When can the order of a Ito integral and a.s. limit be interchanged i.e. assume you got $Y_t^n\rightarrow Y_t$.
When can we say
$$\int\limits_0^t Y_t^n dW_s \rightarrow \int\limits_0^t Y_t dW_s$$
I know dropped a lot of further explenations. But I think it is very clear from the context what I mean. But feel free to assume other conditions on the integrand ect.
Assuming $|Y^{n}|,|Y|\leq M$ or at least dominated $|Y^n_t-Y_t|^2\leq G$ for some $E[G]<\infty$.
By Markov
$$P\left[\int^{T} (Y^n_t-Y_t)dB_t\geq R\right]\leq \frac{1}{R}E\left(\int^{T} (Y^n_t-Y_t)dB_t\right)^{2},$$
and by Itô isometry we have
$$E\left(\int^{T} (Y^n_t-Y_t)dB_t\right)^{2}=\int^{T} E\left(Y^n_t-Y_t\right)^{2}dt.$$
Then we use $$\begin{align*} \int |Y^n_t-Y_t|^2 \, d\mathbb{P} &= \int_{|Y^n_t-Y_t| \leq \epsilon} |Y^n_t-Y_t|^2 \, d\mathbb{P} + \int_{|Y^n_t-Y_t| > \epsilon} |Y^n_t-Y_t|^2 \, d\mathbb{P} \\ &\leq \epsilon^2 + (2M)^2 \mathbb{P}(|Y^n_t-Y_t|>\epsilon). \end{align*}$$
So here you need
$$\lim_{n\to +\infty}\sup_{t\leq T}\mathbb{P}(|Y^n_t-Y_t|>\epsilon)=0.$$
If we just use dominated, the condition is
$$\lim_{n\to +\infty}\sup_{t\leq T}\mathbb{E}G1_{|Y^n_t-Y_t|>\epsilon}=0.$$