What about the not disjoint sets in the definition of measure? I know that in the definition of measure we use the $\sigma$-additivity property, which means if $\left(A_{n}\right)_{n}$ are countable many disjoint sets of $\mathcal{A}$, than $$\mu\left(\cup_{n=1}^{\infty}A_{n}\right)=\sum_{n=1}^{\infty}\mu\left(A_{n}\right).$$
For first glance it was obvious for me, if the $\left(A_{n}\right)_{n}$ are not disjoint sets, then $$\mu\left(\cup_{n=1}^{\infty}A_{n}\right)\leq\sum_{n=1}^{\infty}\mu\left(A_{n}\right).$$
Now I am not really certain about it. However, I have found a statement which claims if $A\subset\cup_{n=1}^{\infty}A_{n}$, then$$\mu\left(A\right)\leq\sum_{n=1}^{\infty}\mu\left(A_{n}\right).$$
My question is if $\mu\left(\cup_{n=1}^{\infty}A_{n}\right)\leq\sum_{n=1}^{\infty}\mu\left(A_{n}\right)$ holds when $\left(A_{n}\right)_{n}$ are not necessary disjoint sets? If it doesn't, then could you give me a counterexample?
 A: Here it is important to note that $\cal A$ is a $\sigma$-algebra, in particular it is closed under set difference. This implies that $\mu$ is monotone: as sets get larger their measure can't get smaller: to see this, let's assume $\mu$ is additive and that $A, B \in \cal A$, with $B \subseteq A$ then also $A \setminus B \in \cal A$ and we have:
$$
\mu(A) = \mu(B) + \mu(A \setminus B) \tag*{($*$)}
$$
(taking $A_1 = A$, $A_2 = B \setminus A$ and $A_n = \emptyset$ for $n > 2$ in the definition of additivity). Hence, for any $A, B \in \cal A$ with $B \subseteq A$, as $\mu(A \setminus B) \ge 0$, we have
$$
\mu(A) \le \mu(B)
$$
Now given an arbitrary sequence $A_n \in \cal A$, not necessarily pairwise disjoint, if we put:
\begin{align*}
B_1 &= A_1 \\
B_n &= A_n \setminus (B_1 \cup B_2 \cup \ldots \cup B_{n-1})
\end{align*}
we have that $B_n \in \cal A$, that $\bigcup_nB_n = \bigcup_n A_n$ and  that $\mu(B_n) \le \mu(A_n)$ by ($*$). So as the $B_n$ are pairwise disjoint and $\mu$ is additive, we get:
$$
\mu\left(\bigcup_n A_n\right) = \mu\left(\bigcup_n B_n\right) = \sum_n \mu(B_n) \le \sum_n \mu(A_n)
$$
A: Let us define a new series of sets $\{B_n\}_{n\ge 1}$ with $B_1 = A_1$ and
$$
B_m = A_m \cap \Big(\bigcup_{n=1}^{m-1}A_n\Big)^c,\quad m=2,3,\cdots
$$
These sets are all in the original sigma algebra, and therefore we may evaluate the measure on them.
It is a simple exercise to show that:
(1) $B_n \subseteq A_n$ for all $n\ge 1$. Consequently, $\mu(B_n)\le \mu(A_n)$ (can be proven using countable additivity).
(2) $\{B_n\}$ is disjoint collection.
(3) $\bigcup_{n=1}^m B_n = \bigcup_{n=1}^m A_n$ for all $m\ge 1$.
Therefore,
$$
\mu\Big(\bigcup_{n=1}^m A_n\Big) = \mu\Big(\bigcup_{n=1}^m B_n\Big) = \sum_{n=1}^m \mu(B_n) \le \sum_{n=1}^m \mu(A_n)
$$
Taking $m$ to infinity requires a little bit of technicality, but it is fine I think.
