Integral over a sequence of sets whose measures $\to 0.$ If $ f \in L_p$ with $1 \leq p \leq \infty $ and ${A_n}$ is a sequence of measurable sets such that $ \mu (An) \rightarrow 0,$ then $ \int_{A_n} f \rightarrow 0$.
Can someone give me a hint? 

My proof might be a bit lengthy but I think it might be a more standard way in dealing with this kind of problem. 
First by $f\in L^{p}$ and DCT ($f_n= |f|^p 1_{\{|f|>n\}}\leq |f|, f_n\rightarrow 0$ a.e. as $f \in L^p$ ), we have 
$$\forall \epsilon >0, \exists N >1 , s.t. \int_{\{|f|>N\}} |f|^p <\epsilon/2.$$
Now since $|f|^p>f$ on $\{|f|>N\}$,$$|\int_{A_n}f |\leq \int_{\{A_n,\; |f|>N\}}|f| + \int_{\{A_n,\;|f|\leq N\}}|f| \leq \int_{\{|f|> N\}}|f|^p +N\mu(A_n)\leq \epsilon/2 + N\mu(A_n).$$
Since $N$ depends on $\epsilon$, you can chose $A_n$ small enough ($\epsilon/2N$)so that 
$$|\int_{A_n}f |\leq \epsilon.$$
