I believe, this problem comes from the definition of a sequence of sets where the limit of the sequence is defined as through indicator function is equivalent to the infimum-supremum definition. I am still not able to argue clearly in this case of the indicator function (multi-definition). So here goes the exact question-
Let ($\Omega$, $\mathcal{F}$, $\mathbb{P}$) be a probability space. We define the indicator function as
$$\textbf{1}_\textit{A}(\omega)= \begin{cases} 1 & if &\omega \in \textit{A} \\ 0 & if & \omega \notin \textit{A} \end{cases} $$ Let $\textit{A}_1, \textit{A}_2 , .... \in \mathcal{F}$. We say that $\textit{A}_n \to \textit{A} $ if $n \to \infty$ if $\textbf{1}_{A_n}(\omega)=\textbf{1}_{A}(\omega)$ for each $\omega \in \Omega$. Show that $\textit{A} \in \mathcal{F}$ and $\mathbb{P}(\textit{A}_n) \to \mathbb{P}(\textit{A})$
I presume you meant to write that $A_n \to A$ iff $\lim_n 1_{A_n}(\omega) = 1_A(\omega)$ for all $\omega$.
If this is the case, the dominated convergence theorem is your friend:
$P A_n=\int 1_{A_n} \to \int 1_A = PA$.