Assume $\sigma:\mathbb{R}^d \times E$ is $C^2$ (on both parameters) and that $X_t=\int_0^t \sigma(X_s,I_s)dW_s$, where $(I_t)_{t \geq 0}$ is some discrete Markov Chain with finite state space $E$ that evolves independently of $X_t$, and $(W_t)_{t \geq 0}$ is a standard Brownian motion.
Is it true that $(X_t)_{t \geq 0}$ is a local martingale? To answer this I would be also pleased to see that this is indeed correct for a process $X_t=\int_0^t \sigma(X_s,s)dW_s$. I know this is true, but yet have not seen any proofs of it. Some references of older stack posts would be ok as well (I haven't myself found any so far).
Using the definition of local martingales I can create some stopping time sequence $\{\tau_n\}_{n \geq 1}$ defined by $\tau_n:= \{\inf t>0 : |X_t| \geq n \}$ that satisfies the first two prerequisites. I can see that continuity of $\sigma$ is important for this sequence to fit in the required conditions, as if implies $X_t$ is continuous (I admit I am not 100% convinced of this).
For the last part however, I'm not sure how to proceed. We are left to check that for all $n \in \mathbb{N}$, $X_{\min(t,\tau_n)}$ is a martingale. How can I see this? Thanks in advance.