# Is it possible for someone who better understand my question to re-edit it?

I have completely redone my question. I doubt users in Math Stack Exchange will answer. If the question is unclear/off-topic but you understand, and know how to improve it, is it possible to re-edit the question?

The only person who helped is @WillieWong and he is on MathOverflow. He understood what I meant by reading the original questions. He might be able to answer this last one but in one of the comments stated: "Now I am really done". I am not sure if he understands everything I require.

I apologize that my posts are so hard to read. I apologize my questions lack rigor but they can become rigorous with help. They are getting better every time. It just seems no one willing to continue helping.

Be sure to check all of @WillieWong's comments and answers here and here.

Edit: I would also like to know why this is unclear.

• In general please consider that lack of brevity can discourage most readers to go into your question. – YCor May 6 at 19:52
• @YCor But why exactly is my question unclear. No one is stating why? – Arbuja May 6 at 19:54
• It looks to me like you are trying to define a type of integral or outer measure, using Lesbesgue measure to define your M. As I read it, M does not have the properties you state, specifically (3). Rather than attempt a rewriting your question, try looking at a book on measure theory (perhaps Halmos's book) to see about defining objects having a countable additive property, and then see if you can make a new sum that has it. Gerhard "Length Is An Issue Here" Paseman, 2020.05.06. – Gerhard Paseman May 6 at 21:10
• @GerhardPaseman Could you give an example proving my properties wrong. Take for instance $M(A_1\cup A_2)$. This is $\frac{\lambda(A_1\cup A_2)}{\lambda(A)}=\frac{\lambda(A_1)+\lambda(A_2)}{\lambda(A)}=\frac{\lambda(A_1)}{\lambda(A)}+\frac{\lambda(A_2)}{\lambda(A)}=M(A_1)+M(A_2)$ – Arbuja May 6 at 21:20
• @GerhardPaseman For property $(1)$ if $A$ is countable then $S=A$ is countable so $M(A)=1$. If $A$ is uncountable and $\lambda(A)=0$ then since $S=A$ Is countable, we have $M(A)=1$. If $\lambda(A)>0$, then $\lambda(A)/\lambda(A)=1$ – Arbuja May 6 at 21:37
• If A is measure 0 and both A and S are uncountable, then M(S) is 1 by your definition. Then taking A to be a Cantor set and S to be "the left half", I get M(S) + M(A\S)=2. I stand by my earlier comment. Gerhard "Cantor Sets Make Good Tests" Paseman, 2020.05.06. – Gerhard Paseman May 6 at 21:58
• @GerhardPaseman Apologies for my earlier comment (I deleted it). When $\lambda(A)>0$, $M$ is countably additive. However, when $\lambda(A)=0$, $M$ is not countably additive. In this case, only when $A$ is uncountable can we remove points from $A$. By definition if the removed points are countable, then $M$ of those countable points is zero. The remaining set substituted into $M$ equals one. – Arbuja May 6 at 22:38
• OK. I think there is a basic misunderstanding going on, which might explain why people can't tell you what is wrong in a way that is useful to you. Even when the measure of A is greater than zero, your M is additive only when the sets involved are Lesbesgue measurable and disjoint. You don't have such a restriction, and you need to make one. I recommend Royden's text Real Analysis and its treatment as well on outer measure. I don't think I can help you further. Gerhard "Averages Do Only So Much" Paseman, 2020.05.06. – Gerhard Paseman May 6 at 23:00
• You were asked, Arbuja, to stop asking so many meta questions on math.stackexchange (math.meta.stackexchange.com/questions/31648/…), and your reaction is to post a meta question on MathOverflow. – Gerry Myerson May 7 at 1:02
• @GerryMyerson I apologize this will be the last question. If not you can downvote. – Arbuja May 7 at 1:18
• I can downvote whether this is the last question or not. I don't expect to, but I can. – Gerry Myerson May 7 at 1:19
• This is a site for research mathematics, and you are not well trained in undergraduate mathematics, so people's willingness to help is limited. If you want to redefine everything, come back when you have more training! Or join us now in areas not requiring so much theoretical background, where you can contribute with more concrete topics. – Matt F. May 7 at 2:54
• @MattF. My worry is if I can’t pass higher level courses, my idea will never see light. Not everyone gets through mathematics. It’s more than doing homework. Lots of students drop out. I have a non-rigorous, simplistic way of absorbing information. I can’t absorb measure theory articles. I have to wait till fall (or spring) to get back to college. I want this done but I don’t want to sacrifice so much. – Arbuja May 7 at 3:12
• @Arbuja You seem to want people who can do mathematics to make your dreams into something substantial when you are not good at doing mathematics. If you have a "non-rigorous, simplistic way of absorbing information" then may I firmly but kindly suggest that you find another outlet for your intellectual enthusiasm. There is more to mathematics than "ideas" and "insights", contrary to a lot of well-meaning popularization; learning techniques, and getting a good command of techniques, is essential. – Yemon Choi May 11 at 1:23
• The academic world really is cold. So many ideas are wasted never to see light. How do you know what is trash and what is not. I am not done till someone understands what I am trying to say. I am not giving up. I am not giving up. I am not giving up. – Arbuja May 11 at 2:11