Statement

For an adapted square-integrable process $(Y_t)$,

$$ M_t=\int_0^t Y_s\,dB_s $$

is a martingale.

Why it holds for simple processes

Each increment has the form

$$ Y^{(k)}(B_{t_{k+1}}-B_{t_k}), $$

where $Y^{(k)}$ is known at time $t_k$ and the Brownian increment is independent of the past with mean $0$.

Therefore,

$$ \mathbb E\bigl[Y^{(k)}(B_{t_{k+1}}-B_{t_k})\mid \mathcal F_{t_k}\bigr]=0. $$

Consequence

In particular,

$$ \mathbb E\left[\int_0^T Y_t\,dB_t\right]=0. $$