Let (X, M) be a measurable space and f a non negative real function, prove that μ(A) = sum x∈A f(x) is a measure. I'm trying to prove that μ(U from n=1 to inf An)= sum from n=1 to inf (μ(An)). But I don't know how I can continue, I have μ(U from n=1 to inf An)= sum (xE (U from 1 to inf An)) f(x)
Thanks
Presumably the $A_n$ are disjoint. $$\mu \left(\bigcup_{n =1}^{\infty} A_n\right) = \sum_{x \in \bigcup_{n=1}^\infty A_n} f(x) \overset{*}{=} \sum_{n=1}^\infty \sum_{x \in A_n} f(x) = \sum_{n=1}^\infty \mu(A_n)$$
The main step is to show the starred equality. In plain words, it is just stating that taking a sum over the elements of $\bigcup_{n=1}^\infty A_n$ is the same as taking the sum of elements in each set $A_n$, and then summing over all the sets; this relies on the fact that the $A_n$ are disjoint.