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Can someone help me understand the difference between a measure, and a function which is measurable?

  1. Sep 26, 2014 #1
    This might not be the right subforum, but I was told that measure theory is very important in probability theory, so I thought maybe it belonged here.


    I am confused about the difference between a measure (which is a function onto [itex]\mathbb{R}[/itex] that satisfies the axioms listed here: https://proofwiki.org/wiki/Definition:Measure_(Measure_Theory) ) and measurability (which is a criteria that a function onto [itex]\mathbb{R}[/itex] can meet. These criteria are listed here: https://proofwiki.org/wiki/Definition:Measurable_Function ). I'm assuming these things are related, but the definitions seems so different so I don't really understand the relation between them. For example, if a function f is sigma measurable, is there automatically some "measure function" we can derive from it? Or similarly, is a measure always sigma measurable?


    Also - I was curious, is there a difference between being "sigma measurable" and just "measurable?" The things I was reading seemed to use the terms interchangeably, so I just wasn't certain. Sorry if these questions are kind of dumb and obvious, I am brand new to measure theory and still trying to wrap my head around these definitions
     
    Last edited: Sep 26, 2014
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  3. Sep 26, 2014 #2

    FactChecker

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    Compare the definitions step by step and you will see that a measure and a measurable function are completely different animals. Most functions you know are measurable. Any continuous function is measurable. Measures are restricted to positive functions on sets whose values are never negative and the measure of any countable union of disjoint sets is the summation of their individual measures. On the real line, think of measures as generalizations of the summation of segment lengths on the real line, not as a function on individual real numbers. All probability functions are measures.
     
  4. Sep 26, 2014 #3

    Stephen Tashi

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    When a functions is called "measureable", it is measureable with respect to some given measure. So, given a measureable function, you already have a measure. You don't need to derive one.

    If we are talking about measureable functions defined on the real numbers. A measure is not a measure.able function. Each element of the domain of a measureable function is a real number. Each element of the domain of a measure is a set.
     
  5. Sep 26, 2014 #4
    I think this is where I'm confused. Based on this definition: https://proofwiki.org/wiki/Definition:Measurable_Function it looks like a function is "sigma measurable" based on some sigma, where a sigma is a sigma-algebra on a set. Where does the measure come in to play? I didn't get the impression that there was always a measure defined when you have a sigma-algebra.
     
  6. Sep 26, 2014 #5

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    I don't know if existence of a measure is guaranteed. The important property of a measurable function is that , given any measure on the sigma algebra, you can define Lebesgue integrals of that function using that measure.
     
  7. Sep 26, 2014 #6

    Stephen Tashi

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    You are correct.

    I'm was thinking of the definition of a "measure space". Your link refers to a "measureable space". The Wolfram site's defiinition of measureable function mentions a specific measure. http://mathworld.wolfram.com/MeasurableFunction.html so it goes beyond the Wikipedia's definition based on a "measureable space". If we use the Wikipedia's definition then I think you have asked a very good question - namely can a measureable function on a "measureable space" be used to define a measure on that space. (It should be clear that the measureable function itself is not a measure since its domain is the universal set X of the measure space, not the sets in the sigma algebra of subsets of X.)
     
  8. Sep 26, 2014 #7

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    Right. And a constant function on a sigma algebra would be measurable. So I don't see how it could help in any way to define a measure on its domain.
     
  9. Sep 26, 2014 #8

    Stephen Tashi

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    Do you mean a constant function on each element of the sigma algebra or function that is constant on each element of the universal set X?

    [itex] \{ \emptyset, X\} [/itex] is sigma algebra and we can define the measure [itex] \mu [/itex] by [itex] \mu( \emptyset) = 0,\mu(X) = 1 [/itex].
     
  10. Sep 27, 2014 #9

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    Oh. Right. I mean a function constant on each element of the universal set X. Because it says nothing, it is no help in constructing a measure on the sigma algebra.
    Thanks for the correction.
     
  11. Sep 27, 2014 #10

    WWGD

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    In the most general sense. given two spaces (X,A_X) and (Y, A_Y) , where A_X, A_Y are sigma-algebras of sets , a function f: X-->Y is measurable if the inverse image of any element of A_Y is in A_X . In the case of the Lebesgue measure on the Real line, this reduces to : the inverse image of an open set is (Lebesgue) measurable, so that, as someone pointed out, continuous functions are measurable , since open sets are Lebesgue measurable.

    A measure is a function that assigns a number --the measure -- to a subset ; an extension to general subsets of the Euclidean idea of length, volume. Still, there are results that not all subsets can be assigned a measure.
     
  12. Sep 27, 2014 #11

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    Regarding the existence of a measure on a sigma algebra, one case that covers a lot of ground is the case where the sigma-algebra contains an atom, A. Consider the function M(s) on the sigma-algebra defined by M(s) = 1 if A [itex]\in[/itex] s and =0 otherwise. It is a probability measure on the sigma-algebra.
     
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