Measure, bewildered by notation

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SUMMARY

The discussion centers on the proof of the theorem regarding the convergence of measures in a σ-algebra, specifically when a sequence of sets \( A_n \) increases to a set \( A \) within the σ-algebra \( \mathcal{A} \). It is established that if \( A_n \uparrow A \), then the measures \( \mu(A_n) \) converge to \( \mu(A) \). The notation \( \uparrow \) indicates that \( \mu(A_n) \) forms an increasing sequence of nonnegative numbers, ultimately converging to \( \mu(A) \). The confusion arises from the notation used, which is not covered in the referenced text, "Rudin Principles of Mathematical Analysis".

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Homework Statement


[tex]\mathcal{A}[/tex] is a [tex]\sigma[/tex]-algebra, [tex]A_n,A\in\mathcal{A}[/tex].

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

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 [tex]\mu (A_n)[/tex]. 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.
 
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Remember that μ(An) is just a nonnegative number. It means {μ(An)} is an increasing sequence converging to μ(A). Since the sets are getting bigger, it makes sense doesn't it?
 
LCKurtz said:
Remember that μ(An) is just a nonnegative number. It means {μ(An)} 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 [tex]\mu(A_n)\rightarrow\mu(A)[/tex] as [tex]n\rightarrow\infty[/tex].
 

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