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I'm reading Real Analysis by Yeh and I'm looking at the following

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(the rest of the proof for part a on the next page essentially just uses telescoping series)

Before reading Yeh's proof I had what I feel is a simpler way to prove part a. My question is whether it still works or if Yeh's way is necessary (for instance, he split the $\mu(E_n)< \infty \; \forall n$ and the $\mu(E_n)= \infty$ for some n cases). In my notes I like to be brief so if a shorter way works I'd like to write that out instead :)
My idea: pick a sequence $(F_n:n \in \mathbb{N})$ in $\frak{A}$ where $F_1=E_1$, $F_n=E_n \setminus ( \cup_{k=1}^{n-1} E_k)$. Then a previous lemma shows $$\cup_{k=1}^n E_k = \cup_{k=1}^n F_k \text{ for } n=1,2,\ldots, \infty$$ Then by the countable additivity of $\mu$: \begin{split} \mu( \cup_{k=1}^\infty E_k) & = \mu( \cup_{k=1}^\infty F_k)= \sum_{k=1}^\infty \mu( F_k)\\ &= \lim_{n \to \infty} \sum_{k=1}^\infty \mu( F_k)= \lim_{n \to \infty} \mu( \cup_{k=1}^n F_k)\\ &= \lim_{n \to \infty} \mu( \cup_{k=1}^n E_k)= \lim_{n \to \infty} \mu( E_n) \end{split} Thanks in advance for any help!

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  • $\begingroup$ Your proof is the same as that in the book. $\endgroup$ – xpaul Feb 27 '17 at 22:05
  • $\begingroup$ @xpaul Thanks for the confirmation. Is it still a good idea to note the $\mu(E_n)= \infty$ case for some n separately? $\endgroup$ – Jason Feb 27 '17 at 22:09
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Your proof is the same (if you note that $\bigcup_1^{n-1}E_k=E_{n-1}$).

The reason to write the case where $\mu(E_{n_0})=\infty$ separately, is to avoid writing sums of infinities. I think that it doesn't hurt in this case, but it can be a source of mistakes.

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  • $\begingroup$ Thank you for the confirmation :) $\endgroup$ – Jason Feb 27 '17 at 22:27

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