Does $\sum_i\mu\,A_i\le M\,\mu(\bigcup_i A_i)$ hold? In some places of proofs of the embedding and "well-definedness" properties of the Sobolev spaces one uses a result of the type of the following
Lemma A. Let $\mu$ be a finitely additive set function defined on a ring of sets with values in $\mathbb R_+=\mathbb R\cap\{t:0\le t\}$. Let $\mathscr A=\{A_1,\ldots\,A_n\}\subseteq\rm{dom\,}\mu$ be finite and such that for every $A\in\mathscr A$ the set $\mathscr A\cap\{B:A\cap B\not=\emptyset\}$ has cardinality at most $M$ where $1\le M<+\infty$. Then $\sum_i\mu\,A_i\le M\,\mu(\bigcup_i A_i)$ holds.
The validity of Lemma A seems to be plausible but I have not been able to find a proper proof. So the question is the following:

How is Lemma A proved, or is it false?

 A: For a subset $I\subseteq \{1,\dots,n\}$ let $B_I=\bigcap_{i\in I} A_i\cap \bigcap_{i\notin I}(A\setminus A_i)$ where $A=\bigcup_i A_i$. Note that $A_i=\bigcup_{i\in I}B_I$ and $B_I\cap B_J=\emptyset$ for $I\ne J$. Hence:
\begin{align*}
\sum_i\mu(A_i)=\sum_i\mu\left(\bigcup_{i\in I}B_I\right)=\sum_i\sum_{i\in I}\mu(B_I)
\end{align*}
Notice that by assumption each $I$ such that $B_I\ne\emptyset$ contains at most $M$ elements. Hence the last sum is $$\leq \sum_I M\mu(B_I)=M\sum_I\mu(B_I)=M\mu\left(\bigcup_{I}B_I\right)=M\mu(A)$$
This also shows that we only need the weaker assumption that every point is contained in at most $M$ sets of the $A_i$.
A: Let $A=\bigcup_{i}A_i$. Consider the function $f = \chi_{A_1}+\dots+\chi_{A_n}$. By assumption we have $f\leq M\chi_A$ (which is actually weaker than what we are given). Integrating gives us $$\sum_i\mu(A_i)=\int f\leq\int M\chi_A=M\mu\left(\bigcup_{i}A_i\right)$$
Note that we don't really have a general integral in this situation because we only have a finitely additive function on a ring of sets and not a measure, but we only needed very basic properties of the integral (for step-functions only). One can go through the usual construction of the integral to see that for this we only need what we are given here.
