Talk:Vitali Set Existence Theorem

Theorem Names
The Wikipedia article of Vitali's Theorem is called '''Vitali Convergence Theorem. ''' This name is already in use here as Vitali's Convergence Theorem, which is about holomorphic mappings in complex function theory.

The set of real numbers which is not Lebesgue measurable which is shown to exist in the Vitali Theorem is referred to in my source (Ernst Hansen : Sandsynlighedregning på målteoretisk grundlag, 2001) as a Vitali set. My source calls the theorem the Vitali Paradox, as the naive assumption that every subset of $\R$ is measurable leads to a paradox. Wikipedia uses Vitali Theorem in its article about Vitali Sets. The German Wikipedia uses (translated) Vitali's Theorem (Measure Theory) , but I find this ambiguous, as Vitali's Theorem is also a statement in measure theory.

If we need a unique name for the Vitali Theorem, my suggestion would be Vitali Set Existence Theorem, but this is not backed up by sources. --Anghel (talk) 22:08, 8 December 2022 (UTC)


 * I went with the latter. I've added a linguistic note confessing to the fact that this name has been made up. Please feel free to contribute to that item. --prime mover (talk) 22:51, 21 August 2023 (UTC)

Mistake in Jech
I believe this is a misprint:
 * $\ds \map \mu {\closedint 0 1} \ge \map \mu {\bigcup \set {M_r: r \in \Q \land 0 \le r \le 1} }$

The correct inequality is:
 * $\ds \map \mu {\closedint 0 2} \ge \map \mu {\bigcup \set {M_r: r \in \Q \land 0 \le r \le 1} }$

I don't see any reason for the former, while the latter is clear, because from $M \subseteq \mathbb I$ follows:
 * $\forall r \in \Q : 0 \le r \le 1 \implies \closedint 0 2 \supseteq M_r$

And so:
 * $\ds \closedint 0 2 \supseteq \bigcup \set {M_r: r \in \Q \land 0 \le r \le 1}$

--Usagiop (talk) 17:00, 22 August 2023 (UTC)


 * Why does $\ds \map \mu {\closedint 0 1} < \map \mu {\bigcup \set {M_r: r \in \Q \land 0 \le r \le 1} }$? I don't understand. --prime mover (talk) 17:25, 22 August 2023 (UTC)


 * I did not say that. I am claiming:
 * $\ds \map \mu {\closedint 0 2} \ge \map \mu {\bigcup \set {M_r: r \in \Q \land 0 \le r \le 1} }$
 * This follows, in view of the monotonicity of $\mu$, from the fact:
 * $\ds \closedint 0 2 \supseteq \bigcup \set {M_r: r \in \Q \land 0 \le r \le 1}$
 * --Usagiop (talk) 18:36, 22 August 2023 (UTC)


 * But if it is not the case that $\ds \map \mu {\closedint 0 1} < \map \mu {\bigcup \set {M_r: r \in \Q \land 0 \le r \le 1} }$, then there is no problem with the statement that $\ds \map \mu {\closedint 0 1} \ge \map \mu {\bigcup \set {M_r: r \in \Q \land 0 \le r \le 1} }$. Sorry, I just want to know why that is a mistake, otherwise I cannot report it as such. --prime mover (talk) 23:45, 22 August 2023 (UTC)


 * OK, I understood the confusing point. I am claiming that, in that context, nobody knows whether the following is true or false:
 * $\ds \map \mu {\closedint 0 1} \ge \map \mu {\bigcup \set {M_r: r \in \Q \land 0 \le r \le 1} }$
 * There is just no clue for that. But, what we can derive from the construction of $\sequence {M_r}_r$ is:
 * $\ds \map \mu {\closedint 0 2} \ge \map \mu {\bigcup \set {M_r: r \in \Q \land 0 \le r \le 1} }$
 * which is sufficient for this proof. --Usagiop (talk) 17:24, 23 August 2023 (UTC)