}^2_c$ by the [Itô Isometry](/theorems/2092) (part (i)). [guided] Why do we localize? The stochastic integral $H^n \cdot M$ is defined for locally bounded previsible integrands, but to use the Itô isometry (which gives $L^2$ control), we need the integrand to belong to $L^2(M)$. The bound $|H^n_s| \leq K_s$ together with $\int_0^t K_s^2 \, d\langle M \rangle_s < \infty$ a.s. guarantees $L^2(M)$-membership only pathwise, not uniformly in $L^1(\mathbb{P})$. The stopping times $T_m$ truncate the integral at a level where the $L^2(M)$-norm is uniformly bounded by $m$, enabling us to take expectations. The choice of $T_m$ is dictated by hypothesis (iii): the finiteness $\int_0^t K_s^2 \, d\langle M \rangle_s < \infty$ a.s. ensures $T_m > t$ for all sufficiently large $m$ (depending on $\omega$). After stopping at $T_m$, we have a uniform bound on the $L^2(M)$-norm: $\int_0^{t \wedge T_m} K_s^2 \, d\langle M \rangle_s \leq m$. [/guided] [/step] [step:Apply the Itô isometry and classical DCT to show $L^2$ convergence of the stopped martingale integral] By the stopping identity ([Itô Isometry](/theorems/2092), part (iii)) and the isometry, \begin{align*} \mathbb{E}\!\left[\left((H^n \cdot M)_{t \wedge T_m} - (H \cdot M)_{t \wedge T_m}\right)^2\right] &= \mathbb{E}\!\left[\left(((H^n - H) \cdot M)_{t \wedge T_m}\right)^2\right] \\ &\leq \mathbb{E}\!\left[\langle (H^n - H) \cdot M \rangle_{t \wedge T_m}\right] \\ &= \mathbb{E}\!\left[\int_0^{t \wedge T_m} (H^n_s - H_s)^2 \, d\langle M \rangle_s\right], \end{align*} where the inequality uses the $L^2$ maximal inequality $\mathbb{E}[Y_t^2] \leq \mathbb{E}[\langle Y \rangle_t]$ for $Y = (H^n - H) \cdot M$ stopped at $T_m$ (which is an equality by the [Itô Isometry](/theorems/2092) applied to the stopped process), and the last equality uses $\langle (H^n - H) \cdot M \rangle_t = \int_0^t (H^n_s - H_s)^2 \, d\langle M \rangle_s$ from the [Bracket of Two Stochastic Integrals](/theorems/2093) (with both integrands and integrators equal). The integrand $(H^n_s - H_s)^2$ satisfies: - $(H^n_s - H_s)^2 \to 0$ for all $s \in [0, t]$ by hypothesis (i), - $(H^n_s - H_s)^2 \leq (|H^n_s| + |H_s|)^2 \leq (K_s + K_s)^2 = 4K_s^2$ by hypothesis (ii), - $\mathbb{E}\!\left[\int_0^{t \wedge T_m} 4K_s^2 \, d\langle M \rangle_s\right] \leq 4m < \infty$. By the classical dominated convergence theorem (applied to the measure $d\mathbb{P}(\omega) \otimes d\langle M \rangle_s(\omega)$ on $\Omega \times [0, t \wedge T_m]$ with dominator $4K_s^2$), \begin{align*} \mathbb{E}\!\left[\int_0^{t \wedge T_m} (H^n_s - H_s)^2 \, d\langle M \rangle_s\right] \to 0 \quad \text{as } n \to \infty. \end{align*} Therefore $(H^n \cdot M)_{t \wedge T_m} \to (H \cdot M)_{t \wedge T_m}$ in $L^2(\mathbb{P})$, hence in probability. [guided] The core of the argument is reducing the stochastic convergence problem to a classical measure-theoretic one. After stopping at $T_m$, the Itô isometry converts the $L^2(\mathbb{P})$ distance between the stochastic integrals into an integral of $(H^n_s - H_s)^2$ against the product measure $d\mathbb{P} \otimes d\langle M \rangle_s$. Why does the classical dominated convergence theorem apply? We need three conditions on the product space $\Omega \times [0, t \wedge T_m(\omega)]$: 1. **Pointwise convergence**: $(H^n_s(\omega) - H_s(\omega))^2 \to 0$ for all $s$, which is hypothesis (i). 2. **Domination**: $(H^n_s - H_s)^2 \leq 4K_s^2$ by the triangle inequality and hypothesis (ii). 3. **Integrability of the dominator**: $\mathbb{E}[\int_0^{t \wedge T_m} 4K_s^2 \, d\langle M \rangle_s] \leq 4m$, which is finite precisely because of the stopping. Without stopping, condition 3 would only give a.s. finiteness (hypothesis (iii)), not $L^1(\mathbb{P})$-integrability. This is why the localization step is essential. The conclusion is $L^2(\mathbb{P})$ convergence of the stopped martingale integrals, which is stronger than convergence in probability. We will then remove the stopping in the next step. [/guided] [/step] [step:Remove the localization and combine with the finite variation part] We have shown: - $(H^n \cdot A)_t \to (H \cdot A)_t$ a.s. (Step 1), - $(H^n \cdot M)_{t \wedge T_m} \to (H \cdot M)_{t \wedge T_m}$ in probability for each $m$ (Step 3). Since $\int_0^t K_s^2 \, d\langle M \rangle_s < \infty$ a.s. by hypothesis (iii), we have $T_m > t$ for all sufficiently large $m$, a.s. On the event $\{T_m > t\}$, $t \wedge T_m = t$ and therefore $(H^n \cdot M)_t = (H^n \cdot M)_{t \wedge T_m}$. For any $\varepsilon > 0$, \begin{align*} \mathbb{P}\!\left(|(H^n \cdot M)_t - (H \cdot M)_t| > \varepsilon\right) &\leq \mathbb{P}\!\left(|(H^n \cdot M)_{t \wedge T_m} - (H \cdot M)_{t \wedge T_m}| > \varepsilon\right) + \mathbb{P}(T_m \leq t). \end{align*} The first term tends to $0$ as $n \to \infty$ (convergence in probability at the stopped level). The second term tends to $0$ as $m \to \infty$ (since $T_m \nearrow \infty$ a.s.). Choosing $m$ large enough that $\mathbb{P}(T_m \leq t) < \varepsilon$, and then $n$ large enough that the first term is less than $\varepsilon$, we conclude $(H^n \cdot M)_t \xrightarrow{\mathbb{P}} (H \cdot M)_t$. Combining both components: \begin{align*} \int_0^t H^n_s \, dX_s = (H^n \cdot M)_t + (H^n \cdot A)_t \xrightarrow{\mathbb{P}} (H \cdot M)_t + (H \cdot A)_t = \int_0^t H_s \, dX_s, \end{align*} since the sum of a sequence converging in probability and a sequence converging a.s. converges in probability. [guided] The final step removes the localization. We have convergence in probability at the stopped level (for each fixed $m$), and we need convergence in probability at the unstopped level. The standard trick is the inequality \begin{align*} \mathbb{P}(|Z_n| > \varepsilon) \leq \mathbb{P}(|Z_n \mathbb{1}_{\{T_m > t\}}| > \varepsilon) + \mathbb{P}(T_m \leq t), \end{align*} applied with $Z_n = (H^n \cdot M)_t - (H \cdot M)_t$. On $\{T_m > t\}$, the unstopped and stopped processes agree: $Z_n = (H^n \cdot M)_{t \wedge T_m} - (H \cdot M)_{t \wedge T_m}$. The two-parameter argument works as follows. Given $\delta > 0$: 1. Choose $m$ large enough that $\mathbb{P}(T_m \leq t) < \delta/2$. This is possible because $T_m \nearrow \infty$ a.s. (which follows from $\int_0^t K_s^2 \, d\langle M \rangle_s < \infty$ a.s.). 2. With $m$ now fixed, choose $n$ large enough that $\mathbb{P}(|(H^n \cdot M)_{t \wedge T_m} - (H \cdot M)_{t \wedge T_m}| > \varepsilon) < \delta/2$. This is possible because convergence in probability at the stopped level was established in the previous step. Then $\mathbb{P}(|Z_n| > \varepsilon) < \delta$, proving $(H^n \cdot M)_t \xrightarrow{\mathbb{P}} (H \cdot M)_t$. For the finite variation part, a.s. convergence implies convergence in probability. Since the sum of a sequence converging in probability (the martingale part) and a sequence converging in probability (the finite variation part, which converges even a.s.) converges in probability, we obtain the stated result: $\int_0^t H^n_s \, dX_s \xrightarrow{\mathbb{P}} \int_0^t H_s \, dX_s$. [/guided] [/step]