Square integrable integral by Brownian motion is martingale

Question

Let $f \in C_b^2(\mathbb{R}_{+} \times \mathbb{R})$. Then my lecturer said that for a Brownian motion $B_t$
\begin{align}
X_t = \int_0^t f(s, B_s) \, dB_s
\end{align}
is a martingale, which should follow from the integral being square integrable, but I can't how to derive this from Ito's formula.

Answer 1

The condition @d.k.o. provided in the comments implies that $E[\int_s^t f(u,B_u)dB_u | \mathscr F_s]=0.$ Then, using the linearity property of the integral, we have $$E[X_t |\mathscr F_s]=E[X_s |\mathscr F_s]+E[\int_s^t f(u,B_u)dB_u | \mathscr F_s]=X_s.$$