Convergence of Stochastic Integral 
Let $f$, $f_n$ be stochastic processes adapted to the filtration generated by Brownian Motion $B(t)$, and $\int^b_aE(f^2) \, dt<\infty$,$\int^b_aE(f^2_n) \, dt<\infty$.
Assume that $\int^b_a|f(t)-f_n(t)|dt\to 0$ almost surely.
Show that $\int^b_af_n(t) \, dB(t)$ converge to $\int_a^bf(t) \, dB(t)$ in probability.

I try to show the convergence in probability by proving the convergence in $L^2$ first, i.e., $E(\int^b_a|f(t)-f_n(t)|dB(t))^2=\int^b_aE(f(t)-f_n(t))^2dt$.
But I have no idea how to show the convergence of the last term by almost surely convergence. Any ideas? Thanks.
 A: The assertion does, in general, not hold true. If we define
$$f_n(t) := \sqrt{n} 1_{[0,1/n]}(t) \qquad f(t) :=0,$$
then the assumptions of your statement are satisfied. However, the sequence of stochastic integrals $$\int f_n(t) \, dB_t = \sqrt{n} B_{1/n}$$ does not converge in probability to $0 = \int f(t) \, dB_t$ as $n \to \infty$; to see this, just note that $\sqrt{n} B_{1/n} \stackrel{d}{=} N(0,1)$ for all $n \in \mathbb{N}$, and therefore $\sqrt{n} B_{1/n}$ does not even converge in distribution to $0$.

If we replace $\int_a^b |f(t)-f_n(t)| \, dt \to 0$ by the stronger assumption $\int_a^b |f(t)-f_n(t)|^2 \, dt \to 0$, then the statement holds true. To prove this note that the stopping times
$$\tau_n := \inf\left\{r \in [a,b]; \int_a^r |f(t)-f_n(t)|^2 \, dt \geq \delta\right\}, \qquad (\inf \emptyset = \infty)$$
satisfy $\tau_n \to \infty$ as $n \to \infty$ for fixed $\delta>0$ and use that
$$\mathbb{P} \left( \left| \int_a^b f_n(t) \, dB_t - \int_a^b f(t) \, dB_t \right| \geq \epsilon \right) \leq \frac{1}{\epsilon^2} \underbrace{\mathbb{E} \left( \int_a^{\tau_n} |f(t)-f_n(t)|^2 \, dt \right)}_{\leq \delta} + \mathbb{P}(\tau_n \leq b).$$
