If $M_t= M_0 + \int_0^t \sigma_s dB_s$ and $E[M_1 | M_0]=M_0$, can we say that $(M_t)$ is a martingale?

Question

If $(B_t)$ is a standard Brownian motion and $M_t= M_0 + \int_0^t \sigma_s dB_s$ for some process $(\sigma_t) \in L^1(B_t)$ and $E[M_1 | M_0]=M_0$, can we say that $(M_t)$ is a martingale on $[0,1]$ ?

Basically, does $E[M_1 | M_0]=M_0$ implies $E[M_t | \mathcal F_s ]=M_s$ for $s<t$ in $[0,1]$ and $\mathcal F_s$ the filtration generated by $(M_s)$ ?

Answer 1

According to a result of R.M. Dudley [Wiener functionals as Itô integrals, Ann. Probability 5 (1977), no. 1, 140–141], given an $\mathcal F_1$-measurable random variable $X$, there is a predictable $(H_t)$ with $\int_0^t H_s^2 ds<\infty$ such that $X=\int_0^1 H_s dB_s$, a.s. Adapting this in the obvious way, one has predictable processes $(H_t)$ and $(K_t)$ such that (i) $H_t=0$ for $1/2<t\le 1$, (ii) $K_t=0$ for $0\le t\le 1/2$, and (iii) $\int_0^1 H_s dB_s =1 =\int_0^1 K_s dB_s$, a.s. The difference $\sigma_t:=H_t-K_t$ is then non-trivial but $\int_0^1 \sigma_s dB_s =0$, a.s. Take $M_0=0$ and you have a counterexample: $M_t=\int_0^t \sigma_s dB_s$ is a continuous local martingale, but not a martingale.