Talk:Countably Additive Function also Finitely Additive

It is straightforward to prove this if $f \left({\varnothing}\right) = 0$ (or, a fortiori, that there exists an element whose measure is finite).

This is shown in Countably Additive Function of Null Set.

But I'm honestly not sure that this holds for a general countably additive set. --Prime.mover 22:33, 24 February 2010 (UTC)