convergence of a sequence of stochastic processes Let $\psi$ be a sequence of progressively measurable processes s.t. for all $t \ge 0$ we have: $\psi_t^n\to0$ a.s. Let B be a standard brownian motion.
Assume also that there exists a process $\phi\in \lambda^2_{loc}$ with $|\psi^n|\le \phi$. Where $\lambda^2_{loc}$ is the set of {$\phi:$ $\phi_s$ is progressive and $\int_0^t\phi_s^2ds< \infty$ for all t}
How could I show that for $t>0$ $$\int_0^t\psi^n_sdB_s\to0$$
in probability?
What is making it hard for me is that $\phi$ is not surely integrable and therefore I cannot apply Lebesgue theorem.
I will provide my attempt, I would really appreciate a feedback.
Let's divide the problem in two cases:
$1)$ assume $E[\int_0^t\phi^2_sds]<\infty$
Then $$\lim_{n \to \infty}E\bigg[ (\int_0^t\psi^n_sdB_s-0)^2\bigg]=\lim_{n \to \infty}E\bigg[ \int_0^t(\psi^n_s)^2ds\bigg]\to0$$
by dominated convergenge theorem, since both $\int_0^t(\psi^n_s)^2ds<\int_0^t(\phi_s)^2ds$ and $(\psi^n_s)^2<\phi_s^2$ and both $\phi_s^2$ and $\int_0^t(\phi_s)^2ds$ are integrable by our initial assumption.
$2)$ Assume now instead that $\tau_k$ is a sequence of stopping time defined as $$\tau_k:inf(t\ge0:\int_0^t\phi^2_sds\ge k)$$
 Now:
For a $\epsilon > 0 $ there exists a $k$ s.t. $P(t > \tau_k)<\epsilon$
For $a >0$
$$\begin{align*} P(|\int_0^t\psi^n_sdB_s - 0|>a) &= P(|\int_0^t\psi^n_sdB_s|>a;t<\tau_k)+P(|\int_0^t\psi^n_sdB_s|>a;t\ge \tau_k) \\ &\le P(|\int_0^t\psi^n_sdB_s|>a;t<\tau_k) + \epsilon \\ &\le
P(|\int_0^{t \land \tau_k}\psi^n_sdB_s|>a) \to0 \end{align*} $$
and the last step follow since, as before we could write:
$\lim_{n \to \infty}E\bigg[ (\int_0^{t \land \tau_k}\psi^n_sdB_s-0)^2\bigg]=\lim_{n \to \infty}E\bigg[ \int_0^{t \land \tau_k}(\psi^n_s)^2ds\bigg]\to0$
and
 $\int_0^{t \land \tau_k}(\psi^n_s)^2ds \le \int_0^{t \land \tau_k}\phi^2_sds$ and $(\psi^n_s)^2<\phi_s^2$ and both $\int_0^{t \land \tau_k}(\phi_s)^2s$ and $\phi_s^2$ (is it right that also $\phi_s^2$ is integrable under the inital assumption of step 2?)  are integrable by our initial assumption.
So what do you think about this proof? is it correct?
 A: We have $\phi\in \lambda^2_{loc}$ which means there exists a localizing secquence $(\tau_m)_{m\in\Bbb{N}}$ s.t. for all $t>0$ $$E\left[\int_0^t(\phi_s\cdot 1_{t\le \tau_m})^2ds\right]< \infty$$
Additionally we have (with $\stackrel{P}{\rightarrow}$ denoting convergence in probability)$$\int_0^t\psi^n_sdB_s\stackrel{P}{\rightarrow}0 \quad \iff \quad \forall m\in\Bbb{N}: \int_0^t\psi^n_sdB_s\cdot 1_{t\le\tau_m} \stackrel{P}{\rightarrow}0\quad\quad(*)$$ as well as 
$$ \int_0^t\psi^n_sdB_s\cdot 1_{t\le\tau_m} = \int_0^t\psi^n_s\cdot 1_{t\le\tau_m}dB_s$$
It follows $$|\psi^n|\cdot1_{t\le \tau_m} \le \phi\cdot 1_{t\le \tau_m}$$ and by Ito's isometry:
$$\begin{align*}E\left[\left(\int_0^t\psi^n_s\cdot 1_{t\le\tau_m}dB_s\right)^2\right] &= E\left[\int_0^t(\psi^n_s\cdot 1_{t\le\tau_m})^2ds\right] \\ &\le E\left[\int_0^t(\phi_s\cdot 1_{t\le\tau_m})^2ds\right] < + \infty\end{align*}$$ 
So we can use Lebesgues theorem to deduce:
$$\begin{align*} 0 &\le \lim_{n\to\infty} E\left[\left(\int_0^t\psi^n_s\cdot 1_{t\le\tau_m}dB_s\right)^2\right] \\ &\le  \lim_{n\to\infty} E\left[\int_0^t(\psi^n_s\cdot 1_{t\le\tau_m})^2ds\right] \\ &= E\left[\int_0^t(0\cdot 1_{t\le\tau_m})^2ds\right] = 0 \end{align*}$$
And we get that for all $m\in \Bbb{N}$ it holds $$\int_0^t\psi^n_s\cdot 1_{t\le\tau_m}dB_s = \int_0^t\psi^n_sdB_s\cdot 1_{t\le\tau_m} \stackrel{P}{\rightarrow} 0$$ so it follows from $(*)$ that $$\int_0^t\psi^n_sdB_s\stackrel{P}{\rightarrow}0$$ as wanted
