Measure, bewildered by notation

[holomorphic]

Homework Statement

\mathcal{A} is a \sigma-algebra, A_n,A\in\mathcal{A}.

Prove that if A_n\uparrow A in \mathcal{A} (i.e., A_1\subseteq A_2 \subseteq ... and \bigcup_{n=1}^\infty A_n = A), then \mu (A_n) \uparrow \mu (A)

The Attempt at a Solution

The up-arrow notation is defined on these sets, but I have no idea what it means in the case of \mu (A_n). I've seen the proof of theorem 11.3 in Rudin Principles of... but I'm having trouble understanding what's going on in this problem.

I'm mostly bewildered by the notation, which isn't in Rudin and hasn't been used in class.

 

[LCKurtz]

Remember that μ(A_n) is just a nonnegative number. It means {μ(A_n)} is an increasing sequence converging to μ(A). Since the sets are getting bigger, it makes sense doesn't it?

 

[holomorphic]

Remember that μ(A_n) is just a nonnegative number. It means {μ(A_n)} is an increasing sequence converging to μ(A). Since the sets are getting bigger, it makes sense doesn't it?

I realized it was as simple as that about 15 minutes after I posted. But I still don't understand the advantage of his notation--in the end, I'm just showing \mu(A_n)\rightarrow\mu(A) as n\rightarrow\infty.