Sequence of integrals of positive function Let $f(x)$ be a function positive almost everywhere on $X$.
Let $A_n$ be a sequence of subsets of $X$ such that $m(A_n) > c> 0$ for all $n$, where $c$ is some constant, and $m$ denotes the Lebesgue measure.
Then, is it possible to have
$$ \int_{A_n} f(x)dx \to 0$$
?
My intutition is that this cannot happen, but I want to prove it rigorously.
 A: 
Then, is it possible to have
$$ c_n:=\int_{A_n} f(x)dx \to 0?$$

Yes when the space is $\sigma$-finite and  has an infinite measure. In this case, you can take an integrable function and a sequence $(A_n)_{n\geqslant 1}$ of pairwise disjoint sets whose measure is greater than $1$. In this case, the sequence $(c_n)_{n\geqslant 1}$ is even summable. 
However, when $X$ has a finite measure, it is not possible. Indeed, for any positive $\delta$, we have 
\begin{align}
  c_n&\geqslant \int f(x)\mathbf 1\{ f\gt \delta\}\mathbf 1_{A_n}\\  
&\geqslant\delta\mu\left(A_n\cap\{ f\gt \delta\} \right) \\ 
&=\delta\mu(A_n)-\delta\mu\left(A_n\cap\{ f\leqslant \delta\} \right) \\
&\geqslant \delta c-\delta\mu\{ f\leqslant \delta\}.
\end{align}
Since $X$ has a finite measure and $f$ is positive almost everywhere, we have  $\lim_{j\to \infty}   \mu\{ f\leqslant 1/j\}$ hence we may choose $\delta$ such that $\mu\{ f\leqslant \delta\}\leqslant c/2$. As a consequence, $c_n\geqslant \delta c/2$ for each $n$. 
