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Homework Help: Sigma-Algebra problem

  1. Sep 6, 2007 #1
    If we define two sets as

    [tex] A = \left\{ (x,y)\in \mathbb{R} \vert x+y<a \right\} [/tex]
    [tex] B = \left\{ (x,y)\in \mathbb{R} \vert x-y<a \right\} [/tex]

    and then define the system of sets,

    [tex] \mathbb{D} = \left\{A\vert a\in \mathbb{R}\right\}\cup\left\{B\vert a\in \mathbb{R}\right\} [/tex]

    and then let [itex] \sigma(\mathbb{D}) [/itex] be the [itex] \sigma [/itex]-algebra generated by the system of set [itex] \mathbb{D} [/itex]. To show that the sets

    [tex] C = \left\{ (x,y)\in \mathbb{R} \vert x+y\leq a \right\} [/tex]
    [tex] D = \left\{ (x,y)\in \mathbb{R} \vert x-y\leq a \right\} [/tex]

    are elements in [itex] \sigma(\mathbb{D}) [/itex] for every [itex] a\in \mathbb{R} [/itex].
    Some questions arises. Can the sets A and B be considered as elements in [itex] \mathbb{D} [/itex] and hence in [itex] \sigma(\mathbb{D}) [/itex] too, or are they just subsets of [itex] \mathbb{D} [/itex]?
    Now I want to express C as a union of A and the set [itex] S = \left\{ (x,y)\in \mathbb{R} \vert x+y=a \right\} [/itex], and similar for D. But then I have to express the new set in some way to show that it is contained in [itex] \sigma(\mathbb{D})[/itex]. (Maybe there are easier approaches?). I was considering to write the new set S as

    [tex] S = \bigcap_{n=1}^{\infty}\left\{ (x,y)\in \mathbb{R} \vert x+y<a+\frac{1}{n} \right\}\cap \left\{ (x,y)\in \mathbb{R} \vert x+y\geq a \right\} [/tex]

    Where the last set is the complement of A. But will this set even converge towards S, or will it just be empty?
    Last edited: Sep 6, 2007
  2. jcsd
  3. Sep 6, 2007 #2


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    I think A and B are not subsets. {A} and {B} are subsets. But, isn't [itex]\sigma[/itex] defined over subsets of [itex] \mathbb{D}[/itex]?

    I think S is empty, but [itex]\bar S[/itex], defined with an < (instead of an <) is non-empty. See http://en.wikipedia.org/wiki/Nested_sequence_of_intervals
    Last edited: Sep 7, 2007
  4. Sep 7, 2007 #3
    So A and B are elements in [itex]\sigma(\mathbb{D})[/itex], right? And yes [itex]\sigma(\mathbb{D})[/itex] is defined as the smallest [itex]\sigma-[/itex]algebra containing [itex]\mathbb{D}[/itex].

    Hmm... Now you have changed your answer of weather S is empty or not. I thought that maybe it would be non-empty because of the way we (in class) defined an interval of the type (a,b] as

    [tex] (a,b] = \bigcap_{n=1}^{\infty}\left(a,\, b+\frac{1}{n}\right) [/tex]

    as a intersection of open intervals.
    But hmm... then again maybe this wasn't even a good inspiration, since it seems different now that I type it...

    But how would you suggest to write S from the 'building blocks' of [itex]\sigma(\mathbb{D})[/itex]?

    The question really is just, will the infinity intersection be

    [tex] \bigcap_{n=1}^{\infty}\left\{ (x,y)\in \mathbb{R} \,\vert\, x+y<a+\frac{1}{n}\right\} = \left\{ (x,y)\in \mathbb{R} \,\vert\, x+y\leq a\right\} [/tex]

    Because that is the set I'm intrested in. And that's why I considered that if (a,b] can be writting in that way, why can't the above equation be true?
    By this expression S becomes completly useless, since I can just set C equal to that expression, and C will then consist of an infinite intersection of A sets. But then we're back to my first question, are A and B elements in the [itex]\sigma[/itex]-algebra even though it is generated by the union of A and B, so that C is can be considered as an element in the algebra?
    Last edited: Sep 7, 2007
  5. Sep 7, 2007 #4
    Ok instead of making yet another edit, I'll make a new reply, getting confusing wit so edits now.
    I think I was confused about [itex] \mathbb{D} [/itex], since I saw it a union of 2 sets, but it's actually a union of 2 system of sets. So the [itex] \mathbb{D} [/itex] is actually,

    [tex] \mathbb{D} = \left\{ A \subset \mathbb{R}^2 \,\vert\, A\in \{A_a\,\vert\, a\in\mathbb{R}\}\quad\mathrm{or}\quad A\in \{B_a\,\vert\, a\in\mathbb{R}\} \right\}[/tex]

    and I think that it is clear from this, that A and B are contained in [itex]\mathbb{D}[/itex] and hence also [itex]\sigma(\mathbb{D})[/itex].
  6. Sep 7, 2007 #5


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    I think I agree with
    [tex] \bigcap_{n=1}^{\infty}\left\{ (x,y)\in \mathbb{R}^2 \,\vert\, x+y<a+1/n\right\} = \left\{ (x,y)\in\mathbb{R}^2\vert x+y\le a\right\} [/tex]

    I think I was right the first time, then confused myself with nested intervals.
    Last edited: Sep 7, 2007
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