# Lebesgue measure of union of disjoint measurable sets

I am wondering if the Lebesgue measure of the union of a countable collection of disjoint measurable sets is equal to the sum of the measure of such sets. I feel that they may be equal. I know that for finite case they are equal. I know that for outer measure, the previous statement is wrong, e.g., (non measurable)Vitali set. How about collections of disjoint measurable set, are they always equal? What if I change the collection into an uncountable collection of Lebesgue measurable sets, are they equal?

• This (countable additivity) is one of the most fundamental results of the theory. Any textbook in the subject will have the proof. With uncountable collections it is no longer true since any set is a union of singletons and every singleton has Lebesgue measure zero. – Umberto P. Apr 25 '15 at 13:02
• @UmbertoP.Thank you. I saw that Royden's book proves subadditivity for outer measure and finite additivity for measurable sets, so can I do the following? As $m(\cup_{k=1}^nE_k) = \sum_{k=1}^nm(E_k)$, so $m(\cup_{k=1}^{\infty}E_k) \ge \sum_{k=1}^nm(E_k)$, then $m(\cup_{k=1}^{\infty}E_k) \ge \sum_{k=1}^{\infty}m(E_k)$. – TH000 Apr 25 '15 at 13:17
• @UmbertoP.I have another questions. Have we defined 0 times uncountably infinity in Real Analysis. I know that union of countably many sets of measure zero has measure zero, but how about uncountably many of them? – TH000 Apr 25 '15 at 13:47