Say $X$ is the solution of a SDE, stopped at the exit time from a bounded domain $D\subset R^d$. This means that given some filtered probability space where $B$ is a $d$-dimensional brownian motion, and $\sigma:\mathbb{R}^d\to \mathbb{R}^{d \times d}$, $b:\mathbb{R}^d\to\mathbb{R}^d$ measurable functions, one has $$ X=x_0+\int_0^t \sigma(X_s) dB_s + \int_0^t b(X_s) ds$$ for all $t \leq \tau$ where $\tau=\inf\{t>0: X_t \notin D\} \lt \infty $ a.s.
Suppose now that $F$ is a $\mathcal{C}^2$ on $D$ and that it satisfies $LF=0$ in $D$, where
$$ LF(x):= \frac{1}{2} \sum_{i,j \leq d} a_{i,j}(x) \frac{\partial^2 F}{\partial{x_i}\partial{x_j}}(x)+\sum_{i=0}^d b_i(x)\frac{\partial F}{\partial x_i}$$ where $A=\sigma \sigma^t$
This is where my problems begin: according to my notes, one can see using Ito's formula that $F(X^{\tau}_t)$ is a local martingale since the other terms vanish because $LF=0$ and $d\langle X^i, X^j \rangle = a_{i,j}(X_t)dt$. I don't understand why the latter equality holds and in general would be grateful to anyone who could show me this computation with Ito's formula.
The latter equality simply holds because, due to the SDE that $X$ satisfies, \begin{align} \langle X^i,X^j\rangle_t&=\left\langle\sum_{k=1}^d\int_0^.\sigma_{ik}(X_s)\,dB^k_s,\sum_{l=1}^d\int_0^.\sigma_{jl}(X_s)\,dB^l_s\right\rangle_t\\ &=\sum_{k,l=1}^d\left\langle \int_0^.\sigma_{ik}(X_s)\,dB^k_s,\int_0^.\sigma_{jl}(X_s)\,dB^l_s\right\rangle_t\\ &=\sum_{k=1}^d\int_0^t\sigma_{ik}(X_s)\sigma_{jk}(X_s)\,ds\\ &=\int_0^ta_{ij}(X_s)\,ds\,. \end{align}