# 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.

\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.