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Sigma Algebra

  1. Aug 24, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove that the σ-algebra generated by the collection of all intervals in Rn coincides with the σ-algebra generated by the collection of all open subsets of Rn.

    2. Relevant equations
    A σ-algebra is a nonempty collection Σ of subsets of X (including X itself) that is closed under the complement and countable unions of its members.

    3. The attempt at a solution
    1. The σ-algebra generated by the collection of all open subsets of Rn would also contain all closed subsets by complementation. It would also contain their unions to generate all partly open subsets. Therefore the σ-algebra generated by the collection of all open subsets of Rn contains all subsets of Rn including ∅.

    2. The σ-algebra generated by the collection of all intervals in Rn contains all intervals of Rn including ∅. Since an interval is a subset, every element of the σ-algebra generated by the collection of all intervals in Rn would coincide with the σ-algebra generated by the collection of all open subsets in Rn.

    How far off am I? Is there a better way to make this argument?
     
  2. jcsd
  3. Aug 24, 2012 #2

    Bacle2

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    You're on the right track, but careful on #2 : there are subsets of R^n that are not measurable.

    It is enough to show that you can generate _all intervals_ by using the open intervals

    alone.
     
  4. Aug 25, 2012 #3

    jbunniii

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    No, this isn't correct. The σ-algebra generated by the open subsets certainly does not contain ALL subsets of Rn.

    Let's introduce some notation to make things easier to express.

    Let I denote the set of all open intervals contained in Rn. Let O denote the set of all open subsets of Rn.

    Let [itex]\Sigma(I)[/itex] be the sigma algebra generated by I, and [itex]\Sigma(O)[/itex] be the sigma algebra generated by O. You need to show that [itex]\Sigma(I) = \Sigma(O)[/itex]. The standard way to prove set inequality is to prove inclusion in both directions:
    [tex]\Sigma(I) \subset \Sigma(O)[/tex]
    and
    [tex]\Sigma(O) \subset \Sigma(I)[/tex]
    Clearly [itex]I \subset O[/itex]. I suggest starting with this to make the argument that [itex]\Sigma(I) \subset \Sigma(O)[/itex]. This is the easier part of the proof.
     
  5. Aug 25, 2012 #4

    Bacle2

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    Just to add that there are infinitely many sets that are not generated by the

    open intervals. These last are "nice-enough" sets, in that they are measurable

    sets, but not "Borel Sets" , i.e., sets in the sigma-algebra generated by open

    intervals. You can show this using a cardinality argument to show that the

    cardinality of the collection of measurable sets is strictly larger than the cardinality

    of the sigma algebra generated by the open set.
     
  6. Aug 27, 2012 #5
    Thanks for the help Bacle2. Here is what I am turning in.

    Proof: Let I be the collection of all intervals of Rn, and Ʃ(I) be the σ-algebra generated by I. Let O be the set of all open subsets of Rn, and Ʃ(O) be the σ-algebra generated by O.

    To show: Ʃ(I) is a subset of Ʃ(O).

    Pick an arbitrary interval i from Ʃ(I). i is either open, closed, or partly open. If i is open, then i is a member of O → i is a member of Ʃ(O). If i is closed, then i is a member of Oc → i is a member of Ʃ(O). If i is partly open, then i can be constructed using the intersections of finite number of elements of O and Oc → i is a member of Ʃ(O).

    Therefore Ʃ(I) is a subset of Ʃ(O).
     
  7. Aug 27, 2012 #6

    jbunniii

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    [edit] - Sorry, my original post assumed that [itex]I[/itex] contained only OPEN intervals, because I misread something. Here is a modified comment:


    In fact, your argument only proves that [itex]I \subset \Sigma(O)[/itex]. This is because [itex]\Sigma(I)[/itex] contains sets that are not intervals. For example, the union of two intervals is not generally an interval.

    You still need an additional step to conclude that [itex]\Sigma(I) \subset \Sigma(O)[/itex]. Hint: what does it mean that [itex]\Sigma(I)[/itex] is the sigma algebra generated by [itex]I[/itex]?
     
    Last edited: Aug 27, 2012
  8. Aug 27, 2012 #7

    Bacle2

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    Just to make sure, ChemEng1: when you talk about open intervals in R^n, are

    you defining these as open "n-cubes" , i.e., sets of the form:

    (x1,x2,...,xn):

    a1< x1< b1 or a1 ≤ x1 <b1, or a

    a2<x2<b2 ,

    ..........

    .........

    an <xn <bn , ....., etc. ?

    i.e., all half-open, half closed combinations of these ?

    EDIT: these should be the components of the open intervals.
     
    Last edited: Aug 27, 2012
  9. Aug 28, 2012 #8
    I got confirmation from the professor that he uses the word "coincide" to mean "equal". So both directions of containment need to be proved. I am also changing the notation from Ʃ(x) to Ʃx to avoid confusion that Ʃ is a function.

    This was my first attempt to show [itex]\Sigma_{G}\subset\Sigma_{I}[/itex], but I don't think that it's right.

    Pick an arbitrary open [itex]G\in\Sigma_{G}[/itex]. Notice [itex]G=\bigcup_{k}G_{k}[/itex] where [itex]G_{k}=G\bigcap I_{k}[/itex]. [itex]I_{k}=[-k,k]\times...\times[-k,k][/itex] for k=1,2,... Thus [itex]G\in\Sigma_{I}[/itex]. Since [itex]\Sigma_{I}[/itex] is a σ-algebra, it contains all intervals. Then [itex]\Sigma_{I}\supset\left\langle\left\{all open sets \right\}\right\rangle=\Sigma_{G}[/itex]. Therefore [itex]\Sigma_{G}\subset\Sigma_{I}[/itex]
     
  10. Aug 28, 2012 #9

    jbunniii

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    This is already problematic, because an arbitrary set in [itex]\Sigma_{G}[/itex] is not necessarily open.

    I assume you are claiming that this follows because [itex]G\bigcap I_{k} \in \Sigma_{I}[/itex]. But you haven't demonstrated why this is true.

    This is true because [itex]\Sigma_{I}[/itex] contains [itex]I[/itex], the set of all intervals, not because [itex]\Sigma_{I}[/itex] is a σ-algebra. There are σ-algebras that don't contain any intervals, for example, the σ-algebra generated by all of the singletons in R^n.

    This equality is false. It is true that

    [itex]\left\langle\left\{all open sets \right\}\right\rangle \subset \Sigma_{G}[/itex]

    but the containment is proper/strict.

    [edit] Actually that depends on the meaning of the notation [itex]\left\langle\left\{all open sets \right\}\right\rangle[/itex]. What does the angle bracket notation mean here? Are you using it as a shorthand to mean "the sigma algebra generated by..."?
     
    Last edited: Aug 28, 2012
  11. Aug 29, 2012 #10
    Yeah. I noticed those problems when I went back to the problem (again). Here's where I am with the problem:

    Let I be the collection of all intervals of Rn and ƩI be the σ-algebra generated by I. Let G be the set of all open subsets of Rn and ƩG be the σ-algebra generated by G.

    Pick an arbitrary interval, i, from Rn. By definition, i is closed. → i is a member of CG. → <{I}> is contained in ƩG. → ƩI is contained in ƩG.

    Pick an artibrary open set, g, from Rn. g is a member of CI. → <{G}> is contained in ƩI. → ƩG is contained in ƩI.

    ƩI is contained in ƩG. ƩG is contained in ƩI. → ƩG equals ƩI.
     
    Last edited: Aug 29, 2012
  12. Aug 30, 2012 #11

    Bacle2

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  13. Aug 30, 2012 #12
    The definition the text gave for an Rn interval was the cross product of: av ≤ xv ≤ bv (v= 1, 2, ..., n). It acknowledged other intervals (open, semi open), but stated that intervals should be assumed to be closed unless specifically mentioned.
     
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