# How do we define this choice function using mathematical notation?

Suppose we have $\omega_1,\omega_2,\omega_3\in\mathbb{N}$. Then for every $\omega_2\in\mathbb{N}$ as $\omega_2\to\infty$ we want all $\left\{G_n\right\}\in\mathcal{F}(A)$ where for all $\left\{H_n\right\}\in\mathcal{F}(A)$ we have $\lim\limits_{\omega_2\to\infty}\frac{\mathcal{P}\left(G_{\omega_1},G_{\omega_1}\cup H_{\omega_2}\right)}{\mathcal{P}\left(H_{w_2},G_{w_1}\cup H_{w_2}\right)}=\inf\left\{|1-s|:s=\lim\limits_{\omega_2\to\infty}\frac{\mathcal{P}\left(G_{\omega_{3}},G_{\omega_3}\cup H_{\omega_2}\right)}{\mathcal{P}\left(H_{\omega_2},G_{\omega_3}\cup H_{\omega_2}\right)},s<1\right\}$, and for all $\left\{H_n\right\}\in\mathcal{F}(A)$ we have $\lim\limits_{\omega_2\to\infty}\frac{\sigma\left\{G_{\omega_1}\right\}}{\sigma\left\{H_{\omega_2}\right\}}\le 1$

I gave background info in the attatchment below. See section 3.1 , 3.7, 3.8

• Martin

This question is quite vague. And it uses the measure P which is not well defined. I guess you are looking to define the average on countable sets using this tool. Is this the case?

• Bharathk98

If we can make P well-defined we can define the average on uncountable sets but I could be wrong. Currently, I wish to convert the statement into a definition.

• Bharathk98

I extended the due date to 31 days so you can answer the previous question, then answer the second question.

• Bharathk98

How come you say measure P is not well-defined? I defined an inner measure and outer measure which form a sigma algebra. Does this work?

• Martin

The properties you listed for P does not uniquely determine it. I have not checked your attempt at the definition yet. I will write about it in my solution to the other problem.

• Martin

The extended due date is fine. But the question is still too vague (more than the previous one). I can try to make it more precise and then answer it, if you are ok with this. There may not be a unique solution, or any solution at all, to what you have in mind; and this depends on how the question is (re)stated in a precise mathematical form. Also, the open nature of the question requires more than usual work, and the current price is not sufficient for that.

• Cmartman

+1^

• Bharathk98

If the properties stated for P does not uniquely determine it what properties can be used to determine a unique P? Also how much more money should I pay?

• Bharathk98

I was hoping if my choice function does not determine a unique P there's a choice function that can be used. (And I hope to apply the result to examples in the previous question.)

• Martin

I am going to use the solution to the previous question and try to construct the choice process you are looking for.

• Bharathk98

Martin, I send another document titled Choosing For Countable A. It gives what the choice function should look like when A is countable.

• Bharathk98

Martin, I sent one more document, this time I'm choosing for general A using Generalized Family of Hausdorff Measures.

• Martin

Ok, I will work with the new documents.

• Bharathk98

Martin, I will try not to upload any more documents. Here is the last one. (I tried explaining with words but a user at math stack exchange couldn't understand.)

• Bharathk98

I keep making edits on Math Stack Exchange but a user keeps stating they can't understand my definitions. Here is the link: https://math.stackexchange.com/questions/4213941/for-countably-infinite-a-can-we-find-a-unique-left-f-n-right-that-gives ...............

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• Martin

I have written the answer for countable sets mainly because the notion of limit for an uncountable "sequence" of real numbers does not make much sense. However the basic problem is also present in that case, specially since an uncountable set of real numbers has infinitely many limit points.

• Bharathk98

I accidentally accepted it, I was looking for a choice function for a subsequence to the average.

• Bharathk98

It’s alright. Thanks for the effort.

• Martin

You're welcome. Let me know if you have questions about it.