Sigma algebra proof in measure theory

[Redsummers]

Homework Statement

Let $$\mathcal{A}$$ be σ-algebra over a set $$X$$, and μ a measure in $$\mathcal{A}$$.

Let $$A_{n} \in \mathcal{A}$$ with $$\sum_{n=1}^{\inf} \mu(A_{n})< \inf$$
Show that this implies
μ ({$$x \in X$$ : $$x \in A_n$$ for infinitely many n}) = 0 .

The Attempt at a Solution

I don't even see how is the measure 0 if the measure of all A_n is a finite number...
I guess that the measure we want to show μ=0 is related with some kind of topology that makes it 0. What do they mean by

μ ({$$x \in X$$ : $$x \in A_n$$ for infinitely many n}) ?

"The measure of the points x in the set X s.t. x is an element of A_n (which the measure of A_n is finite) for infinitely many n."

I can't see how is this measure zero, so if I don't have a minimum intution I can't even attack the problem. Any suggestions?PS: which was the way to write LaTeX so that I can write something along a non-LaTeX text and still seem of the same size? i.e. how to make LaTeX text smaller.

 

[Redsummers]

Ah, nevermind. I got it.
You just prove it by defining A_n as a ring of a circle B_n. That is, B_n is going to converge, so will A_n. (And n goes up to infinity in the union.) Hope that's on the right track.