Why is a Itô integral w.r.t. Brownian motion a martingale? I'm trying to understand why $M_t(\omega)=\int^t_0 f(s,\omega)dB_s$ is a martingale w.r.t. $\mathcal{F}_t$.
I know that for $I_n(t,\omega)=\int^t_0 \phi_n(a,\omega)dB_a(\omega) $ -- the approximation of $M_t(\omega)$, using elementary functions (Riemann sums) -- we have $E[I_n(s,\omega)|\mathcal{F}_t]=I_n(t,\omega)$.
To prove that $M_t(\omega)$ is a martingale, I would need to be able to interchange the $\lim$ and $E()$. 
Which convergence theorem can I use and how?
 A: By "Which convergence theorem can I use and how?" I'd assume you have the monotone convergence, dominated convergence, and Fatou's lemma in mind. I don't think you can prove the martingale property using one of them, because stochastic integrals are not defined as a.s. limits.
Rather, they are defined as the $L^2$ limit of the approximating sequence. I'll assume that the integrand $f$ is square-integrable: $E\int f^2<\infty$. (This is related to the comment about local martingales. If $f$ is only square-integrable a.s., i.e., $\int f^2<\infty$ a.s., then its stochastic integral is only guaranteed to be a local martingale. I'm sure you'll read about this sooner or later.) Then as you said, there is an approximating sequence of simple processes $\phi_n$ of $f$. $\int \phi_n dB$ is a martingale, and $\int\phi_n dB\rightarrow \int f dB$ in $L^2$, which is how we define $\int f dB$.
Now, to show ($\dagger$) $E(\int_0^tfdB|\mathcal F_s)=\int_0^sfdB$, let $A\in\mathcal F_s$ and note
$$
\left|E\left[1_A\left(\int_0^sfdB-\int_0^s\phi_ndB\right)\right]\right|\le \|1_A\|_{L^2}\left\|\int_0^sfdB-\int_0^s\phi_ndB\right\|_{L^2}\rightarrow 0
$$
where the inequality comes from the Cauchy-Schwartz inequality. Similarly, $|E[1_A(\int_0^tfdB-\int_0^t\phi_ndB)]|\rightarrow 0$. Combining the last two results we have
\begin{multline}
\left| E\left(1_A\int_0^sfdB\right)-E\left(1_A\int_0^tfdB\right)\right|\\
\le
\left|E\left[1_A\left(\int_0^sfdB-\int_0^s\phi_ndB\right)\right]\right|
+
\left|E\left[1_A\left(\int_0^tfdB-\int_0^t\phi_ndB\right)\right]\right|
\rightarrow 0,
\end{multline}
or
$$
E\left(1_A\int_0^sfdB\right)=E\left(1_A\int_0^tfdB\right),
$$
which is exactly what we need to assert the martingale property ($\dagger$).
