Let $M$ be any $\sigma$-field and let $m$ be a function such that $m: M \to[0, \infty]$ and for any disjoint sequence $E_n$ in $M$ $m(\cup_n E_n) = \sum_n m(E_n)$ holds.


1) for $A \subset B$ and $A,B\in M$ we have $m(A) \leq m(B)$

2) For $A_n\in M, n\in\mathbb{N}$ (not necessarily disjoint) we have $m(\cup_n A_n) \leq \sum m(A_n)$.


1) Set $E_1 := A$, $E_2 := B\backslash A$, $E_3 := E_4 := \dots := \emptyset$ and see $$ m(A) \leq m(A) + m(E_2) + m(\emptyset) + \dotso = m(\cup_n E_n) = m(B)$$ 2) Define $E_1 := A_1\backslash (\cup_{i=2}^{\infty}A_i), E_2 := A_2\backslash (\cup_{i=3}^{\infty}A_i)$ and so on. This gives you a disjoint partition of $\cup_n A_n$. Thus $$ m(\cup_n A_n) = \sum_n m(E_n) \leq \sum_n m(A_n)$$ where the last step is true by 1), since $E_n \subset A_n$.


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