# Question on the martingale representation theorem

Let $$(X_t)_{0\le t\le 1}$$ be a continuous Markov martingale (with respect to its natural filtration) s.t. $$X_0=0$$ and $$X_1\in\{-1,1\}$$. Can we prove the existence of some measurable function $$\sigma: [0,1]\times \mathbb R\to\mathbb R_+$$ s.t.

$$X_t=\int_0^t\sigma(s,X_s)dW_s,\quad \forall 0\le t\le 1?$$

Here $$(W_t)_{0\le t\le 1}$$ denotes some Brownian motion.

First, the underlying probability space $$(\Omega, \mathcal F)$$ and the natural filtration can be too small for the Brownian motion to exist. Indeed: consider the case when $$X_t$$ is constant for $$t \geqslant \tfrac12$$, and $$\Omega$$ is just the space of paths of $$X_t$$. If $$W_t$$ and $$\sigma$$ existed, then $$\sigma\{W_t : t < \tfrac12\}$$ would contain $$\sigma\{X_t : t < \tfrac12\} = \mathcal F \supseteq \sigma\{W_t : t < 1\}$$, and so $$\sigma\{W_t : t < \tfrac12\} = \sigma\{W_t : t < 1\}$$, a contradiction.
Another, more important, reason is that $$X_t$$ can be too rough to be an Itô integral with respect to a Brownian motion. To see this, consider an arbitrary martingale $$X_t$$, and modify its time using the Cantor's (devil's staircase) function $$\phi(t)$$: the process $$X_{\phi(t)}$$ is again a continuous martingale, but it is piecewise constant on a set of full Lebesgue measure, so the corresponding function $$\sigma$$ would have to be zero almost everywhere, a contradiction.
That said, I believe the following is likely to be true: if $$X_t$$ is a continuous martingale with respect to some Brownian filtration, and additionally $$X_t$$ is a Markov process, then it is an Itô integral with respect to the underlying Brownian motion, with the integrand of the form $$\sigma(s, X_s)$$.