If $x \in X$ belongs to at most $k$ sets $\{A_i\}_{i = 1}^n \subset \mathcal{M}$, then $\sum_{i = 1}^n \mu(A_i) \le k$. I am working on the following question:

If $ (X, \mathcal M, \mu) $ is a measure space with $\mu(X) = 1$, and $A_1, \dots, A_n \in \mathcal M$ are measurable subsets of $X$, then:
If every $x \in X$ belongs to at most $k$ of the sets $A_i$, then $\sum_{i = 1}^n \mu(A_i) \le k$.  Similarly if every $x \in X$ belongs to at least $k$ of these sets, then $\sum_{i = 1}^n \mu(A_i) \ge k$.

It is sort of a generalization of this question.
I understand that the result follows from Fubini's theorem, but I need a solution that only uses first properties of measure.
My thoughts so far:
Consider when $x$ belongs to exactly $k$ of the sets.   For a given $x \in X$, intersect all of the $k$ sets containing it.  There are at most $\binom{n}{k}$ different sets that can be obtained this way and they partition $X$.  Maybe this partition can be used to bound the measure?
I am really stuck so any nudges in the right direction would be appreciated!
 A: Here's an approach that's maybe a little closer to "first properties of measure". Let $f:X\to \mathbb{R}$ be the simple function $$ f = \sum_1^n I_{A_i}, $$
where $I_A$ is the indicator function of $A$. Then $0 \le f \le k$, so $0 \le \int_X f\, d\mu \le k$. The partition you mention can be used to prove that $$ \int_X f\, d\mu = \sum_1^n \mu(A_i)$$
by an argument that isn't Fubini but its discrete version---interchanging the order of summation.
A: Here's an argument that requires nothing beyond the finite additivity of $\mu$ (it's slightly informal, but it's not hard to make it rigorous).
Drawing a picture can help: Imagine the space as a two-dimensional region, and the sets as sub-regions that can overlap. Consider the $m$ distinct nonempty fragments obtained by taking arbitrary intersections of the $A_i$s and their complements (set things up so that the space is the disjoint union of the fragments). Each fragment is measurable. Express each $A_i$ as a disjoint union of fragments. Then expand each summand $\mu(A_i)$ (by finite additivity of $\mu$), and write $$\sum_{1 \le i \le n} \mu(A_i) = \sum_{1 \le j \le m} c_j \mu(B_j)$$ where the $B_j$'s are the fragments, and each coefficient $c_j$ is a nonnegative integer with $c_j \le k$. Thus, $$\sum_{1 \le j \le m} c_j \mu(B_j) \le \sum_{1 \le j \le m} k \mu(B_j) = k \sum_{1 \le j \le m} \mu(B_j) = k.$$
For the other question, just switch the direction of the inequality in the proof.
