Almost every point in a union A of measurable sets $A_n$ belongs to at most k of the sets. Show $m(A)\geq \frac{\sum m(A_n)}{k}$

Question

Let $\{A_n\}$ be a collection of measurable sets in $\mathbb{R}$. If $A$ is the union of the collection and almost every $x\in A$ belongs to no more than $k$ of the $A_n$ then I need to show that $m(A)\geq\frac{\sum m(A_n)}{k}$.

My idea: For each such x, I would like to identify it with a sub-collection of $\{A_n\}$ in such a way that the sub-collections are disjoint. A starting point would be the collection of, say, j sets that $x$ belongs to: $\{A_{x_i}:\ x\in\bigcap_{i=1}^j A_{x_i},\ 1\leq j\leq k\}$. However, I can't see that such sub-collections must be disjoint and I am not sure where to go from here or if this will lead to a solution. Hints would be appreciated!

Answer 1

Hint: Denote $\chi_B$ the characteristic function of a set $B$. Then $\sum_n\chi_{A_n}\leq k\cdot\chi_A$ a.e.