[proofplan] This is a direct application of [Itô's Formula](/theorems/2099). We apply the multidimensional Itô formula to $f(t, X_t)$ where $X$ solves the SDE $dX^i_t = b_i(X_t) \, d\mathcal{L}^1(t) + \sum_k \sigma_{ik}(X_t) \, dW^k_t$. The formula produces a drift term $(\partial_t f + Lf)(t, X_t) \, d\mathcal{L}^1(t)$ and a stochastic integral term. Subtracting the integral of the drift from $f(t, X_t) - f(0, X_0)$ yields $M^f_t$, which equals the stochastic integral part and is therefore a continuous local martingale. [/proofplan]

[step:Apply the multidimensional Itô formula to $f(t, X_t)$] The process $X$ solves $\mathcal{E}_x(\sigma, b)$, meaning \begin{align*} dX^i_t = b_i(X_t) \, d\mathcal{L}^1(t) + \sum_{k=1}^m \sigma_{ik}(X_t) \, dW^k_t, \quad i = 1, \ldots, d. \end{align*} Consider the $(d+1)$-dimensional semimartingale $(t, X_t) = (t, X^1_t, \ldots, X^d_t)$ and apply [Itô's Formula](/theorems/2099) to the function $f \in C^1(\mathbb{R}_+) \otimes C^2(\mathbb{R}^d)$. Since the time component $t$ has zero quadratic variation ($\langle t, t \rangle = 0$) and zero covariation with each $X^i$ ($\langle t, X^i \rangle = 0$, because $t$ is a continuous finite-variation process), Itô's formula gives: \begin{align*} f(t, X_t) - f(0, X_0) = \int_0^t \partial_s f(s, X_s) \, d\mathcal{L}^1(s) + \sum_{i=1}^d \int_0^t \partial_{x_i} f(s, X_s) \, dX^i_s + \frac{1}{2}\sum_{i,j=1}^d \int_0^t \partial_{x_i}\partial_{x_j} f(s, X_s) \, d\langle X^i, X^j \rangle_s. \end{align*} [/step]

[step:Substitute the SDE dynamics and identify the drift with $(\partial_s + L)f$] Substitute $dX^i_s = b_i(X_s) \, d\mathcal{L}^1(s) + \sum_k \sigma_{ik}(X_s) \, dW^k_s$ into the stochastic integral term: \begin{align*} \sum_{i=1}^d \int_0^t \partial_{x_i} f(s, X_s) \, dX^i_s = \sum_{i=1}^d \int_0^t \partial_{x_i} f(s, X_s) \, b_i(X_s) \, d\mathcal{L}^1(s) + \sum_{i=1}^d \sum_{k=1}^m \int_0^t \partial_{x_i} f(s, X_s) \, \sigma_{ik}(X_s) \, dW^k_s. \end{align*} For the quadratic covariation, since $X^i$ has martingale part $\sum_k \int_0^{\cdot} \sigma_{ik}(X_s) \, dW^k_s$, the [Bracket of Two Stochastic Integrals](/theorems/2092) gives \begin{align*} d\langle X^i, X^j \rangle_s = \sum_{k=1}^m \sigma_{ik}(X_s) \sigma_{jk}(X_s) \, d\mathcal{L}^1(s) = a_{ij}(X_s) \, d\mathcal{L}^1(s), \end{align*} where $a_{ij} = \sum_k \sigma_{ik}\sigma_{jk} = (\sigma\sigma^\top)_{ij}$. Collecting all $d\mathcal{L}^1(s)$ terms: \begin{align*} &\int_0^t \partial_s f(s, X_s) \, d\mathcal{L}^1(s) + \sum_{i=1}^d \int_0^t b_i(X_s) \, \partial_{x_i} f(s, X_s) \, d\mathcal{L}^1(s) + \frac{1}{2}\sum_{i,j=1}^d \int_0^t a_{ij}(X_s) \, \partial_{x_i}\partial_{x_j} f(s, X_s) \, d\mathcal{L}^1(s) \\ &= \int_0^t \bigl(\partial_s + L\bigr)f(s, X_s) \, d\mathcal{L}^1(s), \end{align*} where $L = \sum_i b_i \partial_{x_i} + \frac{1}{2}\sum_{i,j} a_{ij} \partial_{x_i}\partial_{x_j}$ is the infinitesimal generator of $X$. [/step]

[step:Identify $M^f_t$ as the stochastic integral part, which is a continuous local martingale]Rearranging the Itô formula: \begin{align*} f(t, X_t) - f(0, X_0) = \int_0^t (\partial_s + L)f(s, X_s) \, d\mathcal{L}^1(s) + \sum_{i=1}^d \sum_{k=1}^m \int_0^t \partial_{x_i} f(s, X_s) \, \sigma_{ik}(X_s) \, dW^k_s. \end{align*} Therefore \begin{align*} M^f_t &= f(t, X_t) - f(0, X_0) - \int_0^t (\partial_s + L)f(s, X_s) \, d\mathcal{L}^1(s) \\ &= \sum_{i=1}^d \sum_{k=1}^m \int_0^t \partial_{x_i} f(s, X_s) \, \sigma_{ik}(X_s) \, dW^k_s. \end{align*} The right-hand side is a sum of stochastic integrals against the continuous local martingales $W^1, \ldots, W^m$. Since $f \in C^2(\mathbb{R}^d)$ and $\sigma$ is continuous, the integrands $\partial_{x_i} f(s, X_s) \sigma_{ik}(X_s)$ are continuous adapted processes, hence locally bounded. By the [Properties of the Semimartingale Integral](/theorems/2094), each integral is a continuous local martingale. A finite sum of continuous local martingales is a continuous local martingale. Therefore $M^f$ is a continuous local martingale.[/step]

[guided]The Martingale Problem result is essentially "reading off" the local martingale part from Itô's formula. When we apply [Itô's Formula](/theorems/2099) to $f(t, X_t)$, we get a decomposition into a finite-variation part and a local martingale part: \begin{align*} \underbrace{f(t, X_t) - f(0, X_0)}_{\text{semimartingale}} = \underbrace{\int_0^t (\partial_s + L)f(s, X_s) \, d\mathcal{L}^1(s)}_{\text{finite variation}} + \underbrace{\sum_{i,k} \int_0^t \partial_{x_i} f(s, X_s) \sigma_{ik}(X_s) \, dW^k_s}_{\text{local martingale}}. \end{align*} The process $M^f_t$ is defined to be the left-hand side minus the finite-variation part, which equals the local martingale part. Why is the operator $L = \sum_i b_i \partial_{x_i} + \frac{1}{2}\sum_{i,j} a_{ij} \partial_{x_i}\partial_{x_j}$ natural here? Because $a_{ij} = (\sigma\sigma^\top)_{ij}$ arises from the quadratic covariation $d\langle X^i, X^j \rangle = a_{ij} \, d\mathcal{L}^1$, and the $b_i$ terms come from the drift of $X$. The operator $L$ is the infinitesimal generator of the diffusion: it captures the local behavior of the process through both its drift ($b$) and its diffusion ($a = \sigma\sigma^\top$). The significance of the Martingale Problem is that it provides a characterization of the SDE solution purely in terms of the generator $L$, without reference to the driving Brownian motion $W$. This is useful for uniqueness arguments (Stroock-Varadhan theory) and for establishing properties of $X$ that depend only on $L$.[/guided]