Let $t>0$ be fixed. Assume that you have a filtered probability space $(\Omega, \mathcal{F},P), \{\mathcal{F}_t\}$. Let $h_n$ be progressively measurable processes such that for every $(\omega,s)$, $|h_n(\omega,s)|<K$ and $h_n(\omega,s)\rightarrow 0$ as $n$ goes to infinity. Let $B$ be a Brownian motion on the probability space.
I want to prove that then
$$\lim\limits_{n \rightarrow \infty} \sup_{s\le t}\left|\int_0^sh_n(\cdot,z)dB_z\right|\rightarrow 0\text{, in probability}.$$
My attempt:
I know that since $h_n$ is bounded by $K$ it is an $L^2$ process and hence we can use the Itô-isometry. Let $\epsilon>0$ be given, I get by the Markov inequality.
$$P\left( \sup_{s\le t}\left|\int_0^sh_n(\cdot,z)dB_z\right|>\epsilon\right)\le \frac{E\left[\sup_{s\le t}\left|\int_0^sh_n(\cdot,z)dB_z\right|^2\right]}{\epsilon^2}.$$
The problem is that the $\sup$ is inside the expecation. If I was able to move the $\sup$ outside the expecation I could use the ordinary dominated convergence theorem(bounded convergence theorem in this case) and I would be done. But how do I handle the $\sup$ inside the expecation?