[proofplan] We prove that the Riemann-sum approximations of $\langle X, Y \rangle$ for continuous semimartingales converge u.c.p. by decomposing $X = X_0 + M + A$ and $Y = Y_0 + N + B$ into local martingale and finite variation parts. The product of increments expands into four terms: $\Delta M \Delta N$, $\Delta M \Delta B$, $\Delta A \Delta N$, and $\Delta A \Delta B$. The martingale-martingale term converges u.c.p. to $\langle M, N \rangle = \langle X, Y \rangle$ by part (iii) of the [Properties of Covariation](/theorems/2086). The three cross terms each involve at least one finite variation factor, and we show they vanish u.c.p. by bounding the sum of absolute values of one factor by its total variation while the maximum oscillation of the other factor tends to zero by uniform continuity on compacts. [/proofplan]

[step:Decompose the semimartingales and expand the product of increments]Write $X = X_0 + M + A$ and $Y = Y_0 + N + B$, where $M, N$ are continuous local martingales with $M_0 = N_0 = 0$ and $A, B$ are continuous finite variation processes with $A_0 = B_0 = 0$. Since $\langle X, Y \rangle = \langle M, N \rangle$ by definition (the covariation depends only on the local martingale parts), it suffices to show \begin{align*} \langle X, Y \rangle^{(n)}_t - \langle M, N \rangle^{(n)}_t \xrightarrow{\text{u.c.p.}} 0. \end{align*} For each $n$ and each subinterval $[(i-1)2^{-n}, \, i \cdot 2^{-n}]$, expand the product of increments: \begin{align*} (X_{i \cdot 2^{-n}} - X_{(i-1) \cdot 2^{-n}})(Y_{i \cdot 2^{-n}} - Y_{(i-1) \cdot 2^{-n}}) &= (\Delta_i M + \Delta_i A)(\Delta_i N + \Delta_i B) \\ &= \Delta_i M \, \Delta_i N + \Delta_i M \, \Delta_i B + \Delta_i A \, \Delta_i N + \Delta_i A \, \Delta_i B, \end{align*} where $\Delta_i M := M_{i \cdot 2^{-n}} - M_{(i-1) \cdot 2^{-n}}$ and similarly for $A$, $N$, $B$. Summing over $i$ from $1$ to $\lfloor 2^n t \rfloor$ gives \begin{align*} \langle X, Y \rangle^{(n)}_t = \langle M, N \rangle^{(n)}_t + R^{(n)}_t, \end{align*} where $R^{(n)}_t$ is the sum of the three cross terms. We must show $\sup_{0 \leq t \leq T} |R^{(n)}_t| \xrightarrow{\mathbb{P}} 0$ for every $T > 0$.[/step]

[guided]The decomposition $X = X_0 + M + A$ is the canonical semimartingale decomposition, which is unique for continuous semimartingales. The covariation $\langle X, Y \rangle$ is defined as $\langle M, N \rangle$ because the finite variation parts $A$ and $B$ contribute zero to the quadratic variation (a continuous finite variation process has zero quadratic variation, by the [Continuous Finite Variation Local Martingale is Identically Zero](/theorems/2081) argument applied to the deterministic case). The Riemann sums, however, do not respect this decomposition: the product $(X_{i \cdot 2^{-n}} - X_{(i-1) \cdot 2^{-n}})(Y_{i \cdot 2^{-n}} - Y_{(i-1) \cdot 2^{-n}})$ mixes the martingale and finite variation parts. Expanding and summing separates the contributions, and the task reduces to showing that the three cross terms (each involving at least one finite variation factor) vanish in the u.c.p. limit.[/guided]

[step:Show that the martingale-martingale Riemann sum converges u.c.p. to $\langle M, N \rangle$] By part (iii) of the [Properties of Covariation](/theorems/2086), the Riemann-sum approximations \begin{align*} \langle M, N \rangle^{(n)}_t = \sum_{i=1}^{\lfloor 2^n t \rfloor} (M_{i \cdot 2^{-n}} - M_{(i-1) \cdot 2^{-n}})(N_{i \cdot 2^{-n}} - N_{(i-1) \cdot 2^{-n}}) \end{align*} converge u.c.p. to $\langle M, N \rangle_t$. This is the only term that survives in the limit. [/step]

[step:Bound each cross term involving a finite variation factor and show it vanishes u.c.p.]We treat the term $\sum_i \Delta_i M \, \Delta_i B$; the terms $\sum_i \Delta_i A \, \Delta_i N$ and $\sum_i \Delta_i A \, \Delta_i B$ are handled identically. Fix $T > 0$. For $t \leq T$, the oscillation of $B$ on dyadic subintervals satisfies \begin{align*} \omega_B^{(n)}(T) := \max_{1 \leq i \leq \lfloor 2^n T \rfloor + 1} |B_{i \cdot 2^{-n}} - B_{(i-1) \cdot 2^{-n}}|. \end{align*} Since $B$ is continuous, each sample path $B(\omega, \cdot)$ is uniformly continuous on $[0, T]$. Therefore $\omega_B^{(n)}(T) \to 0$ almost surely as $n \to \infty$. Bound the cross-term sum for $t \leq T$: \begin{align*} \left|\sum_{i=1}^{\lfloor 2^n t \rfloor} \Delta_i M \, \Delta_i B\right| &\leq \sum_{i=1}^{\lfloor 2^n t \rfloor} |\Delta_i M| \cdot |\Delta_i B| \\ &\leq \omega_B^{(n)}(T) \cdot \sum_{i=1}^{\lfloor 2^n T \rfloor} |\Delta_i M|. \end{align*} The sum $\sum_i |\Delta_i M|$ is bounded above by the total variation of $M$ on $[0, T]$ plus the oscillation error from dyadic endpoints: for any path, $\sum_i |M_{i \cdot 2^{-n}} - M_{(i-1) \cdot 2^{-n}}|$ is a Riemann-sum approximation to the total variation. However, $M$ need not have finite total variation. Instead, we use the estimate \begin{align*} \sum_{i=1}^{\lfloor 2^n T \rfloor} |\Delta_i M| \leq \left(\lfloor 2^n T \rfloor\right)^{1/2} \left(\sum_{i=1}^{\lfloor 2^n T \rfloor} (\Delta_i M)^2\right)^{1/2}, \end{align*} by the Cauchy--Schwarz inequality in $\mathbb{R}^{\lfloor 2^n T \rfloor}$ (with one vector being $|\Delta_i M|$ and the other the constant vector $1$). But this bound grows with $n$, so a different approach is needed. Instead, observe that we can factor the bound differently. For the term $\sum_i \Delta_i A \, \Delta_i N$ (and symmetrically $\sum_i \Delta_i M \, \Delta_i B$), we extract the maximum oscillation of the martingale factor and bound the sum of absolute values of the finite variation factor by the total variation: \begin{align*} \left|\sum_{i=1}^{\lfloor 2^n t \rfloor} \Delta_i A \, \Delta_i N\right| \leq \omega_N^{(n)}(T) \cdot \sum_{i=1}^{\lfloor 2^n T \rfloor} |\Delta_i A| \leq \omega_N^{(n)}(T) \cdot V_A(T), \end{align*} where $V_A(T) := \int_0^T |dA_s|$ is the total variation of $A$ on $[0,T]$, which is finite since $A$ is a finite variation process. Since $N$ is continuous, $\omega_N^{(n)}(T) \to 0$ a.s. by uniform continuity on $[0,T]$, and $V_A(T) < \infty$ a.s., so this cross term vanishes a.s. For the term $\sum_i \Delta_i M \, \Delta_i B$, we extract the maximum oscillation of $B$: \begin{align*} \left|\sum_{i=1}^{\lfloor 2^n t \rfloor} \Delta_i M \, \Delta_i B\right| \leq \omega_B^{(n)}(T) \cdot \sum_{i=1}^{\lfloor 2^n T \rfloor} |\Delta_i M|. \end{align*} The factor $\omega_B^{(n)}(T) \to 0$ a.s. It remains to show that $\sum_i |\Delta_i M|$ does not grow too fast. By the Cauchy--Schwarz inequality in $\mathbb{R}^{\lfloor 2^n T \rfloor}$, \begin{align*} \sum_{i=1}^{\lfloor 2^n T \rfloor} |\Delta_i M| \leq \left(\lfloor 2^n T \rfloor \cdot \sum_{i=1}^{\lfloor 2^n T \rfloor} (\Delta_i M)^2 \right)^{1/2}. \end{align*} The sum $\sum_i (\Delta_i M)^2 = \langle M \rangle^{(n)}_T \to \langle M \rangle_T$ u.c.p., so it is bounded in probability. Thus $\sum_i |\Delta_i M| = O_\mathbb{P}(2^{n/2})$, while $\omega_B^{(n)}(T) \to 0$ a.s. The product $\omega_B^{(n)}(T) \cdot \sum_i |\Delta_i M|$ converges to $0$ in probability because $\omega_B^{(n)}(T)$ tends to $0$ faster than any power. More precisely, since $B$ has finite variation, $\sum_i |\Delta_i B| \leq V_B(T) < \infty$, and we can bound instead: \begin{align*} \left|\sum_i \Delta_i M \, \Delta_i B\right| \leq \max_i |\Delta_i M| \cdot \sum_i |\Delta_i B| \leq \omega_M^{(n)}(T) \cdot V_B(T). \end{align*} Since $M$ is continuous, $\omega_M^{(n)}(T) \to 0$ a.s. by uniform continuity on $[0,T]$, and $V_B(T) < \infty$ a.s. Therefore $\sup_{t \leq T} |\sum_i \Delta_i M \, \Delta_i B| \to 0$ a.s. For the term $\sum_i \Delta_i A \, \Delta_i B$, the same argument gives \begin{align*} \left|\sum_i \Delta_i A \, \Delta_i B\right| \leq \omega_A^{(n)}(T) \cdot V_B(T) \to 0 \quad \text{a.s.} \end{align*} Since all three cross terms converge to $0$ uniformly on $[0, T]$ almost surely (hence u.c.p.), we conclude $R^{(n)}_t \to 0$ u.c.p.[/step]

[guided]The crux of each cross-term estimate is a factorisation: extract the maximum oscillation of the continuous factor and bound the remaining sum by a total variation. The key insight is which factor to choose for the oscillation extraction. For a term like $\sum_i \Delta_i M \, \Delta_i B$, there are two ways to factor: - Extract $\max_i |\Delta_i B|$ (oscillation of $B$) and bound $\sum_i |\Delta_i M|$ (sum of martingale increments). - Extract $\max_i |\Delta_i M|$ (oscillation of $M$) and bound $\sum_i |\Delta_i B|$ by $V_B(T)$ (total variation of $B$). The first approach fails because $\sum_i |\Delta_i M|$ can grow as $2^{n/2}$ (the martingale does not have finite total variation in general). The second approach works: $\max_i |\Delta_i M| \to 0$ by uniform continuity of $M$ on $[0,T]$, and $V_B(T) < \infty$ since $B$ is a finite variation process. The product $\omega_M^{(n)}(T) \cdot V_B(T) \to 0$ a.s. For $\sum_i \Delta_i A \, \Delta_i N$, we extract $\omega_N^{(n)}(T) \to 0$ and bound $\sum_i |\Delta_i A| \leq V_A(T) < \infty$. For $\sum_i \Delta_i A \, \Delta_i B$, either factor works since both have finite total variation. We use $\omega_A^{(n)}(T) \cdot V_B(T) \to 0$. In each case, the convergence is uniform in $t \leq T$ (since the bound does not depend on $t$), giving u.c.p. convergence.[/guided]

[step:Combine to conclude u.c.p. convergence of the semimartingale Riemann sums] From the decomposition $\langle X, Y \rangle^{(n)}_t = \langle M, N \rangle^{(n)}_t + R^{(n)}_t$, the first term converges u.c.p. to $\langle M, N \rangle_t = \langle X, Y \rangle_t$ by part (iii) of the [Properties of Covariation](/theorems/2086), and the remainder $R^{(n)}_t$ converges u.c.p. to $0$ since all three cross terms vanish u.c.p. Therefore \begin{align*} \langle X, Y \rangle^{(n)}_t \xrightarrow{\text{u.c.p.}} \langle X, Y \rangle_t. \end{align*} [/step]