Alternative characterisation of martingales

Question

It is obvious that on a filtrated probability space $(\Omega,\mathcal F,P,(\mathcal F_t)_{t\geq0})$ the following holds $$ M:=(M_t)_{t\geq0}\text{ is a martingale with } M_0=0\implies \forall\,0\leq s\leq t\,\forall\epsilon\geq0: \Bbb EM_t=0\ \land\ \Bbb E(M_{t+\epsilon}-M_t\mid\mathcal F_s)=0. $$ I was wondering if the inverse implication is true as well, i.e. whether a martingale can be characterised by zero mean and zero conditioned increments or not?

EDIT: Thanks to the comment of Alvin Lepik I know that the inverse implication is true if $M$ is adapted. But what about if we weaken the assumption? Is this still true:

Let $M:=(M_t)_{t\geq0}$ be an adapted process with $M_0=0$. Then $$\forall\,0\leq s\leq t: \Bbb EM_t=0\ \land\ \Bbb E(M_{t+\epsilon}-M_t\mid \mathcal F_s)=o(\epsilon) \implies M \text{ is a martingale}, $$ where $f(\epsilon)=o(\epsilon)$ denotes the Landau noation $\lim_{\epsilon\to0}f(\epsilon)/\epsilon=0$.

Answer 1

I found an counterexample which satisfies the conditions and is even a diffusion process but no martingale.

Let $B:=(B_t)_{t\geq0}$ be a Brownian motion and $\mathcal{F}_s$ the filtration generated by $B$. Further let $Z$ be a nonzero random variable with $\Bbb E Z=0$ (e.g. uniformly distributed on $\pm1$) and which is independent of $B$ and $(\mathcal F_t)_{t\geq0}$. We define the stochasic process $\require{cancel} \cancel{\varphi:=(\varphi_t)_{t\geq0}}$ by $$\require{cancel} \cancel{\forall t \geq0:\varphi_t:=Z}$$ and consider the diffusion process $$M_t:=\require{cancel} \cancel{\int_0^t \varphi_s\Bbb dB_s=}ZB_t.$$ The process is a local martingale with satisfies $$\Bbb EM_t=\Bbb E Z\cdot\Bbb EB_t=0\qquad\text{and}\qquad\Bbb E(M_{t+\epsilon}-M_t|\mathcal F_s)=\Bbb E Z\cdot\Bbb E(B_{t+\epsilon}-B_t|\mathcal F_s)=0. $$ But as $$\Bbb E(M_t|\mathcal{F}_s)=\Bbb E(ZB_t|\mathcal{F}_s)=\Bbb E Z\cdot \Bbb E(B_t|\mathcal{F}_s)=0\neq ZB_s=M_s,$$ $M$ is not a martingale.

This answer was motivated by this post, that made me wonder if the example for a diffusion process which is local martinale but no martingale is also possible for integrable stochastic process.

EDIT: Thanks to the comment of Alvin Lepik we know that the example only works as $M$ is not adapted. But this leads also to fact that $M$ can't be a local martingale and that we even cannot declare the stochastic integral $\int_0^t \varphi_s\Bbb dB_s$. So we can forget the part with $\varphi$ and just assume $M$ to be definied as $M_t:=ZB_t$ in the example above. To see why $M$ and also $Z$ is not adapted I give a proof:

Assume $Z$ to be $\mathcal F_t$ measureable. As $Z$ is also independet of $\mathcal F_t$ we get $$ Z=\Bbb(Z|\mathcal F_t)=\Bbb EZ=0, $$ which is a contradiction.