Talk:σ-algebra

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Borel-field

I'm a long way from an expert on this stuff, but I thought that a Borel-field (generated from the family of open subsets of X) is a particular (and very common) example of a sigma-field, but that not all sigma-fields are Borel-fields. Can somebody confirm or deny this? If I am correct, then the article ought not suggest in the first sentence that they are equivalent. —Preceding unsigned comment added by 158.143.65.19 (talk) 10:34, 5 October 2010 (UTC)[reply]

Existence

Is exist a infinite sigma algebra on an set X such that be countable?

No, any sigma-algebra is either finite or uncountable. Prumpf 13:14, 12 Oct 2004 (UTC)

So what does

It is the countable analog of a Boolean algebra, and every σ-algebra is a (represented) Boolean algebra.

mean? And is a σ-field just a variant term, or does it mean something slightly different? Dan Hoey 19:03, 9 March 2007 (UTC)[reply]

I've answered this to the best of my ability, by changing the opening paragraph. Someone who actually knows about this stuff might vet it. I'm also considering changing the example (which is more suited to Boolean algebra) and moving this segment of the discussion down to the bottom of the discussion page.Dan Hoey 19:23, 11 March 2007 (UTC)[reply]

The opening remarks suggest that a sigma algebra satisfies the field axioms - is this true? If so what are the '+' and 'x' operations etc.? --SgtThroat 13:08, 10 Nov 2004 (UTC)

I don't think so. The natural operations are union and intersection, and the identities are trivial, but you hit a problem with the inverses' properties of fields. --Henrygb 01:20, 21 Jan 2005 (UTC)

Font used for denotion in math papers is not important?

The following sentence was deleted: "σ-algebras are sometimes denoted using capital letters of the Fraktur typeface".

Yes, this typeface is not used in this article, but reading math papers I found, that they are usually denoted using it. I did not know, how it is called and how should these letters be read and hoped to find this out in this article, but failed. I found the name of the typeface in other place and I thought this note will be helpful for other people. But it's considered not important...

BTW, no note, that similar constructions which are closed under finite set operations are usually called algebras (this term obviously appeared before σ-algebras). The article does not contain anything more than a definition copied from MathWorld and trivial examples. But other trivial info is irrelevant here... Cmapm 01:07, 3 Jun 2005 (UTC)

Those examples are not trivial. Some of those examples are simple, and such serve to illustrate the concent. Some other examples listed there are actuallly very important, so not trivial either.

The information you inserted is not trivial either. I said it was "not valuable". Please feel free to put it back. It is just when I read the article as a whole, I found that minor point about the font distracting from the overall concent. But there is of course room for disagreement. Oleg Alexandrov 01:43, 3 Jun 2005 (UTC)

Also, you are more than welcome to add content to this article if you feel it is incomplete. Oleg Alexandrov 01:44, 3 Jun 2005 (UTC)

question

For definition 2 of a sigma algebra, it says that for a sigma-algebra X, if E is in X, then so is the complement of E. Does this mean the complement of E in S (i.e. S-E)? Or the complement of E with some universal set?

Thanks!

It means the former. I will now try to make it more explicit in the article. Oleg Alexandrov 02:53, 23 Jun 2005 (UTC)

Here is another question. I know this phrasing is standard, but it is quite confusing to people new to sigma-algebras. Let A be a collection of subsets of X. We often say the following: "The sigma-algebra generated by A contains A". In fact, something like this is mentioned in this article. However, the sigma-algebra generated by A does not actually contain A...afterall, A is a collection of sets. Rather, the sigma-algebra generated by A contains all the elements of A. I know this must be obvious to many, but I found it very confusing when first encountering sigma-algebras...and I know that I was not alone.

In the examples section, it is said: "First note that there is a σ-algebra over X that contains U, namely the power set of X." Again, I think we should be perfectly clear. U is not a member of the power set of X. Rather, all memebers of U are members of the power set of X.

To make this abundantly clear, why not include very trivial examples of a sigma-algebra.

Let X = {1,2,3}

Let C = { {1}, {2} }

Then σ(C) = { {}, {1}, {2,3}, {2}, {1,3}, {1,2,3} }.

This is so very clear and obvious. Also notice, C is not a member of sigma-algebra. So we really should refrain from saying "C is in the sigma-algebra generated by C." It is sloppy even though it is standard.

I think you are confusing "containment" and "membership". A set A is said to contain a set B, iff B is a subset of A, that is iff every member of B is a member of A. So, since as you point out, every member of C is a member of σ(C), σ(C) contains C. Paul August ☎ 05:43, 12 October 2005 (UTC)[reply]

Well, that certainly would explain it! As you can see, I STILL am getting used to this. :-) Either way, it might be helpful, perhaps to people like myself, to see an example like that above. But your comment is correct and now understood by me. "containment" is NOT the same as "membership". Is there a wikipeida entry for "contains"? If so, we probably should link to it to avoid this confusion...as the mathematical usage of the word "contains" is very different from everday usage.

Name and history

Where does the name "sigma-algebra" come from? When were sigma-algebras introduced? -- Tobias Bergemann 13:39, 22 July 2005 (UTC)[reply]

Small sigma and delta are often used the union and intersetion are involved. They seems to be the Greek abbreviation of German words: Summe (sum) and Durchschnitt (intersection). Pura 00:10, 3 October 2005 (UTC)[reply]

Notation

I find it somewhat distracting that the notation used in this article, that in sigma-ideal, and that in measurable function, are not in concordance. I'm also unnerved that the usage of X and S in this article is reversed from the common usage I am used to seeing. That is, I'm used to seeing X be the set, and Σ be the collection of subsets, so that (X, Σ) is the sigma-algebra. Sometimes, S is used in place of Σ. Can I flip the notation used here, or will this offend sensibilities?

Also, do we have any article that defines the notation (X, Σ, μ) as a measure space? (I needed a wikilink for this in dynamical system but didn't find one). linas 13:37, 25 August 2005 (UTC)[reply]

Actually, I want to harmonize the notation in all three articles. But before doing so, we should agree on a common notation. I propose:

X be the set
Σ be the collection of subsets
E and E_n are the elements of Σ (and leave alone A_n for those articles that already use that).
μ is the measure.
(X, Σ) and (Y, T) are the sigma algebras,
(X, Σ, μ) and (Y, T, ν) are measure spaces.

This change will eliminate/replace the use of F, ${\mathcal {A}}$ and Ω in these three articles. Ugh. Measure (mathematics) is not even self-consistent, switching notation half-way through. linas 13:50, 25 August 2005 (UTC)[reply]

Relation to field of sets

Anyone care to wikilink field of sets much earlier in the article, and expound on the difference between that and this? (The difference being that here, the number of intersections & unions is countable)? linas 15:06, 30 August 2005 (UTC)[reply]

Boolean algebra

I was wondering, is a sigma-algebra also a boolean algebra. If so, should this be included in the definition? It seems that we are always using the axioms and results (demorgan) of boolean algebras. --anon

Well, any field of sets is a boolean algebra. So I guess this remark belongs in field of sets rather than the particular case of sigma-algebra. Oleg Alexandrov (talk) 21:46, 15 January 2006 (UTC)[reply]

Families?

"In mathematics, a σ-algebra ... over a set X is a family Σ of subsets of X that is closed under countable set operations..."

Is "family" meant here in the sense of family (mathematics), or is it just a loose way of saying "set"? If family (mathematics) is meant, then a link should be added. If set is meant, why not just say "set"? Dbtfz 06:04, 19 January 2006 (UTC)[reply]

You're right about that. I've changed it to collection. -lethe ^talk 21:25, 25 January 2006 (UTC)

Meaning of terms

I've just read through the current article (having had no knowledge of sigma algebras), and was confused by some terms. It is unclear whether my confusion arises from addressable weaknesses in the article or from lack of prerequisite knowledge on my part.

The problem terms were "countable set operations" (first para), and "(countable) sequence" and "(countable) union" (both in Property 3). I know what a countable set is, and what a set operation is, but the rest of the article leave me unable to guess at the combination.

My guess at the meanings is confounded by an example earlier in this talk page:

Let X = {1,2,3}

Let C = { {1}, {2} }

Then σ(C) = { {}, {1}, {2,3}, {2}, {1,3}, {1,2,3} }.

.. since I had assumed P3 would require {1} U {2} = {1,2} to be included (and hence also its complement {3}). Hv 20:26, 25 January 2006 (UTC)[reply]

As far as I can see, that example was written by a confused person and is wrong. For exactly the reasons you state. Now, as for your confusion... well, I'm not sure what you don't get. You can take the union of two sets, the union of three sets, the union of alef-0 sets, or the union of beth-2 sets. The first three are countable set operations (in fact countable unions), while the last is not. -lethe ^talk 20:33, 25 January 2006 (UTC)

Notice how in the definition, the word "countable" is in parentheses. This is because the notation E₁, E₂, … already implies a countable set of sets. The subscripts give a bijection to N. The union of a countable number of sets is a countable set operation. So is the intersection of a countable number of sets. This is what is meant by the phrase "countable set operation". -lethe ^talk 20:36, 25 January 2006 (UTC)

If the example was wrong that clears up most of my confusion (though I'm surprised nobody pointed it out at the time). I'd be interested to see an example of a subset that does not need to be included in the σ-algebra because it can only be generated as the union of an uncountable number of the subsets that are included. Hv 20:40, 25 January 2006 (UTC)[reply]

What is it you want to see an example of? If the set can only be generated by uncountable operations, then it does have to be explicitly included, since the axioms of a σ-algebra won't get you to uncountable unions. Unless you mean you want an example of a set that isn't in the algebra. I can surely give you an example. Let C be the set of all singleton subsets of R. Then σ(C) is the set of all countable sets of real numbers and their complements. Any uncountable set with uncountable complement, for example (0,1), will not be in σ(C), even though it is generated by union of elements of C, because the union is uncountable. -lethe ^talk 20:48, 25 January 2006 (UTC)

Ah, got it - the last example makes it perfectly clear to me. Thanks very much. Hv 23:27, 25 January 2006 (UTC)[reply]

I'm glad we cleared that up then. Do you have any suggestions for clarifying the language in the article? I looked at it but didn't see anything obvious to change. -lethe ^talk 23:36, 25 January 2006 (UTC)

I think the three phrases I mentioned should all be changed. For "countable set operations" I'd suggest "(a countable number of) set operations"; in P3, I'd replace "(countable) union" with "union" - the current phrase implies that the union must have a countable number of elements, which ain't so.

The tricky one is "(countable) sequence" - it is really just a countable subset of Σ, and the word "sequence" wrongly implies an ordered structure; I suspect the author of those words wanted to avoid confusion arising from that fact that the elements of Σ are themselves sets, but I can't think of a better phrase than "countable subset", and I think better concrete examples would help to clarify for anyone that did get confused as a result.

For examples, I'd suggest X={1,2,3} and generate the minimal SA for {1} - this demonstrates the basic principles and shows that nontrivial SAs exist other than the power set. I'd also suggest a more complex example that brings countability into play: maybe your [C, { r: r in R }] example, or something simpler if anyone can come up with it. These concrete examples should probably go after the trivial SA and powerset, and before the rest (all of which are really too abstract to count as "examples" in the didactic sense). Hv 01:56, 27 January 2006 (UTC)[reply]

I implemented some, but not all, of your suggestions. It turns out that the countable example I gave above is actually already in there, in a someone more general phrasing. -lethe ^talk 03:05, 27 January 2006 (UTC)

Thanks Lethe, I think it is a bit clearer now. The only other thing that seems missing is a mention of what σ-algebras are useful for, and in particular what the restriction to countable operations buys us - does the restriction allow more powerful general theorems to be proved, or is it primarily to ensure that things like Borel algebras are well-defined? And is there another type of algebra defined identically except without the restriction to countable operations? Hv 12:06, 28 January 2006 (UTC)[reply]

Well, the entire raison d'être of the σ-algebra is to define measures. A measure is an extended nonnegative real function on a σ-algebra (that means it assigns to each number either a nonnegative real number or else ∞). These are in turn used to define Lebesgue integrals, a more powerful alternative to the more familiar Riemann integrals. A field of sets is like a σ-algebra, except it's closed only under finite union and intersection. You asked whether there is a name for something that's closed under arbitrary union and intersection. Well, if there is a name for such a thing, I don't know it. As for your last question, what's the gain of allowing countably infinite unions, instead of finite unions; or alternatively, what do you lose by restricting to countable unions, instead of arbitrary unions. Well, I don't know the answer definitively. I suspect that if you restricted to finite unions, you would lose the completeness properties of the Lebesgue integral, that is you wouldn't be able to integrate the limit of a sequence of integrable functions. You wouldn't be able measure sets like the rationals, the cantor set, or even some nice geometric series of sets. And as for what we lose by not allowing arbitrary unions, well, you can't sum more than countably many nonzero real numbers and get a nonzero result, so It would be pretty meaningless to allow more arbitrary unions. Also, I guess the singleton would have to have positive measure, or else all sets would have measure zero. It would be a mess. -lethe ^talk 12:36, 28 January 2006 (UTC)

Definition of Sigma-Algebra

I am confused by the third requirment of a sigma-algebra. Does this just say that the union of countably many member sets is also a member? In any case, this needs to be rewritten to make it more transparent on a first reading. --Njerseyguy 17:13, 27 February 2006 (UTC)[reply]

Well I am the one responsible for the current opaque wording there. Yes indeed, it means that the union of countably many members is a member. I think I wrote it this way in response to a reader who was confused by the term "countable union". Anyway, it can certainly be simplified. See my recent attempt. -lethe ^{talk +} 17:46, 27 February 2006 (UTC)[reply]

The definition would be less confusing if the first requirement "Sigma is non-empty" were replaced with "X belongs to Sigma", which is the way I have always seen it done (Wolfram Mathworld does it this way, for example). The fact that X belongs to Sigma is fairly important and using the definition here, the fact that X belongs to Sigma is nto 100% obvious at first glance and must be proven. Gsspradlin (talk) 21:49, 6 November 2013 (UTC)[reply]

That change sounds fine to me, in and of itself. But ideally we should find a good secondary source that does it that way (meaning definitely not MathWorld, which besides being a tertiary source, has shown itself many times to be idiosyncratic). Rudin is a pretty solid source, so we should find one of equal quality — maybe Folland? I don't have my copy at hand to see what he does. --Trovatore (talk) 21:54, 6 November 2013 (UTC)[reply]

I just consulted Folland and, unfortunately for me, he uses "Sigma is non-empty" as one of his rules. I have a copy of Halsey+Royden's Real Analysis (it appears to be an "international edition" of a fourth edition) in which they include "Sigma contains the empty set" as one of the rules. Oddly, a little later, there is a redundant definition of "sigma-algebra of subsets of R (the real numbers)" that assumes instead that R belongs to Sigma. I think that replacing "Sigma is nonempty" with either "the empty set belongs to Sigma" or "X belongs to Sigma" would be an improvement, for the reasons I give above.Gsspradlin (talk) 23:51, 6 November 2013 (UTC)[reply]

I completely agree with Gsspradlin. At first glance, it might be a good idea to take a minimal or general axiom which "just" says that Sigma contains some set, but actually this is not useful in practice. A similar criticism applies to the subgroup criterion (subgroups should not just assumed to be non-empty, but rather they should contain the neutral element). The requirement "Sigma is non-empty" is also problematic from the perspective of constructive mathematics and from the perspective of general algebra. Regarding sources, Rudin's book "Real and complex analysis" requires that X is contained in Sigma. In my humble opinion, it would be even more natural to require that both X and the empty set are contained in Sigma. In the article, the proof that the definition of a sigma-algebra implies that X is contained, should either be omitted or replaced by a remark that for proving the containment of X and the empty set in Sigma it is enough to prove that Sigma is non-empty. Also the proof on the intersection of sigma-algebras should be adjusted; notice that the current proof already uses X as the test set. -- Martin Brandenburg. --85.181.227.12 (talk) 08:34, 7 April 2015 (UTC)[reply]

The first requirement includes the statement "...X is considered to be the universal set in the following context...". I think universal set (a set containing among other things itself according to Universal_set) should be replaced by something else, perhaps "set universe" as in the article Universe_(mathematics). Joel Sjögren (talk) 22:49, 24 May 2016 (UTC)[reply]

Adjectives and Nouns

I'm not going to make an edit myself, but that lead sentence with "countable set operations" (and others like it) causes problems for people that are not "in the know" (such as myself). Is this article for specialists? I don't think so, personally. Specialists will likely be reading textbooks (of course, if you're not a specialist, why the hell are you reading this!). Personally, I read plenty of textbooks in other areas and find that wikipedia is far better at getting me some basic information on specialized topics then going to, say, a measure theory textbook.

Anyway, we have these words: "that is closed under countable set operations".

They describe the fact that: "if I apply a countable number of set operations to an element of the set, then the result is also an element of the set"

It seems that something like this would remove the ambiguity: "that is closed under the application of a countable number of set operations". Likely, this would require breaking that sentence into two parts.

Regards, Mark

Hi Mark. You're not the first person who's complained about that language recently. I guess we should change it. The problem is that the suggestion "countable number of set operations" is not good: we're still considering only the operations intersection and union. That suggestion was about that same as the last person. I've made an attempt to address your (and the previous guy's) concern. Is it better? -lethe ^{talk +} 23:26, 10 March 2006 (UTC)[reply]

empty set

"The empty set is in Σ": is this really needed in the definition? Closure under complementation and countable unions implies that both the empty set and X are in Σ since:

A in Σ and A^c in Σ imply their union (which is X) is in Σ which implies that X^c (which is the empty set) is in Σ.

Indeed closure under complementation and finite unions is enough to prove that the empty set and X are in sigma. Pramana 19:59, 15 June 2006 (UTC)[reply]

I think you are correct. -lethe ^{talk +} 23:15, 15 June 2006 (UTC)[reply]

I think that's true as long as you assume that the set Σ is nonempty itself. Otherwise there is no way to deduce that the empty set is an element there. Oleg Alexandrov (talk) 23:20, 15 June 2006 (UTC)[reply]

I guess that axiom precludes the empty set from being a sigma algebra. So whether or not you want to list the axiom explicitly depends on whether you want the empty set to be a sigma algebra. Well, I checked Halmos and Royden; they don't use this axiom, so I felt OK removing it from the article. -lethe ^{talk +} 23:27, 15 June 2006 (UTC)[reply]

I don't believe any authors allow for empty sigma algebras. Are you sure? Oleg Alexandrov (talk) 03:03, 16 June 2006 (UTC)[reply]

I said that these authors do not use the axiom "a sigma algebra contains the empty set and the whole space". I never said that anyone allows empty sigma algebras, though I guess dropping the axiom does open the door for empty sigma algebras. Presumably these authors require nonempty subsets then. I will modify the article to clarify this point. -lethe ^{talk +} 03:28, 16 June 2006 (UTC)[reply]

It is possible that someone, somewhere, does want empty sigma-algebras. In any event, I think the original definition is the traditional one, for what it's worth. Not that it's a major issue. —Vivacissamamente 04:52, 16 June 2006 (UTC)[reply]

I can clearly say that the set of people who does not want empty sigma algebras is non empty! Oleg Alexandrov (talk) 05:18, 16 June 2006 (UTC)[reply]

Perhaps we can consider the following points: 1. Halmos and Royden dont use this axiom (stated above). 2. The main use of sigma algebras is in measure and integration for which we actually need a "rich" collection of subsets of X. I am currently unaware of any "deep" results because of including this axiom in the definition. I agree that this is not a major issue, but i thought definitions should be "lean" and Halmos and Royden are good enough for me. Pramana 06:09, 16 June 2006 (UTC)[reply]

Halmos (in my edition, at least) mentions " ... a non-empty class of sets ..." when defining algebras and sigma-algebras, and this is pretty much equivalent to having the axiom that includes the empty set. Rudin ("Real and Complex") insists that the whole space is in the sigma-alebra, Doob and S. Lang both insist that the empty set is in there, so it seems that Royden is the only one who actually allows empty sigma-algebras. I suggest we keep the axiom in one form or another, to aggree with the majority of serious authors. Madmath789 06:47, 16 June 2006 (UTC)[reply]

So there are two choices for our definition here: that the sigma algebra be nonempty or that the sigma algebra include the empty set and the whole space (or just one of the two). The latter is what we had in this article before I changed it. Now we have the former. I prefer the first choice, it seems sleeker to me. They are of course equivalent. A third possibility is to take neither, which allows empty sigma algebras, and is probably not standard. -lethe ^{talk +} 11:24, 16 June 2006 (UTC)[reply]

I am happy with either version (slight preference for current) - but don't really like including empty sigma-algebras. Madmath789 11:31, 16 June 2006 (UTC)[reply]

If you read further in Royden, it becomes clear that he implicitly assumes the empty set in the σ-algebra. Daren Cline (talk) 15:27, 8 April 2015 (UTC)[reply]

I prefer the current version too.Pramana 12:48, 16 June 2006 (UTC). And if we decide to stick with the current version, we need to correct "from 2. and 3. it follows.......".Pramana 12:58, 16 June 2006 (UTC)[reply]

Surely the fact that Σ is closed under countable unions implies that the union of zero sets is in Σ, which in turn implies that the empty set must be in Σ. So there's no need for a separate axiom demanding that Σ is nonempty, it follows from 3. Bat020 14:53, 28 March 2007 (UTC)[reply]

I remarked elsewhere that it would be less confusing to replace the condition that the sigma-algebra be nonempty with the condition that the entire set belong to the sigma-algebra (arguing by authority, this is what Rudin does), or perhaps that the empty set belong to the sigma-algebra (as Doob and S. Lang do). The way it stands, it is not 100% obvious at first glance that the empty set and the whole set belong to the sigma-algebra, and it requires a little work to show it, which seems pointless when you could just make this one of the requirements. I am assuming that no one likes empty sigma-algebras (someone commented above that Royden does, but I have never heard of anyone else doing this). Gsspradlin (talk) 22:04, 6 November 2013 (UTC)[reply]

@Bat020 : what I propose is not an extra assumption that the empty set or the whole set belong to Sigma, but replacing the requirement that Sigma be nonempty with either (i) the empty set belongs to Sigma or (ii) the whole set belongs to Sigma. They are all equivalent, of course, but I think it's probably important that the empty set and the whole set belong to Sigma, and in the current definition, those facts are not immediately obvious and need to be proven. Rudin requires that the whole set belong to Sigma, and Royden requires that the empty set belongs to Sigma. Unfortunately for me, I admit that Folland's definition uses "Sigma is nonempty". I have not seen a definition anywhere that allows Sigma itself to be empty, and I would guess that very few people are interested in this possibility (but with mathematicians, you cannot rule it out). — Preceding unsigned comment added by Gsspradlin (talk • contribs) 23:59, 6 November 2013 (UTC)[reply]

The only reason I can imagine that mathematicians could want the empty set to be a σ-algebra would be to have a null σ-algebra. But ${\emptyset, X$ } already is the null σ-algebra as it contains zero information in terms of identifying proper subsets of X; X and ∅ being identifiable already by default. I also agree, especially for the typical reader, that there is little motivation for making non-empty part of the definition when saying $X \in Σ$ is equivalent and more to the point. It also makes it quickly clear that ${\emptyset, X$ } is not only a σ-algebra but also the smallest σ-algebra. Daren Cline (talk) 15:22, 8 April 2015 (UTC)[reply]

I agree, completely;

{\emptyset, X

} is the smallest σ-algebra; no one doubts it, I believe. Boris Tsirelson (talk) 17:12, 8 April 2015 (UTC)[reply]

"... if and only if X is uncountable. "

The examples states: "The collection of subsets of X which are countable or whose complements are countable is a σ-algebra, which is distinct from the powerset of X if and only if X is uncountable." If X is N, then this is false - there are subsets of N which are countable and whose complements is also countable. (67.102.227.19 19:50, 25 September 2006 (UTC))[reply]

Yes, but the point is that in your case, the sigma algebra of countable and co-countable subsets IS the same as the powerset. The statement is saying that the countable and c-countable sigma algebra is not the same as the powerset iff X is uncountable. I.e. the article is correct. Madmath789 20:00, 25 September 2006 (UTC)[reply]

D'oh! I misinterpreted "countable/co-countable" as "finite/co-finite" in the original statement. Nevermind. 67.102.227.19 00:55, 26 September 2006 (UTC)[reply]

Sigma homomorphism

Can someone go over and give a bit more context to Sigma homomorphism? It now links back here but my measure theory is way to rusty to give the proper context. --Chrispounds 03:50, 9 October 2006 (UTC)[reply]

A source: Sect. 15.C in book "Classical descriptive set theory" by A.Kechris. Boris Tsirelson (talk) 18:56, 7 September 2015 (UTC)[reply]

Is the power set always a sigma-algebra?

I'm confused by the statement that the power set of X is always a sigma-algebra. If the power set of the reals is the set of all subsets of reals, then I assume it contains the Vitali sets. I thought the aim was to avoid such sets because they're not measurable. LachlanA 03:41, 31 October 2006 (UTC)[reply]

You are right, the power of all reals is a very poor sigma-algebra. One can't define the Lebesgue measure on it. But the power of reals is still a sigma-algebra, as it satisfies all the properites, and one can define some measures on it, although they are not helpful. Oleg Alexandrov (talk) 03:52, 31 October 2006 (UTC)[reply]

Intersections of Sigma Algebras

In the section on generated sigma-algebras, one of the steps involves taking the intersection of multiple sigma algebras. Does this mean that the intersection of two sigma-algebras is always another sigma-algebra? What about countable and uncountable intersections? Calumny 17:38, 27 August 2007 (UTC)[reply]

Yes, yes and yes: the intersection of an arbitrary family of sigma-algebras on a given set will yield a sigma-algebra. Size does not matter. --Mark H Wilkinson ^{(t, c)} 17:52, 27 August 2007 (UTC)[reply]

Sigma-fields

Could one write something about sigma-fields? Here it states that those are somewhat interchangable... could this be clarified? —Preceding unsigned comment added by 83.6.96.216 (talk) 17:07, 31 August 2008 (UTC)[reply]

I have changed this to reflect usage of terms both as given in the article you cite, and as explained to me by my statistics teacher. I'm just learning this stuff, so I hope I have not introduced an error, but if so, I would appreciate any explanation anyone can give that will clear this us for me. BearMachine (talk) 19:09, 17 February 2010 (UTC)[reply]

"explicit description"

The text currently asserts that

in general there is no explicit description of the σ-algebra generated by a given collection

but nothing is said about what would constitute an "explicit" description. I would argue that, just taking the example of the Borel sets of the reals for concreteness, the following description is explicit: the Borel sets are the interpretations of the Borel codes, where a Borel code is a gadget of the sort described at infinity-Borel set, but where the ordinal height of the tree thereby generated is countable.

The burden is on those who would claim that there is no explicit description to say what is meant by that and prove that none exists. Therefore I am re-removing the text. --Trovatore (talk) 18:01, 20 November 2008 (UTC)[reply]

Whoops, slight self-correction: The distinction is not that the height is countable, but that the nodes are countably splitting.

That is, every Borel set can be described by starting with a bunch of open sets as the leaves of a tree. Sometimes you go up one level from a node of the tree (my trees have their roots at the top) by taking the complement of the set at the lower node. Sometimes you have countably many nodes, and the node just above them represents the union of the sets at the lower nodes. You proceed up the tree, and when you get to the root, you have the Borel set you wanted to describe.

It's really hard for me to see what's "non-explicit" about this, unless you want to say you can't explicitly describe even the closed sets (after all, every singleton is closed in the reals, and you can't necessarily "explicitly" describe the real in the singleton). The only remaining argument that this is not explicit seems to boil down to discomfort with transfinite recursion. --Trovatore (talk) 18:39, 20 November 2008 (UTC)[reply]

Difference

What differs Sigma-algebra from Sigma-ring? Only first axiom of whole set to be member of sigma-algebra or something more? —Preceding unsigned comment added by 212.87.13.75 (talk) 02:58, 6 December 2008 (UTC) Ok, I see something: complement vs. relative complement. One could write something about differences and when one concept is used and when other (as with delta-ring and sigma-ring). 212.87.13.75 (talk) 03:01, 6 December 2008 (UTC)[reply]

There should be a definition or at least a link to "sigma-ring" Gsspradlin (talk) 21:55, 6 November 2013 (UTC)[reply]

A question ...

Is it the case that any sigma-algebra over X can be understood as the power set of X' where X' is X with elements lumped together (ie, with some sort of equivalence relation)?

To put it another way, does any sigma-algebra over X define an equivalence relation over X, by virtue of it having lumps that are not split into subsets by any intersection operation in the algebra? —Preceding unsigned comment added by Paul Murray (talk • contribs) 00:31, 13 January 2009 (UTC)[reply]

No, not even close. The smallest "interesting" sigma-algebra is usually considered to be the collection of all Borel sets (if X is a topological space), and every singleton is Borel (provided X is "reasonable") so there aren't any lumps that don't get split — and yet, it's certainly not true that every element of the powerset of X is Borel. --Trovatore (talk) 00:50, 13 January 2009 (UTC)[reply]

Oh, except I suppose I should say that your "put it another way" is true — you can indeed define an equivalence relation based on whether two points cannot be split by sets in the sigma-algebra. In the typical case, though, that equivalence relation will be just the identity, and it won't get you what you asked for in your first sentence. --Trovatore (talk) 01:29, 13 January 2009 (UTC)[reply]

On the other hand, the idea about "elements lumped together" works for sigma-algebras on a countable set. Yes, I know, this case is too simple for measure theory. However, undergraduates usually see only such simple probability spaces, and accordingly they are puzzled, why the fuss with sigma-algebras. Boris Tsirelson (talk) 16:25, 20 January 2010 (UTC)[reply]

countable union

Why countable? Does the theory break in some way if uncountable unions are allowed? 207.241.239.70 (talk) 04:02, 8 March 2009 (UTC)[reply]

If uncountable unions are allowed non-measurable sets would be included which goes against the motivation for introducing sigma-algebras. For example, the Borel sigma-algebra contains the singletons and if uncountable unions are allowed it would also contain all subsets of R. But the Borel sigma-algebra does not contain all subsets of R (e.g. Vitali sets or other non-measurable sets) Zubbuzzub (talk) 04:03, 13 January 2011 (UTC)[reply]

Well, it depends on what you mean by "allowing uncountable unions". If you allow arbitrary unions, then sure. But suppose you just mean, say, unions of

\aleph _{1}

-many sets.

In that case, if the continuum hypothesis holds, you get back arbitrary sets of reals. But if CH fails and, for example, Martin's axiom holds, then the union of

\aleph _{1}

-many measurable sets is still measurable. --Trovatore (talk) 07:38, 13 January 2011 (UTC)[reply]

(Oh, by the way, we really shouldn't be discussing this on this page. Talk pages are for discussing improvements to the article. A better venue for asking the question would have been the mathematics reference desk, located at Wikipedia:Reference desk/Mathematics.) --Trovatore (talk) 07:40, 13 January 2011 (UTC)[reply]

relationship to probability

No thanks to wikipedia or any other resources online, I only recently figured out how these connect to probability (ie that a probability space is a set, along with a sigma algebra and a probability measure). A particularly important (and probably obvious to most of the editors but not to a beginning undergrad looking through wikipedia's math articles) insight was that the probability spaces commonly encountered in undergrad probability have as a sigma algebra nothing more than the power set of the sample space. —Preceding unsigned comment added by 128.174.230.125 (talk) 15:43, 22 May 2009 (UTC)[reply]

It is a pity that you did not see Probability space#Discrete case; the point is explained there. Being interested in probabilistic application you should first look there, not here. Boris Tsirelson (talk) 19:28, 20 January 2010 (UTC)[reply]

Too technical?

This page has had the "Too technical" template added to it. Since I saw no discussion on this page about the reasons for the template's addition, and since, in my opinion, the article is at about the right technical level for the likely or intended audience, I removed the template. But my edit was reverted, so I'm attempting to start a discussion here. In particular I'd like to know other editor's opinions on this article, the reasons why this article is thought too technical, and for suggestions for how this article might be made usefully less technical. Paul August ☎ 15:12, 20 January 2010 (UTC)[reply]

Maybe some informal explanations (in probabilistic context) given in Citizendium could give some ideas useful here? Boris Tsirelson (talk) 16:18, 20 January 2010 (UTC)[reply]

This suggestion lets us recall that there are other articles already on wikipedia that are closely connected, including probability space and maybe others, that need to be considered. Melcombe (talk) 16:45, 20 January 2010 (UTC)[reply]

My reason for replacing the "Too technical" template was, as stated in the edit descriptor, that there is not even an attempt at having an understandable lead section as required by WP:LEAD. I think that, if an essentially non-mathematical lead section can be constructed, then the template could be removed immediately, otherwise a general reader has no chance of understanding what the article is trying to cover. To progress I state an off-the-cuff version of a start to what is required:

A σ-algebra (or sigma-algebra) is a type of object required in mathematics in order to provide a generally applicable basis for measure theory and probability theory. In simple cases where one starts with a finite list of items, a corresponding sigma-algebra would be a list of all the possible collections of items from this list, including those containing no items and all items. The eventual objective is to be able to assign to each item in this constructed list a value or "measure", in such a way that that the overall set of such values obey certain important rules. For general use this concept is extended to apply in cases where the initial list of items can be replaced and applied in much more complicated settings.

In probability theory and statistics, the sigma-algebra for a given scenario provides the basis for allowing the definition of the probabilities involved since it describes the most general types of outcomes to which probabilities must be assigned.

The existing start, with its mathematics, could be made into a "Introduction" section, perhaps moving some lines up into the lead. As you can probably tell, I am not well up on this stuff, so do feel free to ignore the above suggestion. Melcombe (talk) 16:40, 20 January 2010 (UTC)[reply]

I think I understand Melcombe's point. The issue is not that the lead is too difficult or too advanced, but that it is too technical. As written, the lead is quite clear and understandable, but it does not illuminate the subject because it gives no sense of what a sigma-algebra is for or why it is defined as it is. —Dominus (talk) 16:50, 20 January 2010 (UTC)[reply]

I've added a new section called "Motivation" (I did this before noticing any the above suggestions or remarks, and haven't thought yet how to roll any of these in.) Does this help at all? Paul August ☎ 17:04, 20 January 2010 (UTC)[reply]

I think it's a big improvement; thanks. —Dominus (talk) 18:04, 20 January 2010 (UTC)[reply]

I rearranged the introduction a little to put a slightly more intuitive part ahead of the formal definition. Feel free to revert if you think it's not agreeable. Ray^Talk 18:07, 20 January 2010 (UTC)[reply]

Algebra of sets

Shouldn't there be some link to Algebra of sets at the beginning of the article since it belongs to this type of algebra? --kupirijo (talk) 12:00, 5 October 2010 (UTC)[reply]

I agree, and I've replaced a link to Boolean algebra to a link to Algebra of sets (the latter is a special case of the former). Boris Tsirelson (talk) 13:45, 5 October 2010 (UTC)[reply]

Notation and naming

Names with words should always refer to the same concept. Anytime notation and/or naming has been abused (which is to be avoided), it has to be made clear. As an example, in the paragraph that starts with "Elements of the σ-algebra are called measurable sets..." (section Sigma-algebra#Definition_and_properties), it is stated that "A function between two measurable spaces is called measurable function..." I guess that a measurable function f should be either f:X→X' or f:Σ→Σ', but not f:(X,Σ)→(X',Σ') as stated in the text. A related point: in section Talk:Sigma-algebra#Notation (X,Σ) is called the sigma-algebra, while in the main article it is Σ the sigma-algebra and (X,Σ) is the measurable space. Names and underlying concepts should be unified. —Preceding unsigned comment added by Sanchin s (talk • contribs) 02:43, 7 December 2010 (UTC)[reply]

In principle I agree. But let me note that the notation f:(X,Σ)→(X',Σ') is a usual and convenient abuse in math literature. Boris Tsirelson (talk) 05:48, 7 December 2010 (UTC)[reply]

I don't see it as an abuse. These are the morphisms in the relevant category; (X,Σ) and (X',Σ') are the objects that start and end the arrow (I forget what categorists call those). It's a problem only if you insist on interpreting the notation as though it were a function from one set (X,Σ) to another set (X',Σ'), but why would you do that? --Trovatore (talk) 18:40, 8 December 2010 (UTC)[reply]

Well, assuming that everyone is acquainted with the language of morphisms... Boris Tsirelson (talk) 21:20, 8 December 2010 (UTC)[reply]

I don't think it's too much of an issue. If someone complains about it, we can give them Trovatore's explanation, and they will learn something. Most people will find it clear what the meaning is supposed to be, and won't complain or even notice that it's an issue. It's not like we're writing a whole paragraph about category theory; we're just using (what I think is) standard notation. It's no different than talking about an isometry as a function from a metric space (A, d₁) to a metric space (B, d₂). — Carl (CBM · talk) 21:29, 8 December 2010 (UTC)[reply]

Motivation Section

The explanation of measurable sets could do with a bit of clarification and wikification. For example, exactly which operations should we expect the privileged family of sets to be closed under. There should be links for every technical term, down to the level of "subset", including "operation" and "closure". —Preceding unsigned comment added by 74.176.113.247 (talk) 23:20, 29 December 2010 (UTC)[reply]

I found the following to be vague in this section:

"If the subsets of X in Σ correspond to numbers in elementary algebra, then the two set operations union (symbol ∪) and intersection (∩) correspond to addition and multiplication. The collection of sets Σ is completed to include countably infinite operations."

Should I take it to mean that there is only on possible Homomorphism between a sigma-algebra and elementary algebra or instead is the intent to say that a sigma algebra is analogous to elementary algebra? Further, if it is analogous to elementary algebra then in it would be helpful to know in what ways is it analogous? S243a (talk) 19:01, 24 October 2013 (UTC)[reply]

Yes, somewhat vague. I guess, the only point is that in elementary algebra we deal with numbers and two operations (addition and multiplication), and analogously in the algebra of sets we deal with sets and two operations (union and intersection). Does this help to beginners? I do not know. (I am far not a beginner.) In what ways is it analogous? Well,

(a+b)c=ac+bc

, and similarly,

(A\cup B)\cap C=A\cap C\,\cup \,B\cap C.

Not very deep... Boris Tsirelson (talk) 20:38, 24 October 2013 (UTC)[reply]

pi-system and d-system?

I don't see mentioned in the article, but I seem to recall that sigma algebras are both pi-systems and d-systems. If that's true, can we get that added to the article at an appropriate place? 70.247.162.18 (talk) 00:02, 3 November 2013 (UTC)[reply]

Relation to sigma-ring

There is a section entitled "Relation to sigma-ring", but there is no definition of what a sigma-ring is or even a link to a definition. I think the term "sigma-ring" is much less well-known than sigma-algebra, even to mathematicians, and there should be a link or definition, or the section should simply be omitted. Gsspradlin (talk) 21:53, 6 November 2013 (UTC)[reply]

I added a wikilink to sigma-ring and a book ref giving the definition of a sigma ring, but the section could still use sourcing for the particular assertions made there. --Mark viking (talk) 00:48, 7 November 2013 (UTC)[reply]

Construction

The Borel σ-algebra cannot be constructed in a very real sense, unless you want to think of it as a limit of ever-increasing collections of sets. To do that, you start with (for example) the open sets, then include all countable intersections of those sets, then the next step is to include all countable unions of the collection form the first step, then countable intersections from the second step, and so on. "ad infinitum" is a perfectly apt description of this process.

One common misconception about Borel sets is that they can all be represented in some fashion (for example, constructed in terms of unions of intervals). Unfortunately, this is nowhere close to the truth.

It is really the word "construction" that I object to. Perhaps there is another way to explain it without (at least at this point in the article) referring to "transfinite" operations. Daren Cline 19:06, 4 March 2014 (UTC) — Preceding unsigned comment added by Darencline (talk • contribs)

OK, so here's my issue. I'm going to introduce some standard notation; please don't be offended if you already know it, it'll save time for me to go ahead and write it down rather than waiting to find out whether you know it or not.

The open sets are called

\Sigma _{1}^{0}

; their complements, that is the closed sets, are

\Pi _{1}^{0}

. Countable unions of closed sets are

\Sigma _{2}^{0}

, and their complements are

\Pi _{2}^{0}

, and so on.

So it's faster to describe if we just skip the pi's altogether, and just say that

\Sigma _{n+1}^{0}

is all countable unions of complements of

\Sigma _{n}^{0}

sets.

Now, the way you describe it, it sounds like you just take

\Sigma _{1}^{0}

,

\Sigma _{2}^{0}

,

\Sigma _{3}^{0}

, and so on, "ad infinitum"; well, that sounds like you mean you just take the union of all the

\Sigma _{n}^{0}

for n a positive natural number.

But are you done after that? You are not. ~~The union of all the $\Sigma _{n}^{0}$ is called $\Sigma _{\omega }^{0}$ ,~~ and it is not closed under complements. So you can take all the countable unions of complements of

\Sigma _{\omega }^{0}

sets, which gives you

\Sigma _{\omega +1}^{0}

, then you do it again to get

\Sigma _{\omega +2}^{0}

, and so on. The union of all the

\Sigma _{\omega +n}^{0}

gives you

\Sigma _{\omega \cdot 2}^{0}

, and you can keep going.

This process finally closes off at

\Sigma _{\omega _{1}}^{0}

, where

\omega _{1}

is the smallest uncountable ordinal. You can keep on going if you like, nothing stops you, but you never get anything new after that.

Emending this after Daren helpfully pointed out my blunder here.

\Sigma _{\omega }^{0}

is not the union of all the

\Sigma _{n}^{0}

. A set in

\Sigma _{\omega }^{0}

is a union of countably many sets, each of which is from some

\Sigma _{n}^{0}

(not necessarily the same n for each of the sets). --Trovatore (talk) 19:11, 7 March 2014 (UTC) [reply]

Again, sorry for the elementary exposition if you already knew it.

So anyway, I just think that ad infinitum is a bit misleading; by saying something but not everything, it might make people think that the process closes off at ω, and in fact it's not even close to done at that point. I think if we're going to say anything about this, then we need to say something about iterating through countable ordinals. --Trovatore (talk) 19:53, 4 March 2014 (UTC)[reply]

I understand your notation, though analysts and probabilists have used other notation. For example, see Borel set, which would be an apt link here.

And I agree that "ad infinitum" may be a little glib. I was partly assuming that since the article is introductory at this point it just needs to be accurate, if still imprecise. (This isn't even what has provoked me most to think about contributing.) But I was also mis-remembering the theorem. Still, I don't see that $\Sigma _{\omega }^{0}$ is not closed under complements. I do see why it is not closed under countable unions.

Frankly, I tend to think that any discussion of a construction of a σ-algebra can lead to misconceptions. And so I like to just say to students that there is no representation for an arbitrary Borel set; we can only describe the collection in terms of classes of sets that generate the Borel σ-algebra. It also helps to bring home the point that proofs have to be done on the simpler classes and then "bootstrapped" to the Borel σ-algebra. (Which is why I also agree with the comment above that this article should refer to pi-system, lambda-system and the pi-lambda theorem, and their relevance to theory.)

My real concern here is accessibility of this article to graduate students (for example, in statistics) who know very little about analysis, let alone transfinite induction, but nevertheless do have to learn and understand σ-algebras and measure theory in order to do probability. Jumping from a finite example to constructing the Borel class tells them right away that this article isn't likely to help them much.

Indeed, the article is pretty much void of probability applications. So perhaps I can remedy that and then modify the introduction enough to show that aspect. I would also like to see more examples near the top of the article.Daren Cline 00:10, 5 March 2014 (UTC) — Preceding unsigned comment added by Darencline (talk • contribs)

OK, so a couple of points that we shouldn't dwell on on this talk page, but feel free to ping me on my talk page:

\Sigma _{\omega }^{0}

actually is closed under countable unions (at least, given enough of the axiom of choice to prove that the union of countably many countable sets is countable). It's not closed under complements, because a set in

\Sigma _{\omega }^{0}

might be, say

A_{1}\cup A_{2}\cup A_{3}\cup \ldots

, where each

A_{n}

is in

\Sigma _{n}^{0}

. But then if you take the complement, that gives you an infinite intersection of sets in the corresponding

\Pi _{n}^{0}

, and there's no reason that has to be in

\Sigma _{\omega }^{0}

.

Saying that there's "no" representation for an arbitrary Borel set is not quite right. You can come up with a reasonably canonical scheme for representing them, given an arbitrary real parameter. (That's the best you can hope to do, because for any real number r, {r} is a Borel set.) It is admittedly probably something you wouldn't throw at first-year grad students in statistics.

More on topic — I think the idea of adding more probability stuff, and mentioning it in the lead, is a good one; please go ahead. --Trovatore (talk) 01:08, 5 March 2014 (UTC)[reply]

Relevance to Probability

As mentioned above, the article currently gives the basics of the definition and few examples. There is some effort to discuss the use of σ-algebras for measure theory, but not much is said about probability. I would like to incorporate (or see incorporated) a number of points, including the following.

more examples, such as
- σ-algebra generated by a partition of X
- σ-algebra generated by the cylinder sets on the spaces $\scriptstyle \mathbb {R} ^{\infty }$ and $\scriptstyle \{0,1\}^{\infty }$ (with special attention to the latter as it would be easier to follow and is relevant for coin toss experiments)
a section on the π-λ theorem, which could be short as this topic is covered elsewhere
a section specifically on uses for probability, including
- σ-algebra generated by a random variable or random process
- use of σ-algebras in the definition of conditional expectation
- filtrations of σ-algebras and their use in martingales and Markov processes

--Daren Cline (talk) 00:47, 6 March 2014 (UTC)[reply]

On filtrations, we have Natural filtration. Boris Tsirelson (talk) 06:35, 6 March 2014 (UTC)[reply]

Thanks. I can use help with links to other pages. I don't intend for this to be lengthy; just to show more uses for σ-algebras. Daren Cline (talk) 14:25, 6 March 2014 (UTC)[reply]

I have begun to draft some ideas for inclusion which may be accessed from my user page. Comments are very welcome!

Also, I have a couple thoughts about rearranging the current material. First, the current last paragraph of the Motivation section would seem to be more appropriately placed in the Definition section. Indeed some of the same ideas are already there. Second, the two subsections about generating σ-algebras don't quite fit the idea of "properties", so I propose a new section just on that topic. It could include some of the more general examples I had in mind as well. Or, it could be a subsection of the Examples section. --Daren Cline (talk) 20:27, 10 March 2014 (UTC)[reply]

(The draft is here: User:Darencline/sigma algebra.)

I added two concepts to the Motivations section. The first, limit of sets, is supposed to indicate one simple reason why closure under countable unions and intersections is necessary. However, in order to do this I felt it necessary to define set limits. Maybe that's straying a bit off topic, but it is also true that the definition is not usually known very well, and can be at odds with one's intuition based on limits of points. (Is there an article for this specific topic?)

Yes, a short one: Set-theoretic limit. Boris Tsirelson (talk) 14:48, 12 March 2014 (UTC)[reply]

Short is right! My subsection here could be copied there, and expanded. Another item for my list ... Daren Cline (talk) 15:47, 12 March 2014 (UTC)[reply]

The second, use of sub σ-algebras, is extremely important to probability and this just presents the idea which will be expanded on in a section specific to probability uses of σ-algebras. --Daren Cline (talk) 14:33, 12 March 2014 (UTC)[reply]

"A useful property is the following: for two functions f and g defined on X, σ(f) ⊂ σ(g) if and only if there exists a measurable function h such that f(x) = h(g(x))." — First, this is often called the Doob-Dynkin lemma. Second, I doubt: does it hold as written, or rather, modulo null sets? Moreover, working with σ-algebras on a probability space, I (and many others) always include all null sets in every σ-algebra (by assumption or by construction). Then we have a one-to-one correspondence between σ-algebras and operators of conditional expectation. Otherwise we have too many different σ-algebras. Boris Tsirelson (talk) 07:20, 13 March 2014 (UTC)[reply]

No, really, I should not doubt: the claim holds as written. However, for now it is not clear which kind of functions are considered. For an arbitrary (Y,B) the claim definitely fails. For Rⁿ with the Borel σ-algebra it holds.Boris Tsirelson (talk) 08:46, 13 March 2014 (UTC)[reply]

Admittedly, I was a bit hasty and imprecise. But, from a set theoretic perspective, the claim seems to me to be a tautology. (Null sets are not relevant - no measure is involved.) What am I missing? Do you have a counterexample? - you say it definitely fails for arbitrary (Y,B). Daren Cline (talk) 01:38, 14 March 2014 (UTC)[reply]

Not a tautology. Consider a perforated interval. Its identity map to itself is not a measurable function of its identity map into the [0,1]. Boris Tsirelson (talk) 06:13, 14 March 2014 (UTC)[reply]

Ok, revised the statement to the real vector-valued case, though I'm sure it applies much more generally. But I'm not following your example. It sounds like you are saying you have f(x) = x maps Y = perforated interval to Y and g(x) = x maps Y to [0,1]. I presume that the measurable space for Y is any that contains both the Borel sets and Y. But is σ(f) contained in σ(g)? It looks to me more like σ(g) is contained in σ(f), in which case g = g(f) is a measurable function of f. — Preceding unsigned comment added by Darencline (talk • contribs) 18:04, 15 March 2014 (UTC)[reply]

Right, σ(g) is contained in σ(f), too; they are equal, in fact (and they both are equal to the whole σ-algebra of Y). But this does not invalidate my example. Boris Tsirelson (talk) 18:45, 15 March 2014 (UTC)[reply]

About "it applies much more generally": well, it applies (at least) to all standard Borel spaces. This is indeed "much more" in the sense that it is sufficient for most applications. On the other hand, up to isomorphism, it is just a single space (plus finite and countable spaces, of course). About the astonishing phenomenon of absence of the needed measurable map (that seems to be tautology), see also here. Boris Tsirelson (talk) 06:20, 16 March 2014 (UTC)[reply]

Alright. It is fixed now, with a reference to Kallenberg's probability book. --Daren Cline (talk) 15:50, 17 March 2014 (UTC)[reply]

Nice. Boris Tsirelson (talk) 16:47, 17 March 2014 (UTC)[reply]

Discrete σ-Algebra?

In the examples section currently is the statement that the power set is also known as the discrete σ-algebra. I've been wondering about this as it was new to me. In fact, I think the σ-algebra generated by the discrete topology is something quite different: the collection of countable subsets and their complements. Does anyone have thoughts on this? Should the wording be changed? --Daren Cline (talk) 20:09, 28 March 2014 (UTC)[reply]

No, why? The discrete topology means, all sets are open, right? Not only countable/cocountable. Boris Tsirelson (talk) 16:35, 29 March 2014 (UTC)[reply]

Another case is, the topology such that finite sets are closed, and only finite (and the whole space). This one indeed generates the countable/cocountable σ-algebra. Boris Tsirelson (talk) 16:39, 29 March 2014 (UTC)[reply]

Yes, that's what I was thinking of: x_n can only converge to x if there exists N such that x_n = x for all n ≥ N, right? ~~I guess I've forgotten the term for that – will have to look it up.~~ (correction: that kind of convergence does imply singleton sets, and thus every set, is open.) --Daren Cline (talk) 16:53, 29 March 2014 (UTC)[reply]

By the way: at most countable space is standard if all subsets are measurable (which is often meant); otherwise the corresponding quotient space (still, at most countable, of course) is standard. Boris Tsirelson (talk) 16:43, 29 March 2014 (UTC)[reply]

Of course, but I would expect it to be assumed. --Daren Cline (talk) 16:53, 29 March 2014 (UTC)[reply]

Integral at end of Dynkin's pi-lambda theorem

I'm pretty sure that F() appearing in that integral is the probability density function (PDF), not the cumulative density function (CDF). (We integrate PDFs to get probabilities; we take differences of CDFs to get probabilities.) Could someone who *knows* the correct answer either correct the integral or add a note about how this unusual use of the CDF is correct? -- 216.137.30.166 (talk) 00:59, 6 September 2015 (UTC)[reply]

Do not be so sure. The integral with the PDF (if exists) would be

\int _{A}F(x)\,dx

rather than

\int _{A}\,F(dx).

Boris Tsirelson (talk) 05:16, 6 September 2015 (UTC)[reply]

The notation is quite standard and usual:

\int _{A}\,F(dx)

is a Lebesgue-Stieltjes integral. (The sentence itself says so.) Such integrals are used, for example, because one does not wish to make any assumption about the type of the distribution (discrete, continuous, etc.). In this case, the statement and expression would be correct even if

X

was a random vector. Daren Cline (talk) 13:08, 7 September 2015 (UTC)[reply]

@216.137.30.166: There is no such thing as a "cumulative DENSITY function". The words "cumulative" and "density" contradict each other. The letters "CDF" stand for "cumulative distribution function." Michael Hardy (talk) 18:37, 23 February 2019 (UTC)[reply]

merging separable sigma algebra into sigma-algebra

The consensus is to merge separable sigma algebra into sigma algebra. Cunard (talk) 00:08, 1 August 2016 (UTC)

The following discussion is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.

hi guys,

want to put out a feeler for the appetite to merge separable sigma algebra into sigma algebra. i feel the former could be slotted in under a subsection of the latter.

hopefully this is not too radical? the page for separable sigma algebra is quite bare and i think merging into sigma algebra will improve its exposure.

thoughts? 174.3.155.181 (talk) 21:40, 29 June 2016 (UTC)[reply]

I think the existing content at separable sigma algebra could easily be merged here, yes. If there is enough to say about them to justify a separate article, then anyone who wants to write that article should feel free to undo the merge at that time. --Trovatore (talk) 21:46, 29 June 2016 (UTC)[reply]

thank you. i have done the merge.... properly... i think? :P 174.3.155.181 (talk) 22:03, 29 June 2016 (UTC)[reply]

Endorse the merge as the obviously right thing to do (the target article is not unduly long, nor is the section that was merged, and the content surely fits well there). Summoned by bot; I suggest that the next person the RfC-bot summons close this RfC; I do not think it was a necessary step to open it in the first place.

The merge was properly done; the important thing to do was to say where it was merged from in the edit summary (because our license requires attribution). I added the wiki-templates merged-to and merged-from (they are good to use, though not compulsory). Tigraan^{Click here to contact me} 09:28, 22 July 2016 (UTC)[reply]

Well, but is it an example? Would you put "Continuous functions" into "Examples" of "Function"? Boris Tsirelson (talk) 17:46, 22 July 2016 (UTC)[reply]

No, because continuous function is a large article already. If it was included in function (mathematics), it would qualify for a WP:SIZESPLIT.

If your argument is that mentioning SSA in SA is undue weight, despite its small size, then a fortiori it should not have a standalone article. Tigraan^{Click here to contact me} 09:24, 25 July 2016 (UTC)[reply]

I rather mean that this is an example of a useful property of sigma-algebras; "Simple set-based examples" are indeed examples of sigma-algebras, but separable σ-algebras are just a wide class of sigma-algebras. Boris Tsirelson (talk) 15:39, 25 July 2016 (UTC)[reply]

OK, I misunderstood you. Well, to be honest, I do not feel competent enough to appreciate how the target page is organized, but maybe "examples" should be retitled "particular cases" or something like that? Be bold and edit, if you know what to do. Tigraan^{Click here to contact me} 16:58, 25 July 2016 (UTC)[reply]

Yes I did. Boris Tsirelson (talk) 17:04, 25 July 2016 (UTC)[reply]

The discussion above is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.

Separable sigma algebra

The last sentence in this section says "If the measure space is separable, it can be shown that the corresponding metric space is, too." But in the previous paragraph, "separability" of a measure space was defined to be separability of the corresponding metric space. Perhaps there's some other (equivalent) way of defining it, and this got lost when the separable sigma algebra page was merged into the sigma algebra page? — Preceding unsigned comment added by 73.114.17.117 (talk) 15:45, 4 October 2016 (UTC)[reply]

Yes, you are right. And the equivalent way is this: a sigma-algebra is separable if and only if it is generated by some countable subset together with all sets of measure zero. At least, this is the case for finite measures; full generality may be more complicated.

Some more weak points of this section. First, the notion of a separable sigma-algebra is well-defined only if a measure is given (or at least, an equivalence class of measures, which really means, the sigma-ideal of null sets). Second, the measure need not be finite, while a metric is, by definition, finite. Thus, one should take not just the measure of the symmetric difference, but rather, the minimum of 1 and that measure. Boris Tsirelson (talk) 17:28, 4 October 2016 (UTC)[reply]

π-system generating the join

It is stated that:

A π-system that generates the join is

{\mathcal {P}}=\left\{\bigcap _{i=1}^{n}A_{i}:A_{i}\in \Sigma _{\alpha _{i}},\alpha _{i}\in {\mathcal {A}},\ n\geq 1\right\}.

Shouldn't this π-system be defined using the union of $A_{i}$ instead? — Preceding unsigned comment added by Sopasakis p (talk • contribs) 11:52, 13 August 2017 (UTC)[reply]

Surely not. The join must contain both the union of

A_{i}

and the intersection of

A_{i}

; but such unions are not a π-system, while such intersections are a π-system. Boris Tsirelson (talk) 17:03, 13 August 2017 (UTC)[reply]

Is intersection needed to generate?

In "σ-algebra generated by an arbitrary family", it says the generated algebra is made from the elements of F via a countable number of complement, union, and intersection operations. But intersection can be defined in terms of union and complement: A intersection B = (A' union B')' via De Morgan's laws. Plus, the sigma algebra itself is defined as being closed under compliment and union, so why add intersection to this section? Vityou123 (talk) 08:26, 14 September 2023 (UTC)[reply]

Good faith "edit", but ... [maybe not needed; AND!] grammatical errors

Not vandalism

I am confident that the most recent edit (see the diff listing) was made with good intentions, but the grammar is faulty, and some topics, "such as" ... one topic that may be relevant here ... that is, the issue of

whether or not the definition given in the 'lede' of this article should explicitly state certain things even if they are already 'implied' (or "required") by some other parts of that definition -- (e.g., that in order for a nonempty collection Σ of subsets of X to be called a σ-algebra, Σ must contain X itself)

have already been discussed -- at length -- on this very "Talk:" page.

This section of this "Talk:" page is not intended to address the above mentioned (in a "bullet" item) "issue". Instead, this section is intended mainly to call attention to some grammatical errors in the first sentence of this article, after the (currently) most recent edit.

Grammatical errors

The addition of "is", ... without saying "that is"
IMHO a school teacher would correct this kind of wording. Failing to fix this would not be a [good] option.
The addition of "contain X itself" ... which uses the [plural] verb "contain", while the noun -- the word "collection" -- that serves as the "subject" for that "predicate" is singular. This violates a grammatical rule about 'subject' / 'predicate' "agreement" in number.

Also, the addition of "contain X itself" ... [if not reverted] should perhaps be moved to a separate sentence. This "addition" strikes me as somehow failing to be 'worded' correctly, in order to maintain consistency with the ['requirement'] items already in that list ... which are [consistently] adjectival phrases. (Would it be inconsistent only if we fixed "is" to say "that is"? That doesn't help, as leaving it saying "is" instead of "that is" ... would be incorrect.) I think it -- (the phrase "contain X itself") -- should be removed (that is, I think the addition of it should be reverted); but if it is not, then ... moving it to a separate sentence would afford an opportunity to mention the fact that this situation -- (the fact that Σ must "contain X itself") -- while TRUE, is already 'implied' (or "required") by some other rules in the definition.

What should be done, to fix this?

Personally, I would be fine with just reverting the most recent edit ... the one which created the "Latest revision as of 17:31, 25 February 2024" version of the article.

However, I was hesitant to do so (without discussing it here first) partly because there might not have been enough space in the edit comment to express the ideas included here ... in order to make it clear that

I realized that it was probably a "good faith edit",
and that
my ignorance about the customs within math about how to explain things -- (and preferably, to do so the best way) -- might be nonzero, and it might exceed your ignorance -- ('if any') -- on that topic, but I did not completely ['100%'] disregard (I did not ignore) those customs.

Any comments?

Thanks for listening.
If there are no objections, then ... I intend to revert the most recent ["17:31, 25 February 2024"] edit. Mike Schwartz (talk) 18:57, 27 February 2024 (UTC)[reply]

Discussion is good. But a four-hundred-word essay with three subheadings for a grammar complaint about a single sentence might be overdoing it a tiny bit. Revert if you like, or fix the grammar if you think the nontriviality condition should be explicit, and we can revisit the latter point. --Trovatore (talk) 20:32, 27 February 2024 (UTC)[reply]

Thank you ... for your comments, and for your patience. I reverted the edit of 25-Feb-2024, and now the first sentence of the article is back to saying -- (as it did 'as of' 17:52, 11 January 2024') --

In mathematical analysis and in probability theory, a σ-algebra (also σ-field) on a set X is a nonempty collection Σ of subsets of X closed under complement, countable unions, and countable intersections.

I would be OK with having the phrase "that is" -- (but not the word "is", by itself) -- appear, just before the only occurrence of the word "closed" in that lede sentence. But that phrase is probably not needed at all, there.

I do not know what should be done -- maybe nothing -- about the fact (implied by other parts of the definition, but not explicitly stated in that lede sentence) that, if Σ is to be a Σ-algebra, then Σ must "contain X itself". --Mike Schwartz (talk) 03:30, 29 February 2024 (UTC)[reply]