Talk:Σ-algebra

message: 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)
thread_title: Borel-field

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

message: 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)
thread_title:  Relation to [[field of sets]] 

message: 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)
thread_title: Boolean algebra

message: "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)
replies: text: You're right about that.   I've changed it to collection. -lethe talk 21:25, 25 January 2006 (UTC)
thread_title: Families?

message: 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)
replies: text: 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)

text: 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)
thread_title:  Definition of Sigma-Algebra 

message: 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)
thread_title:  Adjectives and Nouns 

message: "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)
replies: text: I think you are correct.  -lethe talk [ +] 23:15, 15 June 2006 (UTC)
replies: text:  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)
replies: text:  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)

text: 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)
replies: text:  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)
replies: text:  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)

text: 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)
replies: text:  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)
replies: text: 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)

text:  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)

text: I agree, completely; } is the smallest σ-algebra; no one doubts it, I believe. Boris Tsirelson (talk) 17:12, 8 April 2015 (UTC)
thread_title:  empty set 

message: 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)
replies: text: 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)
replies: text:  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)
thread_title:  "... if and only if X is uncountable. " 

message: 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)
replies: text: A source: Sect. 15.C in book "Classical descriptive set theory" by A.Kechris. Boris Tsirelson (talk) 18:56, 7 September 2015 (UTC)
thread_title: Sigma homomorphism

message: 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)
replies: text: 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)
thread_title: Is the power set always a sigma-algebra?

message: 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)
replies: text: 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)
thread_title:  Intersections of Sigma Algebras 

message: 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)
replies: text: 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)
thread_title:  Sigma-fields 

message: 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)
thread_title:  Difference 

message: 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)
replies: text: 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)
replies: text: 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)

text: 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)
thread_title:  A question ... 

message: 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)
replies: text: 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)
thread_title:  relationship to probability 

message: 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)
replies: text: 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)
thread_title:  Algebra of sets 

message: 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)
replies: text: 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)
replies: text: 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)
replies: text: Well, assuming that everyone is acquainted with the language of morphisms... Boris Tsirelson (talk) 21:20, 8 December 2010 (UTC)
replies: text: 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)
thread_title:  Notation and naming 

message: 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)
replies: text: 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)
thread_title:  Motivation Section 

message: 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)
thread_title:  pi-system and d-system? 

message: 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)
replies: text: 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)
thread_title:  Relation to sigma-ring 

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

text: 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)

text: 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)
thread_title: Discrete σ-Algebra?

message: 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)
replies: text: 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)

text: 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)
thread_title:  Integral at end of Dynkin's pi-lambda theorem 

message: It is stated that:
:A π-system that generates the join is
::\mathcal{P}=\left \{\bigcap_{i=1}^nA_i:A_i\in\Sigma_{\alpha_i},\alpha_i\in\mathcal{A},\ n\ge1 \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)
replies: text: 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)
thread_title:  π-system generating the join 

message: 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)
thread_title:  Is intersection needed to generate?