Talk:Lusin's theorem

General statement wrong
what does sup |f(x)| mean, if f takes values in an arbitrary 2nd countable topological space ?! -- — Preceding unsigned comment added by 62.47.39.14 (talk) 17:44, 17 February 2020 (UTC)

Untitled
I am relatively novice student of Measure Theory, but I am confused by the moving of Lusin's theorem to Luzin's theorem as every reference I can find refers to "Lusin's theorem". Could you explain the reason for moving it to "Luzin's theorem"? Thanks.


 * these are just different English spellings of the same Russian name (are there other Lusin's who worked/works in related areas?), e.g. Lusin/Luzin, Tychonoff/Tihonoff, etc. Mct mht 22:20, 29 April 2006 (UTC)

To clarify: there are at least two types of romanizations current, one that is adapted to French speakers (probably more common before 1940), and one that would be used now for English speakers. These days we should definitely use 'Luzin'. The only exceptions really should be for things like Tchaikovsky (not Chaikovskii, which no one would recognise). Charles Matthews 19:52, 30 April 2006 (UTC)

Lusin-Menchoff theorem
I had a vague idea that Lusin-Menchoff theorem, which is requested, is just another name for this theorem. Can anyone confirm or deny this? Via strass 17:15, 19 July 2006 (UTC)

The proof does not prove the full theorem
The hypothesis in Lusin's theorem is that the function be measurable on [a,b]. But the proof given assumes -- without any comment! -- that f is in L^1, and apparently does not work in the general case. To my mind the easiest fix is to remove the proof (there is already a link to a proof available online), so I have done so. But perhaps someone else will want to replace it with a complete proof.... Plclark 12:20, 7 October 2007 (UTC)Plclark

Okay, the proof can be (and has been) rather easily fixed. Good. (But there was no point in having a 90% complete proof...)

Also
What is meant by "another" form of Littlewood's second principle? What was the first form? Plclark 12:22, 7 October 2007 (UTC)Plclark

Is the converse true?
It seems like it should be. Timhoooey 00:56, 15 October 2007 (UTC)

Yes, it is true. Plclark 16:16, 15 October 2007 (UTC)Plclark

Spelling of Lusin's name
In the original Comptes Rendus paper where he proves his theorem, the author spells his last name as "Lusin". Similarly, most references I have seen are to "Lusin's Theorem." Might it not be better to have the official name be "Lusin's Theorem", even if the link is to the mathematician "Luzin"? Plclark 16:16, 15 October 2007 (UTC)Plclark


 * I agree. The only spelling I have ever seen is Lusin. Timhoooey 01:19, 16 October 2007 (UTC)


 * I think that was unnecessary. We don't have to use the common name in every case. As is explained above, "Lusin" is what you'd expect in a French journal. This is like Tychonoff versus Tikohov, and it is antiquarian to insist on the first. Charles Matthews 21:57, 20 October 2007 (UTC)

I believe you that the spelling of the author's name in the paper in question is contingent on the time and place of publication. The question is: given that most literature up until the present day continues to refer to "Lusin's theorem", should the spelling be changed to be consistent with how his descendants would have their name rendered in English today? It might well be antiquarian to stick with the same old spelling that everyone knows. Whether being "antiquarian" is something to avoid in mathematics, I'm not sure: we are after all studying the work of someone who lived one hundred years ago. There are a lot of names in mathematics that make a lot less sense to contemporary eyes than "Lusin": e.g. since we think of the x-coordinate as coming before the y-coordinate in an ordered pair, it would seem to make more sense to reverse "sine" and "cosine", and even given this, who decided that the reciprocal of the sine would be called the cosecant -- medieval category theorists?!? But we stick with this terminology because it's the terminology that everyone is familiar with and it is not a serious impediment to the understanding of the subject. If you change the terminology to "Luzin's Theorem", then any reader who has heard it the other way will find it jarring. There's nothing wrong with "Lusin's Theorem", so I think it's easiest not to change it. By the way, everyone still does write "Tychonoff", including the wikipedia entry on Tychonoff's theorem (which by chance I rewrote, but not for linguistic reasons). 72.152.92.55 (talk) 09:35, 29 November 2007 (UTC)Plclark

Theorem not proven
I may be missing something here, but to prove the theorem you need to find a compact set E, and Egorov's Theorem only provides us with a measurable set of arbitrarily small measure off of which the functions converge uniformly. This measurable set is not necessarily open. So I think there is a little more fiddling that needs to be done to get a compact set.Robomb162 (talk) 04:48, 18 October 2008 (UTC)

Is the formulation correct?
I'm probably missing something obvious, but I can't see how you can restrict the function $$1_{\mathbb{Q}}$$ on [0,1] to a compact non-null subset E of [0,1] such that the result is continuous. --SuneJ (talk) 20:37, 5 December 2009 (UTC)


 * I can't see one right now either, so it's probably some "ugly" set. It's surely there, though, just not intuitive. 92.237.83.191 (talk) 17:54, 20 April 2010 (UTC)


 * What about this set? Let $$\{x_n; n=1,2,\dots\}$$ be any enumeration of Q. Put $$G_n=(x_n-\varepsilon/2^n,x_n+\varepsilon/2^n)$$ and $$E:=[0,1]\setminus\bigcup_{n=1}^\infty G_n$$. --Kompik (talk) 11:04, 2 October 2010 (UTC)


 * Very nice. This should be copied into the article as an example, as it is marvelously unintuitive. 67.198.37.16 (talk) 06:38, 6 December 2023 (UTC)
 * And so I did copy it into the article. It seemed entirely appropriate. We'll see if anyone complains. 67.198.37.16 (talk) 07:06, 6 December 2023 (UTC)
 * $$E:=[0,1]\setminus\bigcup_{n=1}^\infty G_n$$ does not work since $$\mathbb{Q}$$ is dense in $$[0,1]$$ for every $$G_n=(x_n-\epsilon/2^n, x_n+\epsilon/2^n)$$ there is an $$x_k \in G_n$$ with $$x_n \neq x_k$$. Then $$\{G_n\}_{n\in \mathbb{N}}$$ would be a cover of $$[0,1]$$ and $$ \mu (E) = 0$$. Am I missing something maybe?
 * Blueee A (talk) 18:23, 24 April 2024 (UTC)

Assessment comment
Substituted at 02:18, 5 May 2016 (UTC)