0
$\begingroup$

I am following the proof for closed sets being measurable by in Stein and Shakarchi (2009).

Consider $Q_j^s$ being almost disjoint closed cubes. Why is it the case the authors claim the LHS also holds in the limit as $N$ tends to $\infty$? I think by that, the authors mean the tail of this sum is something negligible, so we can throw away? But is it true in this case? If not, why is their claim valid?

$\sum\limits_{j=1}^Nm_*(Q_j)\leq\epsilon\rightarrow\lim\limits_{N\rightarrow\infty}\sum\limits_{j=1}^Nm_*(Q_j)\leq\epsilon$

Reference: $\textit{Real Analysis: Measure Theory, Integration, and Hilbert Spaces}$. Elias M. Stein, Rami Shakarchi. Princeton University Press, 2009.

$\endgroup$

1 Answer 1

4
$\begingroup$

It's just the fact that if you have some convergent sequence $(a_n)_{n\in\mathbb{N}}$ such that $a_n\leq \varepsilon$ for every $n$, then $\lim_{n\to\infty}a_n\leq \varepsilon$ (say, because $(-\infty,\varepsilon]$ is a closed set).

$\endgroup$
3
  • $\begingroup$ Thank you!!!!!! $\endgroup$ Commented Sep 19, 2019 at 18:09
  • $\begingroup$ Why does the fact that $(-\infty,\varepsilon)$ being a closed set justify that? $\endgroup$ Commented Sep 19, 2019 at 18:11
  • $\begingroup$ A subset $A$ of a metric space $M$ is closed if and only if for every sequence $(a_n)_{n\in\mathbb{N}}\subseteq A,$ which is convergent in $M$, we have $\lim_{n\to\infty} a_n\in A$. $\endgroup$ Commented Sep 20, 2019 at 10:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .