Q:

Property of a $\sigma$-additive function

Let $X$ be a set and ${\cal A}\subset 2^X$ a ring.  If $\mu:{\cal A}\to [0,+\infty]$ is a $\sigma$-additive function and $(A_i)_{i\in\Bbb N}$ is a family in $\cal A$ such that $\bigcup_{i=1}^\infty A_i$ belongs to $\cal A$, then does the relation
$$\mu\left(\bigcup_{i=1}^\infty A_i\right)\leq \sum_{i=1}^\infty\mu(A_i)$$
hold?
I know that $\mu\left(\bigcup_{i=1}^n A_i\right)\leq \sum_{i=1}^n\mu(A_i)$ for all $n$, so that $\mu\left(\bigcup_{i=1}^n A_i\right)\leq \sum_{i=1}^\infty\mu(A_i)$ for all $n$.  But I can't go further.

A:

Define $A'_1 = A_1$, $A'_2 = A_2 \backslash A_1$,$ A'_3 = A_3 \backslash ( A_1 \cup A_2)$ and 
in general $A'_k= A_k \backslash ( \cup_{i=1}^{k-1} A_i)$. Clearly $A'_i \subset A_i$. Moreover, $A'_i \cap A'_j = \emptyset$ whenever $i \ne j$ and also 
$$\bigcup_{i=1}^\infty A_i =\bigcup_{i=1}^\infty A'_i$$
(these two statements require some thinking).
Therefore
$$\mu(\cup_{i=1}^\infty A_i) = \mu(\cup_{i=1}^\infty A'_i ) = \sum \mu(A'_i) \le \sum \mu(A_i)$$ 

