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I have a question regarding how to show the collection of sets that satisfies the Outer Measurability (i.e. being Caratheodory Measurable, say denoted the set as $M^{*}$) is a sigma algebra.

The Caratheodory Measurability condition is this:

a set is $\mu^{*}$ measurable ($\mu^{*}$ measurable is same thing as being Caratheordory Measurable) if

$\mu^{*}(B) = \mu^{*}(B \cap M) + \mu^{*}(B \cap M^{c})$ for all $B\subseteq S$, where $M$ is a set in $M^{*}$, i.e. $M \in M^{*}$.

Basically $M^{*}$ is a sigma algebra (and I am trying to understand the proof that shows it is a sigma algebra).

The proof that I read has all the steps (on Page 8 where it says Part III in this document here https://www.ma.utexas.edu/users/gordanz/notes/measures.pdf) . It shows first $M^{*}$ is an algebra, and try to show $M^{*}$ is closed under countable disjoint union, so that $M^{*}$ actually satisfies the condition of being sigma-algebra.

The part that confuses me is here:

$\mu^{*}(B) \geq \mu^{*}(B\cap M^{c}) + \sum_{k \in N} \mu^{*}(B\cap M_k)$ $\geq \mu^{*}(B\cap M^{c}) + \mu^{*}(\bigcup\limits_{k} (B \cap M_k)) + \mu^{*}(B \cap M^{c})$.

I don't quite understand how the author goes from the first inequality to the second inequality.

The $M_k$ are assumed to be pair-wise disjoint and being Caratheodory-measurable, so each $M_k$ satisfies the condition of the above formula.

Could someone gives me some hints as to how the inequalities is arrived at?

Thank you for your time

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I think that there is a typo in the lectures notes. In the preceding lines, the author shows that, for every, $n \in \mathbb{N}$, $$ \mu^*(B) \geq \mu^{*}(B \cap M^c) + \sum_{k=1}^{n}{\mu^*(B \cap M_k)} $$ Conclude from the confrontation lemma that $$ \mu^*(B) \geq \mu^{*}(B \cap M^c) + \sum_{k=1}^{\infty}{\mu^*(B \cap M_k)} $$ Finally, it follows from the subadditivity of $\mu^*$ (page 7) that $\sum_{k=1}^{\infty}\mu^*(B \cap M_k) \geq \mu^*(\cup_{k}(B \cap M_k))$. Therefore, \begin{align*} \mu^*(B) &\geq \mu^{*}(B \cap M^c) + \mu^*(\cup_{k}(B \cap M_k)) \\ &= \mu^{*}(B \cap M^c) + \mu^*(B \cap (\cup_{k}M_k)) \\ &= \mu^*(B \cap M^c) + \mu^*(B \cap M) \end{align*}

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  • $\begingroup$ Hi madprop, thanks a lot for your detailed steps of explanations. But I just have one more questions which always bothers me also in this proof. My question is this: is it always true that if an inequality holds for some n, then it automatically holds for all n in $\mathbb{N}$? Because the author shows that for a particular $n\in$ \mathbb{N}, the inequality $\mu^{*}(B) \geq \sum_{k=1}^{n} \mu^{*}(B \cap M_k) + \mu^{*}(B\cap M_c)$ is true. Then AUTOMATICALLY says $\mu^{*}(B) \geq \sum_{k \in \mathbb{N}} \mu^{*}(B \cap M_k) + \mu^{*}(B\cap M_c)$. $\endgroup$ – john_w Nov 7 '17 at 2:47
  • $\begingroup$ Because the entries on the right of the inequality are all minimum of zero or greater. $\mathbb{N}$ includes a lot lot more terms than say n = 100 or n = 7000, or n = 3500000. My question is, is it IN GENERAL always true that if I can shows an inequality that seems to hold for certain n, then it will automatically holds for all $n \in \mathbb{N}$, and hence can have the statement that the inequality holds for all $k \in \mathbb{N}$. Because in your summation from k = 1 to $k=\infty$, the infinity in the upper sum seems to be because you assumed the inequality holds for all $k \in \mathbb{N}$. $\endgroup$ – john_w Nov 7 '17 at 2:58
  • $\begingroup$ basically I want to know if during a proof, if we can show an inequality that can holds for certain n, then does it automatically means it holds for all the n in the natural number? because the set of Natural Number is a much bigger set. And when the inequality involves non-negative terms indexed by the natural number, it seems if we keep adding something positive in the summation, and it bothers me a little whether the inequality still holds, if we keep adding something positive on the right hand side, but without changing the left hand side. Thank you. $\endgroup$ – john_w Nov 7 '17 at 3:04
  • $\begingroup$ If, for every $n \in \mathbb{N}$, $a \geq b_{n}$ and $\lim_{n \rightarrow \infty} b_{n}$ is defined, then $a \geq \lim_{n \rightarrow \infty} b_{n}$. $\endgroup$ – madprob Nov 7 '17 at 3:46
  • $\begingroup$ thanks a lot madprob. Your answer is very clear, and thanks for pointing out the typo. $\endgroup$ – john_w Nov 9 '17 at 3:24

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