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


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*}

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.