Consider a Brownian Motion, $B(t)$, which are all $\mathcal F_t$ measurable. Let $f,f_n$ be in $L_{ad}^2\{[a,b] \times \Omega\}$, which are the set of stochastic processes that are adapted to the filtration $\{\mathcal F_t\}$ and satisfy $\int_a^bE\left(f(t)^2\right)dt < \infty$. Suppose that $\int_a^b|f(t)-f_n(t)|dt\to0$ almost surely. Prove that $\int_a^bf_n(t)dB(t) \to \int_a^bf_n(t)dB(t)$ in probability.
I am trying to show this two different ways. First by using $L^2$ convergence which will then imply convergence in probability. What I have for this way is the following:
Since $f$ and $f_n$ are in $L_{ad}^2$ then the ito integral $\int_a^bf(t)-f_n(t)dB(t)$ exists and $E\left|\int_a^bf(t)-f_n(t)dB(t)\right|^2=\int_a^bE\left|f(t)-f_n(t)\right|^2dt$, so it will suffice to show that the right hand side of this equal sign converges to $0$ which i'm unsure how to do.
Second, I am try to show this directly using the definition of convergence in probability.
$P\left(\left|\int_a^bf(t)-f_n(t)dB(t)\right|>\epsilon\right)\le\frac{1}{\epsilon}E\left|\int_a^bf(t)-f_n(t)dB(t)\right|$, hence if I could show that $\int_a^bf(t)dB(t) \to \int_a^bf_n(t)dB(t)$ in $L^1$ then I would be done.
Similarly,
$ P\left(\left|\int_a^bf(t)-f_n(t)dB(t)\right|>\epsilon\right)\le\frac{1}{\epsilon^2}E\left|\int_a^bf(t)-f_n(t)dB(t)\right|^2= \frac{1}{\epsilon^2}\int_a^bE\left|f(t)-f_n(t)\right|^2dt$ so if we could show this last term converged to $0$ then we would be done.