Limit and expected value interchange for stochastic integral We know that $f\in L^2[0,\infty]$
Prove that we can interchange limit and expectation
$$\lim_{n\rightarrow\infty}\mathbb EB_t\int_0^nf(s)dB_s=\mathbb EB_t\int_0^\infty f(s)dB_s.$$
$B_s$ is Brownian motion.
I had ideas about The Bounded Convergence Theorem, but got stuck.
 A: \begin{align*}
&\left|{\mathbb E}\left[B_{t}\int_{0}^{n}f(s){\rm d}B_{s}\right]-{\mathbb E}\left[B_{t}\int_{0}^{\infty}f(s){\rm d}B_{s}\right]\right| \\
&=\left|{\mathbb E}\left[B_{t}\left(\int_{0}^{n}f(s){\rm d}B_{s}-\int_{0}^{\infty}f(s){\rm d}B_{s}\right)\right]\right| \\
&\leq {\mathbb E}[|B_{t}|^{2}]^{1/2}{\mathbb E}\left[\left|\int_{0}^{n}f(s){\rm d}B_{s}-\int_{0}^{\infty}f(s){\rm d}B_{s}\right|^{2}\right]^{1/2} \\
&=t^{1/2}\lim_{m \uparrow \infty}{\mathbb E}\left[\left|\int_{0}^{n}f(s){\rm d}B_{s}-\int_{0}^{m}f(s){\rm d}B_{s}\right|^{2}\right]^{1/2} \\
&=t^{1/2}\lim_{m \uparrow \infty}{\mathbb E}\left[\left|\int_{0}^{m \wedge n}f(s){\rm d}B_{s}-\int_{0}^{m}f(s){\rm d}B_{s}\right|^{2}\right]^{1/2} \\
&=t^{1/2}\lim_{m \uparrow \infty}{\mathbb E}\left[\left|\int_{0}^{m}{\bf 1}_{[0,n]}(s)f(s){\rm d}B_{s}-\int_{0}^{m}f(s){\rm d}B_{s}\right|^{2}\right]^{1/2} \\
&=t^{1/2}\lim_{m \uparrow \infty}{\mathbb E}\left[\left|\int_{0}^{m}{\bf 1}_{(n,\infty)}(s)f(s){\rm d}B_{s}\right|^{2}\right]^{1/2} \\
&=t^{1/2}\lim_{m \uparrow \infty}{\mathbb E}\left[\int_{0}^{m}{\bf 1}_{(n,\infty)}(s)f(s)^{2}{\rm d}s\right]^{1/2} \\
&=t^{1/2}{\mathbb E}\left[\int_{0}^{\infty}{\bf 1}_{(n,\infty)}(s)f(s)^{2}{\rm d}s\right]^{1/2} \rightarrow 0 \quad \text{ as } \quad n \uparrow 0.
\end{align*}
The third is obtained by the Cauchy–-Schwarz inequality and the last is obtained by the dominated convergence theorem noting $f \in L^{2}([0,\infty))$.
The fourth may require some conditions.
Please check the uniform integrability or the martingale convergence theorem.
