Talk:Radon–Nikodym theorem

The printable version is no longer supported and may have rendering errors. Please update your browser bookmarks and please use the default browser print function instead.

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)[reply]

Untitled

So presumably not just a function in the statement, but integrable? Charles Matthews 11:16, 11 Oct 2003 (UTC)

Yes, I'll make this explicit in the article. Pete 12:39, 11 Oct 2003 (UTC)

Another question is: f is not unique (actually, it is unique almost sure), as far as I know (but I may be perfectly wrong). If this is the case I think it is worth mentioning it (something like the equality holds for f and g whenever f=g a.s.). Pfortuny 12:48, 11 Mar 2004 (UTC)

The second LaTeX formula bothers me. First of all, what is A supposed to be? I would guess a random variable rather than a set (like above). Also, shouldn't it rather be

E_{Q}(X)=E_{P}\left({\frac {dQ}{dP}}X\right)

(E_P and E_Q switched?). Not sure though, else I would change it myself. DrZ 13:50, 16 Mar 2004 (UTC)

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)

But what is the expectation operator applied to a set supposed to mean? DrZ 14:50, 16 Mar 2004 (UTC)

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)

proof

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)[reply]

measures on an algebra (not sigma)

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)[reply]

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)[reply]

it looks like itaj is saying the following: <<< is a continuity condition that's equivalent to the usual absolte continuity <<, for countably additive measures. the notation ||.|| looks like it means the total mass of a measure. a suitable f is then obtained by approximation using simple functions. question is: when the assumption that $L^{1}$ being complete is invoked, aren't you implicitly extending to a σ-algebra? Mct mht 03:25, 15 June 2006 (UTC)[reply]

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)[reply]

well, that's the question. when you use the assumption that $L^{1}$ is complete, implicitly you're extending to a sigma algebra anyway, no? in other words, the support of functions in $L^{1}$ might as well be addded to the algebra to make it a sigma algebra. so maybe it's not so much of an improvement. Mct mht 21:06, 15 June 2006 (UTC)[reply]

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)[reply]

Proof for signed measure

The article says, "If ν is a signed measure, then it can be Hahn–Jordan decomposed as ν = ν+−ν− where one of the measures is finite". Yes, this is true. However, how can we be sure that the non-fininite measure is σ-finite? Is it always true? If yes, then any nonnegative measure is σ-finite? Jackzhp 02:34, 10 December 2006 (UTC)[reply]

You are right, one should have mentioned that ν was σ-finite. I fixed that now. And no, not every measure is sigma-finite. Oleg Alexandrov (talk) 03:13, 15 February 2007 (UTC)[reply]

In the "Radom-Nikodym derivative" section the article states

   [...] if μ is a nonnegative σ-finite measure, and ν is a finite-valued signed
   or complex measure such that  $|\nu |\ll \mu$ , there is μ-integrable real- or
   complex-valued function g on X

It is however sufficient for the signed measure to be σ-finite. In this case, g will not be μ-integrable in general, because it is not necessarily finite. Maybe the following would be more accurate:

   [...] if μ is a nonnegative σ-finite measure, and ν is a σ-finite signed or
   complex measure such that  $|\nu |\ll \mu$ , there is a μ-(quasi-)integrable real- or
   complex-valued function g on X

By quasi-integrable I mean that the integral exists, but it is not necessarily finite.--Drizzd 17:39, 23 February 2007 (UTC)[reply]

"Financial mathematics uses the theorem extensively" ?

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)[reply]

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)[reply]

as to complex measures...

In the section on the Radon-Nikodym-derivative, it is stated that the theorem also holds in case

ν is a finite-valued signed or complex measure such that $|\nu |\ll \mu$

However, I just caught the theorem (as noted by user:Drizzd above) stating that $\nu$ also needs to be $\sigma$ -finite (which is, as far as I know, not the same as finite-valued...).

I also noted somewhere else that (at least in this setting), $|\nu |\ll \mu \iff \nu \ll \mu$ (please correct me if I'm wrong with this!). Wouldn't therefore, replacing the total variation measure $|\nu |$ by just $\nu$ make this entry easier to use?

--Björn

Wrong definition of Renyi divergence

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)[reply]

Radon-Nikodym derivative is not in general integrable

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)[reply]

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)[reply]

Ah, I didn't see that, thanks for pointing it out. 21:39, 14 March 2012 (UTC) — Preceding unsigned comment added by Paul Laroque (talk • contribs)

What is meant by Y?

> If $Y$ is a Banach space and the generalization of the Radon–Nikodym theorem also holds, mutatis mutandis, for functions with values in $Y$ , then $Y$ 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)[reply]

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)[reply]

Reverting proof to include Hahn decomposition

I've reverted the following edit "Simplified the proof by removing the necessity of a Hahn decomposition" from 18 Aug 2020.

While the Hahn decomposition may perhaps be too strong for what we need, I think the suggested alternative is too weak. In particular, it relies on the claim that

\int _{X}\left(g+\varepsilon 1_{B}\right)\,d\mu \leq \nu (X)<+\infty ,

implies $g + ε 1 B \in F$ for the set $B$ that the edit constructs. But there may be a measurable subset $A$ of $B$ for which

\int _{A}\left(g+\varepsilon 1_{B}\right)\,d\mu >\nu (A),

which would disqualify $g + ε 1 B \in F$ . Invoking the Hahn decomposition (as in the unedited version) avoids that possibility.

I'm not a measure theorist, so would welcome someone who is to check this out. Can we cite a source for this proof approach? It might help future editors in judging what kind of changes might be appropriate.

NeilOnWiki (talk) 11:13, 24 August 2020 (UTC) (Clarified what $B$ referred to - NeilOnWiki (talk) 19:59, 28 August 2020 (UTC))[reply]

An IP editor has changed this back to a version that avoids invoking the Hahn decomposition in the Proving equality part of the proof, with the following edit summary:

I'm reverting to a previous edit, I cannot see anything wrong with this previous proof. Whilst it may be the case that

\nu -\varepsilon \mu

is a signed measure, the Hahn decomposition here appears unecessary.

It isn't clear if this is an original simplification of the proof or based on an attested source: either way my own reservations remain, which it might be helpful to describe more fully.

The simplified proof constructs a set $B$ with $ν 0 (B) > ε μ (B)$ for some $ε > 0$ and then claims that

\int _{X}\left(g+\varepsilon 1_{B}\right)\,d\mu \leq \nu (X)<+\infty

implies $g + ε 1 B \in F$ .

But the definition of $F$ requires that we have

\int _{A}\left(g+\varepsilon 1_{B}\right)\,d\mu \leq \nu (A)

for every measurable set $A$ in $Σ$

Unfortunately, the construction of $B$ doesn't rule out the possibility that there may be a measurable subset $A$ of $B$ with $ν 0 (A) < ε μ (A)$ , in which case the integral inequality breaks down and the assertion that $g + ε 1 B \in F$ is untrue (or at best unproven).

Now consider the (potentially) signed measure $ν 0 - ε μ$ , where we've chosen $ε > 0$ so that $ν 0 (C) > ε μ (C)$ for some measurable $C$ , hence $(ν 0 - ε μ)(C) > 0$ for $C$ . I think one way we could rescue the proof while avoiding a fully-fledged Hahn decomposition would be via a weaker lemma that if a signed measure has positive value for some measurable set $C$ then there exists a measurable subset $B$ with positive measure, all of whose subsets $A$ have non-negative measure (so $B$ is a positive set). Applying this to $ν 0 - ε μ$ would fix the above issue for $A$ ; but would involve an extra proof step.

Personally, I find the weaker result casts more light on what's happening, so is more intuitively satisfying than wheeling in the full Hahn (assuming I've got this right!). Aside from the need for rigour, I think a good proof ideally provides revelation and understanding, which seems more in keeping with Wikipedia's aim to inform (I'm still trying to get my head round this). If the Hahn decomposition offends, I wonder whether a comment in the original Hahn-based proof could be added to the effect that we need a positive set for the logic to go through and that a Hahn decomposition is a convenient way of producing one.

Whatever we do, I've been keen to explain my continued concern about the validity of the simplified, edited proof as it stands.

NeilOnWiki (talk) 09:24, 8 October 2020 (UTC)[reply]

To avoid locking in what seems (to me) to be an error, I've just reluctantly re-reinstated the original proof involving the Hahn decomposition, because the most recent version without it didn't show how $\int _{A}\left(g+\varepsilon 1_{B}\right)\,d\mu \leq \nu (A)$ for all $A$ in $Σ$ , as required by the definition of $F$ . I've also added some explanatory comments in the Hahn-based proof to indicate how the decomposition helps us, as discussed above. I'd welcome any further comment on this, either confirming or refuting the approach I've taken. NeilOnWiki (talk) 10:05, 27 October 2020 (UTC)[reply]

Hi Anon, not sure if you'll get to read this, but I noticed that you modified the proof yesterday and then retracted it. I do agree with you that Hahn decomposition is overly strong and would be happy to talk this over with you (though I'm less sure if sidestepping it is necessarily an improvement from Wikipedia's point of view). Not sure if this is helpful to you, but I find it useful when authoring maths stuff to have a sandbox drafting area to develop the content over time, rather than get it right immediately. One way to do this is as a logged-in user (not sure if IP-only users have sandboxes); creating your own account like this also makes discussions easier.

For my own satisfaction, I did concoct a proof of a weaker lemma than Hahn that seemed to work (though still for signed measures); and found this non-Hahn-based proof on-line, which might be of interest to you. I'm also fairly convinced that you can define

g

in terms of simple functions in a way that would make the proof approach more widely understandable to scientists and other non-mathematicians. Unfortunately, home-based proofs don't necessarily sit well with Wikipedia: see eg. WP:No original research and WP:WikiProject Mathematics/Proofs (I'm still trying to get my head round the implications of these texts). I've no immediate plans to edit the existing proof, but do have good intentions to make the main article more accessible to more people, so will be looking at these pages for a while if you do wish to get in touch. NeilOnWiki (talk) 20:55, 15 December 2020 (UTC)[reply]

Dear NeilOnWiki,

I'm the anonymous editor (although I do plan to make an account soon). Thank you for diligently corrected my error - it certainly was an oversight. I also apologise for my previous reversions with limited discussion, which were rather careless. I now agree that using a sandboxing area is better for the reliability of Wikipedia articles.

I did also manage to find a way to overcome the issue, but this seemed to use a similar approach to the approach used in the proof of the Hahn decomposition theorem (see [1]). From a cursory reading, I think the link you uploaded also uses a similar approach. As a result, I think I have a deeper understanding now of why the the Hahn decomposition is used, which I didn't have earlier (I hope not at the expense of the understanding of some unfortunate Wikipedia readers). It also seems to be standard in other textbooks (at least in the book by Bogachev).

Thanks again for your efforts!

Dear Anon (or possibly non-Anon now)

Thanks for getting in touch — and also your interest in the article. I'm still finding my own way around and the sandbox has worked well for me so far. I'm glad you're feeling more reconciled with the role of Hahn decomposition. I have a lot of sympathy with your questioning it; though I also find the conceptual shift in recognising

ν 0 - ε μ

as a signed measure adds a useful insight. The overall leap in intuition to signed measures is quite small (and seems fairly natural), though I guess ~~[it]~~ the signed theory does need extra effort for a rigorous approach.

Funnily enough, I found your error quite useful because it forced me to understand this more fully, as I was reluctant to undo the work you'd put in without a strong reason (you're my first serious Revert), so this particular Wikipedia reader is okay about it. It's also good to know there's an extra pair of eyes interested in the article, in case something I do needs checking or correcting. Feel free to say hello here or on my talk page if you do sign up. NeilOnWiki (talk) 15:31, 18 December 2020 (UTC)[reply]

Should this be raised to at least Mid Importance?

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)[reply]

Making the introduction more accessible

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 $ν$ ; 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)[reply]

Hard-to-spot edit in the proof

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 $g$ is an extended valued function, as constructed, then it can't be in $F$ as defined in the unedited version. This definition required each of its members to be a measurable function from $X$ to $[0, \infty)$ , ie. to the finite non-negative Reals; so I think the parts of the proof that depended on $g \in F$ failed. Examining various university lecture notes from a web search, an accepted approach seems instead to include extended valued functions in $F$ . As far as I can tell, the subsequent construction of $g$ in the text still goes through okay (for the monotone convergence theorem, etc), so I've edited the range to $[0, \infty]$ 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 $F$ 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)[reply]