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Measurable Functions - Any Help Appreciated. (Very appreciated)

  1. Nov 3, 2011 #1
    1. The problem statement, all variables and given/known data

    I'm new to Measure Theory, and to be honest, I'm having a really hard time making any sense of it at all. My prof is a nice guy, but his approach to teaching involves giving zero worked solutions. This doesn't work for me. Personally I need to see solutions to get an idea of how to work these problems out, and get an idea of what everything means.

    I have a number of problems where I've been asked to find the smallest σ-algebras of subsets of ℝ where f:ℝ→ℝ is a measurable function. The first question is given below:

    2. Relevant equations

    f(x) = -1 if x <= 0, and f(x) = 1 if x > 0

    3. The attempt at a solution

    Don't laugh - as I said I really don't know what I'm doing here...

    I can see that sub-intervals of X = ℝ: (-∞, 0) and [0, ∞) are in one-to-one correspondence with the singleton set 2 = {-1,1}. So there are |P(2)| = 2^2=4 sets in the σ-algebra.

    Would the answer be: {∅,(-∞,0),[0,∞),ℝ} ?

    Am I even close to being on the right track here?
     
  2. jcsd
  3. Nov 8, 2011 #2
    Is this even slightly right?
     
  4. Nov 8, 2011 #3
    Spitz, to answer your question, you have to first ask whether [itex]\left\{\emptyset,(-\infty,0),[0,\infty),\mathbb{R}\right\}[/itex] is a sigma-algebra. If it is a sigma-algebra, check whether f is measurable with respect to this sigma-algebra. If yes, can you reduce the set such that the reduction is still a sigma-algebra and that f is still measurable with respect to it?

    To do these, you have to know the definition of a sigma-algebra and function measurability. You can consult your notes.
     
  5. Nov 9, 2011 #4
    My notes are terrible. I've looked at all the definitions, but if somebody would actually go through an example it would be easier to understand.
     
  6. Nov 9, 2011 #5
    Spitz, the definition of sigma-algebra is what you need first. If you can't find it in your notes, you can search wikipedia or wolfram for this. If you want to make sure, you can present the definition here so we can check it.

    So, I'll just comment about measurability here. If you're working with a real-valued function without any specific mention of sigma-algebras associated with the real space, then, in standard practice, one assumes that the sigma-algebra associated with it is a Borel sigma-algebra, denoted by B. The principal content of Borel sigma-algebra is standard open sets (open interval). A standard open set in real space can be regarded as the union of open intervals.

    Anyway, in this case, where we assume that B is the sigma-algebra of the real space (the one which is the range of f, not the domain of f), the definition of a measurable function is equivalent with the following.

    Suppose [itex]f:X\rightarrow\mathbb{R}[/itex] is real-valued function. Let M be the sigma-algebra of X. Then, f is M-measurable iff
    [tex]f^{-1}((a,\infty))\in M,\forall a\in\mathbb{R}.[/tex]

    It should be easy to check whether f is [itex]\left\{\emptyset,(-\infty,0),[0,\infty),\mathbb{R}\right\}[/itex]-measurable afterwards, assuming that [itex]\left\{\emptyset,(-\infty,0),[0,\infty),\mathbb{R}\right\}[/itex] is a sigma-algebra.
     
  7. Nov 9, 2011 #6
    Well, for each element of the set, the complement is also in the set. The union of all the elements, R, is also in the set. Am I wrong in assuming its a sigma-algebra?

    It's the easy part (f is measurable) that I'm having trouble with.
     
  8. Nov 9, 2011 #7
    Do you mean by R as the element of the sigma-algebra? I'll assume so. Yes, you're correct for those two conditions. You're missing one though. A sigma-algebra M of a space X must include the emptyset and X itself.

    So, have you shown whether [itex]\left\{\emptyset,(-\infty,0),[0,\infty),\mathbb{R}\right\}[/itex] is a sigma-algebra or not?

    Oh, and also, you also asked about the limits of set sequence in another thread a while ago? Have you found the answer?
     
  9. Nov 9, 2011 #8
    Well, X = R, so the empty set and X are also elements. Isn't that all that is required? I assume the only smaller sigma algebra would be the trivial one, which doesn't work.

    Yeah, I figured out the other question. I just hadn't thought of the events in terms of occurring infinitely often and every time.
     
  10. Nov 9, 2011 #9
    Hmm, let me get this straight. If M is a sigma-algebra of R (real space), then, M must be a subset of the power set of R. That is, M is of the form {emptyset, R, and some other subsets of R satisfying the former two conditions} (M is a set of sets, while R is a set of numbers). Is this what you have in mind?
     
  11. Nov 9, 2011 #10
    I think so. A = {∅,ℝ} being the smallest, and A = P(ℝ) being the largest. Sorry If this is in kindergarten English, but I'm doing five thing at the same time. I was assuming that f sends every x≤0 to -1, and every x>0 to 1. I'm assuming the above is a σ-algebra (and also the smallest one), I'm just confused about the measurable function part.
     
    Last edited: Nov 9, 2011
  12. Nov 9, 2011 #11
    Ok, then. I believe you're good on track. Just work on f measurability now.
     
  13. Nov 9, 2011 #12
    Can I just ask you one last question:

    To check if f is measurable, do I just show that {x ε (-∞,o]:f(x) > λ} for every x ε (-∞,o]. And so on for every element of the σ-algebra?

    Or do I have to show {f > λ} for every x ε ℝ?
     
  14. Nov 9, 2011 #13
    Sorry, I don't think I can make sense of what you've just said. Anyway, to show that f is measurable with respect to {emptyset, (-infinity,0), [0,infinity), R}, you have to show that:

    For every lambda, {x:f(x)>lambda} is a member of {emptyset, (-infinity,0), [0,infinity), R}, i.e., you have to show that {x:f(x)>lambda} is either empty, R, (-infinity,0), or [0,infinity), for every lambda.
     
  15. Nov 9, 2011 #14

    micromass

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    Your answer is totally correct. I have no idea why canis is being so difficult.
     
  16. Nov 9, 2011 #15
    oh. Well thanks for both of your "helps" anyway.

    Can I just say that:

    [tex]f^{-1}(1)=(0,\infty)[/tex], which is measurable.

    [tex]f^{-1}(-1)=(-\infty,0][/tex], which is also measurable.

    So f is a measurable function for my σ-algebra?
     
  17. Nov 9, 2011 #16
    Spitz, as micromass have told you, you've already had the right answer from the beginning. Sorry for being so difficult. It maybe partially due to me having a flu (damn the weather!). Anyway, I was trying to help you prove that it is the right answer by yourself. Anyway, cheers.

    Oh, and forgot to mention.

    Yes, you're right. This way, f is measurable w.r.t your sigma-algebra.
     
    Last edited: Nov 9, 2011
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