Does $$Y\int_a^b f(s,\cdot )dW_s=\int_a^b Yf(s,\cdot )dW_s \ \ ?$$ whenever $Y$ is a bounded random variable ?
I could prove it whenever $Y$ is $\mathcal F_a$ as follow :
If $f$ is predictable, the solution is obvious.
If $f$ is not predictable but progressively measurable s.t. $\mathbb E\int_s^t |f|^2<\infty $, there is a sequence of progressively measurable $f_n\to f$ in $L^2(\Omega \times [s,t])$. Then $(Yf_n)$ is also predictable for all $n$ and s.t. $Yf_n\to Y f$ in $L^2(\Omega \times [s,t])$. Then, (all limit are taken in $L^2$, $$\int_s^t Y f(u)dW_u=\lim_{n\to \infty }\int_s^t Yf_n(u)dW_u=\lim_{n\to \infty }Y\int_s^t f_n(u)dW_u=Y\int_s^t f(u)dW_u.$$
Q1) Does it work ?
Q2) But what if $Y$ is not supposed to be $\mathcal F_a$ measurable ? I guess it's true, but I couldn't find a counter example. Any idea ?