Talk:Radon–Nikodym theorem

message: The stement is vague 
"The Radon–Nikodym theorem essentially states that, under certain conditions, any measure ν can be expressed in this way with respect to another measure μ on the same space."
Certain conditions??
The second measure is said to be absolutely continuous with respect to the first one if the second measure of a set is always zero when the first one is. The radon nykodym theorem states that any measure is the sum of an absolutely continuous part and a remainder called the singular part. The absolutely continuous part always has a density with respect to the first measure.  — Preceding unsigned comment added by Izmirlig (talk • contribs) 04:29, 12 May 2023 (UTC)

message: So presumably not just a function in the statement, but integrable?
Charles Matthews 11:16, 11 Oct 2003 (UTC)
replies: text: Yes, I'll make this explicit in the article. Pete 12:39, 11 Oct 2003 (UTC)

text: E_Q(X) = E_P\left( \frac{dQ}{dP} X \right) 
(EP and EQ switched?). Not sure though, else I would change it myself. DrZ 13:50, 16 Mar 2004 (UTC)

text: I changed the last bit, presumably A is supposed to be same arbitrary measurable set as from the first formula? Pete/Pcb21 (talk) 14:10, 16 Mar 2004 (UTC)
replies: text: But what is the expectation operator applied to a set supposed to mean? DrZ 14:50, 16 Mar 2004 (UTC)
replies: text: That was wrong. I think now it is right. There is no expectation operator applied to a set, so you were right in your concern. Pfortuny 15:15, 16 Mar 2004 (UTC)
thread_title: Untitled

message: heh, article claims theorem is from functinal analysis yet doesn't give a (more elegant) functional analytic proof. Mct mht 11:49, 10 June 2006 (UTC)
thread_title:  proof 

message: definition: ν <<< μ : for all e>0 exists d>0 for all A in algebra (μ(A)<d -> ν(A)<e)
lemma: (ν <<< μ) -> (ν << μ)
lemma: for sigma additive ν, μ on a sigma-algebra (ν << μ) -> (ν <<< μ)
definition: simple-function : finite linear combination of indicator functions of measurable sets.
there's also a theorem that states:
let ν <<< μ positive meaures on an algebra, let e>0 then exists a simple-function f s.t. ||ν-fdμ||<e.
with completeness of L1(μ), the regular Radon-Nikodym theorem follows. --itaj 00:22, 15 June 2006 (UTC)
replies: text: I don't understand what you are trying to say. But either way I don't think this belongs in this article, rather a new article containing this should be created. Oleg Alexandrov (talk) 03:07, 15 June 2006 (UTC)

text: what i meant was to discuss that theorem i described. to see if it's to be added here or maybe make another article. and the last comment says that this theorem together with the theorem that L^1 is complete can prove the regular radon-nikodym theorem. --itaj 20:59, 15 June 2006 (UTC)

text: what i was saying is this: the theorem i described is a version of radon-nikodym for an additive measure over an algebra (not sigma) and is weaker. and this one does not follow from regular radon nikodym (as far as i know, not in a canonical way). but the regular radon nikodym (for sigma-additive measures over sigma-algebra) does follow from this one (using this theorem and also using the theorem that L1 of sigma-additive measure over sigma-algebra is complete). --itaj 17:00, 14 February 2007 (UTC)
thread_title:  measures on an algebra (not sigma) 

message: The article doesn't explain why this should be.
Does it become particularly important when treating stochastic processes to work with tools defined for probability measures on arbitrary sets, rather than probability densities on real numbers?   Jheald 12:34, 4 March 2007 (UTC)
replies: text: I suspect so.  The Wiener process is used in financial mathematics, and certainly that's not a distribution on the real line. Michael Hardy 01:31, 5 March 2007 (UTC)
thread_title:  "Financial mathematics uses the theorem extensively" ? 

message: The article defines Renyi divergence of order α as
:: D_{\mathrm{\alpha}}(\mu\|\nu) = \frac{1}{1-\alpha} \log \left( \int_X  \left( \frac{d \mu}{d \nu} \right)^{1-\alpha} \; d\mu. \right) \!
I think the parametrisation is off. It should rather be:
:: D_{\mathrm{\alpha}}(\mu\|\nu) = \frac{1}{\alpha-1} \log \left( \int_X  \left( \frac{d \mu}{d \nu} \right)^{\alpha-1} \; d\mu. \right) \!
(Compare Renyi divergence.)
Ohjaek33 (talk) 12:43, 2 September 2009 (UTC)
thread_title:  Wrong definition of Renyi divergence 

message: In the "Radon-Nikodym derivative" section, the article says "\mu-integrable...function g on X such that..."
The function  g is only guaranteed to be \mu-measurable.  Consider for instance \mu=\nu= Lebesgue measure.  Then g=1 is not Lebesgue integrable on  \mathbb{R}^n.  I will change in a few days unless someone can explain or provide a reference.  By the way, the statement at the beginning of the article is correct.Paul Laroque (talk) 23:26, 13 March 2012 (UTC)
replies: text: Ah, but in the last sentence of that section, the setting is restricted to σ-finite μ and finite ν.  Your example with Lebesgue measure on the real line fails the second condition. Sullivan.t.j (talk) 21:02, 14 March 2012 (UTC)
thread_title:  Radon-Nikodym derivative is not in general integrable 

message: > If  is a Banach space and the generalization of the Radon–Nikodym theorem also holds, mutatis mutandis, for functions with values in , then  is said to have the Radon–Nikodym property. All Hilbert spaces have the Radon–Nikodym property.
Which Y? It isn't mentioned anywhere else. --Svennik (talk) 15:42, 8 May 2020 (UTC)
replies: text: This slightly threw me on first reading as well, so I've reworded it. But I'm not sure the original was wholly correct! The Radon–Nikodym property section under the Bochner Integral article states "The Banach space B has the Radon–Nikodym property if B has the Radon–Nikodym property with respect to every finite measure." To work mutatis mutandis, wouldn't we need σ-finiteness? The two entries seem inconsistent. NeilOnWiki (talk) 13:05, 20 August 2020 (UTC)
thread_title:  What is meant by Y? 

message: I wonder if this article should be raised from Low to at least Mid Importance. This would bring it on a par with eg. Absolute continuity (classed as Mid), with which it's closely linked. I'm reluctant to do this unilaterally for various reasons: it's been unchanged as Low Importance since 2012 and without wider input there might be a danger of an unjustified upward ratchet effect.
In favour of the change, Wikipedia:WikiProject Mathematics/Wikipedia 1.0 suggests that a Mid Importance subject "Adds important further details within its field, with some impact beyond it". There are parallels in the statement of this theorem in Lebesgue integration (classed as High) and Measure theory (High) with that of the Fundamental theorem of calculus (Top) for the Integral (Top), so it can add important insight for a student or reader of these areas. Also, it plays a key role in the foundations of Probability theory (Top), especially Conditional expectation (High), and has applications in Mathematical finance (Mid).
NeilOnWiki (talk) 15:33, 31 August 2020 (UTC)
thread_title:  Should this be raised to at least Mid Importance? 

message: I've tried to follow the Wikipedia:Manual of Style/Mathematics#Article introduction, which says "The lead should, as much as possible, be accessible to a general reader, so specialized terminology and symbols should be avoided.". It therefore seemed preferable to move the previous description of the theorem into a formal statement section in the main body, raising the question of what an accessible explanation might look like.
To do this, I've taken the following as some guiding assumptions:
* A reader may have little or no knowledge of measure theory, but is likely to have an intuition for it through motivating examples.
* The reader has probably come across integration (though not necessarily Lebesgue's), so an integral such as \nu(A) = \int_A f \, d\mu will make some sense.
* It's relatively easy to imagine that the above integral defines a measure ; which then provides a handle onto the Radon-Nikodym theorem as a converse.
* Readers are likely to have come across probability theory and probability density functions.
One might then ask two separate questions. Namely, whether these are reasonable working assumptions (given the nature of the topic)? And if so, how well the text expresses the result? It's hard for me to judge either.
If these edits seem to be working, I might reword the end paragraph about the Radon–Nikodym property of Banach spaces and move it into a main-body Generalizations section, as it seems to be more specialist.
NeilOnWiki (talk) 11:00, 7 September 2020 (UTC)
thread_title:  Making the introduction more accessible 

message: I've just made a possibly hard-to-spot 1-character edit in the proof, to fix what seemed to be a small logic error. (This issue applies to both the Hahn-decomposition-based and non-Hahn versions of the proof discussed earlier.)
If  is an extended valued function, as constructed, then it can't be in  as defined in the unedited version. This definition required each of its members to be a measurable function from  to , ie. to the finite non-negative Reals; so I think the parts of the proof that depended on  failed. Examining various university lecture notes from a web search, an accepted approach seems instead to include extended valued functions in . As far as I can tell, the subsequent construction of  in the text still goes through okay (for the monotone convergence theorem, etc), so I've edited the range to  accordingly.
See eg. https://www.cmi.ac.in/~prateek/measure_theory/2010-10-13.pdf
An alternative might have been to keep the definition of  unchanged and bring the Restricting to finite values step forward, but I didn't find any evidence for that approach.
NeilOnWiki (talk) 20:56, 7 November 2020 (UTC)
thread_title:  Hard-to-spot edit in the proof