[proofplan] We approximate the integrand $H$ by piecewise-constant step processes $H^{(m)}$ built from the partition. Left-continuity of $H$ ensures pointwise convergence $H^{(m)}_s \to H_s$, and boundedness of $H$ provides a uniform dominating process. The stochastic dominated convergence theorem then yields convergence of the stochastic integrals in probability. Since each step-process integral equals the corresponding Riemann sum by the linearity property of the stochastic integral, the result follows. [/proofplan] [step:Construct step-process approximations $H^{(m)}$ from the partition] For each $m \geq 1$, define the step process \begin{align*} H^{(m)}: \Omega \times [0, t] &\to \mathbb{R} \\ (\omega, s) &\mapsto \sum_{i=1}^{n_m} H_{t^{(m)}_{i-1}}(\omega) \, \mathbb{1}_{(t^{(m)}_{i-1},\, t^{(m)}_i]}(s). \end{align*} Each $H^{(m)}$ is a simple previsible process: it is left-continuous by construction, and for each interval $(t^{(m)}_{i-1}, t^{(m)}_i]$ the coefficient $H_{t^{(m)}_{i-1}}$ is $\mathcal{F}_{t^{(m)}_{i-1}}$-measurable since $H$ is adapted. [guided] Our goal is to express the Riemann sum as a stochastic integral of a simple process, and then show that these stochastic integrals converge to $\int_0^t H_s \, dX_s$. The natural candidate is the step process that is constant on each partition interval, taking the value of $H$ at the left endpoint. For each $m \geq 1$, define \begin{align*} H^{(m)}: \Omega \times [0, t] &\to \mathbb{R} \\ (\omega, s) &\mapsto \sum_{i=1}^{n_m} H_{t^{(m)}_{i-1}}(\omega) \, \mathbb{1}_{(t^{(m)}_{i-1},\, t^{(m)}_i]}(s). \end{align*} Why is $H^{(m)}$ previsible? Each coefficient $H_{t^{(m)}_{i-1}}$ is $\mathcal{F}_{t^{(m)}_{i-1}}$-measurable because $H$ is adapted, and the indicator $\mathbb{1}_{(t^{(m)}_{i-1}, t^{(m)}_i]}$ is a deterministic left-continuous function of $s$. The product is therefore measurable with respect to the predictable $\sigma$-algebra $\mathcal{P}$, so $H^{(m)}$ is previsible. [/guided] [/step] [step:Identify the stochastic integral of $H^{(m)}$ with the Riemann sum] By the linearity of the stochastic integral for simple previsible processes, \begin{align*} \int_0^t H^{(m)}_s \, dX_s = \sum_{i=1}^{n_m} H_{t^{(m)}_{i-1}} \bigl(X_{t^{(m)}_i} - X_{t^{(m)}_{i-1}}\bigr). \end{align*} This is precisely the Riemann sum on the right-hand side of the claimed identity. [/step] [step:Verify pointwise convergence $H^{(m)}_s \to H_s$ using left-continuity] Fix $\omega \in \Omega$ and $s \in (0, t]$. For each $m$, the point $s$ belongs to some interval $(t^{(m)}_{i_m - 1}, t^{(m)}_{i_m}]$, so $H^{(m)}_s(\omega) = H_{t^{(m)}_{i_m - 1}}(\omega)$. Since $\max_i |t^{(m)}_i - t^{(m)}_{i-1}| \to 0$, we have $t^{(m)}_{i_m - 1} \uparrow s$ (the left endpoints of the partition intervals containing $s$ converge to $s$ from below). Because $H$ is left-continuous at $s$, \begin{align*} H^{(m)}_s(\omega) = H_{t^{(m)}_{i_m - 1}}(\omega) \to H_s(\omega) \quad \text{as } m \to \infty. \end{align*} [guided] Why does $t^{(m)}_{i_m - 1} \to s$? The point $s$ lies in $(t^{(m)}_{i_m - 1}, t^{(m)}_{i_m}]$, so $0 \leq s - t^{(m)}_{i_m - 1} \leq t^{(m)}_{i_m} - t^{(m)}_{i_m - 1} \leq \max_i |t^{(m)}_i - t^{(m)}_{i-1}| \to 0$. Therefore $t^{(m)}_{i_m - 1} \to s$ from below. Left-continuity of $H$ at $s$ means $\lim_{r \uparrow s} H_r(\omega) = H_s(\omega)$, so \begin{align*} H^{(m)}_s(\omega) = H_{t^{(m)}_{i_m - 1}}(\omega) \to H_s(\omega) \quad \text{as } m \to \infty. \end{align*} Note that we used left-continuity rather than right-continuity: the step process evaluates $H$ at the left endpoint $t^{(m)}_{i_m-1}$, which approaches $s$ from below. If $H$ were only right-continuous, this argument would fail. [/guided] [/step] [step:Apply the stochastic dominated convergence theorem to pass to the limit] Since $H$ is bounded, there exists a constant $K > 0$ such that $|H_s(\omega)| \leq K$ for all $(\omega, s) \in \Omega \times [0, t]$. The step-process approximations inherit this bound: \begin{align*} |H^{(m)}_s(\omega)| \leq K \quad \text{for all } m, \, s, \, \omega. \end{align*} We now apply the [Stochastic Dominated Convergence Theorem](/theorems/2096). The hypotheses are: 1. $H^{(m)}_s \to H_s$ for all $s \in (0, t]$, a.s. — verified in the previous step. 2. $|H^{(m)}_s| \leq K$ for all $m$ and $s$ — verified above. 3. The dominating process $K_s \equiv K$ satisfies $\int_0^t K^2 \, d\langle X \rangle_s < \infty$ a.s. and $\int_0^t K \, |dA_s| < \infty$ a.s., where $X = M + A$ is the semimartingale decomposition of $X$ into a continuous local martingale $M$ and a continuous finite variation process $A$. Both conditions hold because $\langle M \rangle_t < \infty$ a.s. (since $M$ is a continuous local martingale) and $\int_0^t |dA_s| < \infty$ a.s. (since $A$ has continuous paths of finite variation on $[0, t]$). The stochastic dominated convergence theorem concludes that \begin{align*} \int_0^t H^{(m)}_s \, dX_s \xrightarrow{\mathbb{P}} \int_0^t H_s \, dX_s \quad \text{as } m \to \infty. \end{align*} [guided] The stochastic dominated convergence theorem is the stochastic analogue of the classical dominated convergence theorem: if a sequence of integrands converges pointwise and is uniformly dominated by an integrable process, then the stochastic integrals converge in probability. Why do the integrability conditions on the dominating process hold? Write $X = M + A$ where $M$ is a continuous local martingale and $A$ is a continuous finite variation process. The constant process $K_s \equiv K$ satisfies: - $\int_0^t K^2 \, d\langle M \rangle_s = K^2 \langle M \rangle_t$. Since $M$ is a continuous local martingale, $\langle M \rangle_t$ is finite a.s. for each fixed $t$, so this integral is finite a.s. - $\int_0^t K \, |dA_s| = K \int_0^t |dA_s|$. Since $A$ is a continuous process of finite variation on $[0, t]$, this is finite a.s. With these conditions verified, the stochastic dominated convergence theorem gives convergence in probability: \begin{align*} \int_0^t H^{(m)}_s \, dX_s \xrightarrow{\mathbb{P}} \int_0^t H_s \, dX_s \quad \text{as } m \to \infty. \end{align*} [/guided] [/step] [step:Combine the identification and the limit to conclude] From the second step, $\int_0^t H^{(m)}_s \, dX_s = \sum_{i=1}^{n_m} H_{t^{(m)}_{i-1}}(X_{t^{(m)}_i} - X_{t^{(m)}_{i-1}})$. From the fourth step, $\int_0^t H^{(m)}_s \, dX_s \xrightarrow{\mathbb{P}} \int_0^t H_s \, dX_s$. Combining these two facts: \begin{align*} \sum_{i=1}^{n_m} H_{t^{(m)}_{i-1}}\bigl(X_{t^{(m)}_i} - X_{t^{(m)}_{i-1}}\bigr) \xrightarrow{\mathbb{P}} \int_0^t H_s \, dX_s \quad \text{as } m \to \infty. \end{align*} This is the claimed Riemann sum approximation. [/step]