TITLE: Do finite additivity and countable subadditivity imply countable additivity?
QUESTION [5 upvotes]: Given a Measure Space and f a positive set function on the sigma-algebra of the space (not identically infinite), how could I prove that f is a measure given the hypothesis above? 
I've tried both by contradiction and induction, but I struggle a bit with the intuition.

REPLY [6 votes]: Yes. 
Finite additivity implies the next inequality:
$$
\mu \left(\bigsqcup\limits_{i = 1}^{\infty} A_i \right) \geq \sum\limits_{i = 1}^{\infty} \mu(A_i)
$$
Indeed, 
$$
\mu \left(\bigsqcup\limits_{i = 1}^{\infty} A_i \right) = \mu \left(\bigsqcup\limits_{i = 1}^{N} A_i \right) + \mu \left(\bigsqcup\limits_{i = N+1}^{\infty} A_i \right) \geq \mu \left(\bigsqcup\limits_{i = 1}^{N} A_i \right)  = \sum\limits_{i = 1}^{N} \mu(A_i)
$$
Taking limits of both sides of the inequality finishes the proof.
Obviously, countable subadditivity (or also called $\sigma$-subadditivity) implies the opposite inequality. 
Combining the ideas above, it can be concluded that a finite additivity and a countable subadditivity do imply a countable additivity.