Let $(X,\mu)$ be a probability measure space. Suppose $f_n \to f$ pointwise and $f_n$ is dominated by some $\mu$-integrable function. Let $(A_n)_n$ be a sequence of measurable sets converging to some measurable $S$ with $\mu(S) = 1$.
Question. Is it true that $\int_X f_n1_{A_n}\,d\mu \to \int_X f\,d\mu$ ?
A half-baked idea
Define $g_n := f_n1_{A_n}$. In case,
Conjecture. $1_{A_n} \to 1_S$ pointwise,
which I suspect to be true, then $|g_n|$ is dominated by a $\mu$-integrable function and so the Dominated Convergence Theorem gaurantees that $\int_X g_n\,dP \to \int_X f1_S\,dP = \int_X f\,dP$ since $P(S)=1$.
Another way to show this is a $2\epsilon$-argument. First note that for any integrable function $f$, we have that $$\int_X fd \mu=\int _S fd\mu+\int_{X\setminus S}fd\mu=\int _Sfd\mu.$$ As $X\setminus S$ has measure zero. Then we have that \begin{align*}|\int _Xf_n1_{A_n}d\mu-\int _Xfd\mu|&=|\int _Xf_n1_{A_n}d\mu-\int _Xf1_{S}d\mu|\\ &=|\int _Xf_n1_{A_n}d\mu-\int _Xf1_{S_n}d\mu+\int_Xf_n1_{S}d\mu-\int_Xf_n1_{S}d\mu|\\ &\leq |\int _Xf_n1_{A_n}d\mu-\int_Xf_n1_{S}|+|\int _Xf1_{S}d\mu-\int_Xf_n1_{S}d\mu|\\ &\leq |\int _Xf_n(1_{A_n}-1_{S})d\mu|+|\int _X(f_n-f)1_{S}d\mu|\\ &\leq\mu(A_n-S)\cdot||f_n||_{\infty}+||1_S||_{\infty}||f_n-f||_{L^1} \to0 \end{align*} As $f_n\to f$ pointwise and $f$ is uniformely bounded, and $A_n\to S$.