Talk:Sigma-algebra

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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)

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)

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)

1 Font used for denotion in math papers is not important?
2 question
3 Name and history
4 Notation
5 Relation to field of sets
6 Boolean algebra
7 Families?
8 Meaning of terms
9 Definition of Sigma-Algebra
10 Adjectives and Nouns
11 empty set
12 "... if and only if X is uncountable. "
13 Sigma homomorphism
14 Is the power set always a sigma-algebra?

[edit] 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)

[edit] 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)

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.

[edit] Name and history

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

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)

[edit] 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)

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)

[edit] 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)

[edit] 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)

[edit] 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)

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

[edit] 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)

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)

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)

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)

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)

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)

[edit] 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)

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)

[edit] 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)

[edit] 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)

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

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)

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)

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

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)

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)

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)

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)

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)

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)

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)

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)

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)

[edit] "... 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))

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)

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)

[edit] 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)

[edit] 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)

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)

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