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:

Isn't a measure subadditive, or something, such that the measure of the union of subsets of a set is less than or equal to the measure of the set itself? Can't be bothered to go look for it now, I just thought it was a theorem, or a definition of a measure, or something. Not my area of expertise, it bores me. --prime mover (talk) 17:42, 23 August 2023 (UTC)

Yes, as I wrote above,it is clear that:

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

hence, the inequality of measures, too. And, what is in no way clear is, whether:

$\ds \closedint 0 1 \supseteq \bigcup \set {M_r: r \in \Q \land 0 \le r \le 1}$ or not

So, I want to correct $\closedint 0 1$ to $\closedint 0 2$. Is it OK for you that I fix it and add some lines of explanations how this follows? --Usagiop (talk) 18:52, 23 August 2023 (UTC)

No it's not, because I hve yet to be convinced that it *is* wrong as it is.

If you like I'll raise it on StinkExchange. Sorry but I trust Jech knows what he's talking about. And unless I've drastically misinterpreted the notation, it's intrinsically obvious that it's true. --prime mover (talk) 23:31, 23 August 2023 (UTC)

Then please ask on StackExchange, if you need more inputs. I am sure that this proof is invalid. It is not about the correctness of that inequality, that appears in the middle of an indirect proof. You need to make a contradiction step by step, where each step must be logically reasoned. I am pointing out that one step cannot be reasoned. --Usagiop (talk) 16:55, 24 August 2023 (UTC)

My take on this is that you're wrong. MSE will soon demonstrate this. --prime mover (talk) 17:08, 24 August 2023 (UTC)

... except nobody on that forum seems to care enough to answer. Have to do it myself. --prime mover (talk) 21:40, 28 August 2023 (UTC)

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

which is sufficient for this proof. --Usagiop (talk) 17:24, 23 August 2023 (UTC)

E.g. if we pick $1/2$ from $\eqclass {1/2} \sim$, we have $3/2 \in M_1$. But since we pick a representative from $\closedint 0 1$ and add a non-negative rational between $0$ and $1$ we get something inbetween $0$ and $2$. Caliburn (talk) 21:19, 23 August 2023 (UTC)

OK, you are right, it is in fact clear, that the following is in general wrong:

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

Thus it is in no way clear, whether:

$\ds \map \mu {\closedint 0 1} \ge \map \mu {\bigcup \set {M_r: r \in \Q \land 0 \le r \le 1} }$ or not

because in this context there is no possibility to prove this. --Usagiop (talk) 23:09, 23 August 2023 (UTC)

Now I can derive this:

$\ds \map \mu {\closedint 0 1} \ge \map \mu {\bigcup \set {M_r: r \in \Q \land 0 \le r \le 1} }$

Consider:

$F_r := \paren {M_r \cap {\hointr 0 1} } \cup \paren {\paren {M_r \cap {\hointr 1 2} } - 1}$

for $r \in \closedint 0 1 \cap \Q$.

Then $\sequence {F_r}_{r \in \hointr 0 1 \cap \Q}$ are disjoint subsets of $\closedint 0 1$ in view of the particular construction of $M$.

Thus:

$\ds \closedint 0 1 \supseteq \bigcup \set {F_r: r \in \Q \land 0 \le r \le 1}$

Furthermore:

\(\ds \map \mu {F_r}\)

\(=\)

\(\ds \map \mu {M_r \cap {\hointr 0 1} } + \map \mu {\paren {M_r \cap {\hointr 1 2} } - 1 }\)

\(\ds \)

\(=\)

\(\ds \map \mu {M_r \cap {\hointr 0 1} } + \map \mu {M_r \cap {\hointr 1 2} }\)

\(\ds \)

\(=\)

\(\ds \map \mu {M_r \cap {\hointr 0 2} }\)

\(\ds \)

\(=\)

\(\ds \map \mu {M_r}\)

as $\map \mu {\set 2} = 0$

So:

\(\ds \map \mu {\closedint 0 1}\)

\(\ge\)

\(\ds \map \mu {\bigcup \set {F_r: r \in \Q \land 0 \le r \le 1} }\)

\(\ds \)

\(=\)

\(\ds \sum_{r \in \Q \land 0 \le r \le 1 } \map \mu {F_r}\)

\(\ds \)

\(=\)

\(\ds \sum_{r \in \Q \land 0 \le r \le 1 } \map \mu {M_r}\)

\(\ds \)

\(=\)

\(\ds \map \mu {\bigcup \set {M_r: r \in \Q \land 0 \le r \le 1} }\)

However, I don't think this was the author's intention. --Usagiop (talk) 19:10, 29 August 2023 (UTC)

Translation invariance --prime mover (talk) 19:42, 29 August 2023 (UTC)

What exactly do you mean? I don't see how the translation invariance helps here. --Usagiop (talk) 20:20, 29 August 2023 (UTC)

The intervals in question are $1$ long, are they not? So shuffle them down so they coincide with $\closedint 0 1$ and the job's done. --prime mover (talk) 21:24, 29 August 2023 (UTC)

I guess your idea is to shuffle individual $M_r$s inside $\map \mu {\bigcup \set {M_r: r \in \Q \land 0 \le r \le 1} }$. I can't do this. You can only shift the whole set like $\map \mu {A + r} = \map \mu A$. My proof above strongly uses the curious structure of $M$, in order for $F_r$s to still be disjoint. Such thing is in general wrong and really curious. --Usagiop (talk) 23:10, 29 August 2023 (UTC)

There's one remaining puzzle: why the proof of this theorem is on the talk page. --prime mover (talk) 05:09, 30 August 2023 (UTC)