Q: Stochastic convergence (product) Consider the measurable space $(\Omega,\mathcal{A},\mu)$. Let $f,f_1,f_2,\ldots$ be measurable functions and $A,A_1,A_2,\ldots\in\mathcal{A}$. Let $(f_n)_{n\in\mathbb{N}}$ be convergent stochastically to $f$ and $(1_{A_n})_{n\in\mathbb{N}}$ be convergent stochastically to $1_A$.
My question is if then $(f_n1_{A_n})$ converges stochastically to $f1_A$.
If yes how can I see that using the following definition:
In our reading, we defined stochastic convergence as follows:


Let $(\Omega,\mathcal{A},\mu)$ be a measurable space.
    A series $(f_n)_{n\in\mathbb{N}}$ converges locally stochastic to the measurable function $f$, if for all $A\in\mathcal{A}$ with $\mu(A)<\infty$ and for all $\varepsilon>0$ it is
    $$
\lim\limits_{n\to\infty}\mu(A\cap\left\{\omega\in\Omega: \lvert f_n(\omega)-f(\omega)\rvert >\varepsilon\right\})=0.
$$
    Is this for $A=\Omega$, too, we call it stochastic convergence.



A: Hint: triangles inequality
Consider something like $f1_{A}-f1_{A_n}+f1_{A_n}-f_n1_{A_n}$
Edit. So i was wrong the correct argument is this, thanks to a similar question asked about someone else
$\{\omega \in\Omega : f|1_A-1_{An}|>\epsilon \}\subset \{\omega\in\Omega:|1_A-1_{An}|=1\}$ which we know has measure going to 0. 
