I have repeatedly come across the statement that "a process with a drift cannot be a martingale". Is this true also for stochastic drifts?
Suppose I have a process with a stochastic drift:
$$X_t=X_0+\int_{h=0}^{h=t}\left(f_1(W_h)\right)dh+\int_{h=0}^{h=t}\left(f_2(W_h)\right)dW_h$$
Above, $f_1$ and $f_2$ are some functions of $W_t$.
What if the expected value of the stochastic drift is zero? i.e.:
$$\mathbb{E}[X_t]=X_0+\int_{h=0}^{h=t}\left(\mathbb{E}\left[f_1(W_h)\right]\right)dh+0=\\=X_0+\int_{h=0}^{h=t}0dh+0=X_0$$
I know the above is not a sufficient condition for $X_t$ to be a martingale, but my intuition tells me it should at least ensure that $X_t$ "has a shot" at being a martingale, right?
Edit: the example below serves as a "counterexample" (the drift has zero expectation, but the process is not a martingale) (thank you to @AntoineConze). So I wonder if its possible after all to have a process with a stochastic drift that is a martingale?
Example 1:
Let $X_t$ be defined as:
$$X_t=X_0+\int_{h=0}^{h=t}W_hdh + \int_{h=0}^{h=t}1dW_h$$
NTS: $\forall 0\leq s < t$: $\mathbb{E}\left[X_t | \mathcal{F}_s \right]=X_s$
Now:
$$\mathbb{E}\left[X_0+\int_{h=0}^{h=t}W_hdh + \int_{h=0}^{h=t}1dW_h|\mathcal{F}_s \right]=\\=X_0+\mathbb{E}\left[\int_{h=0}^{h=s}W_hdh + \int_{h=s}^{h=t}W_hdh +W_s +W(t-s)|\mathcal{F}_s \right]=\\=X_0+\int_{h=0}^{h=s}W_hdh + \int_{h=0}^{h=s}1dW_h+\int_{h=s}^{h=t}\mathbb{E}[W_h|\mathcal{F}_s]dh + \mathbb{E}[W(t-s)]_{=0}=\\=X_s+\int_s^tW_sdh\neq X_s$$