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Falling back on the Lebesgue measure from the abstract theory?

  1. Sep 26, 2007 #1

    quasar987

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    I am studying the abstract theory of measure and I was wondering how the Lebesgue case for real functions of a real variable arises. But I did not find it.

    In the original theory of Lebesgue, a function f:E-->R was said to be measurable if for every real constant b, the preimage of [itex]]-\infty, b][/itex] by f was measurable. Let the collection of all measurable sets be denoted [tex]\mathcal{L}_{\mathbb{R}}[/tex] (the Lebesgue sigma-algebra). The pair [tex](\mathbb{R},\mathcal{L}_{\mathbb{R}})[/tex] is a measurable space.

    In the abstract theory, we consider a function f btw two measurable spaces:

    [tex]f:(X_1,\mathcal{T}_1)\rightarrow (X_2,\mathcal{T}_2)[/tex]

    and say that it is measurable if, given a family of subsets of X_2 [itex]G_2[/itex] that generates the sigma-algebra [itex]\mathcal{T}_2[/itex] (i.e. [itex]\mathcal{T}(G_2)=\mathcal{T}_2[/itex]), we have

    [tex]f^{-1}(G_2)\subset \mathcal{T}_1[/tex]

    If I set X_1 = E a subset of R and X_2 = R, I am trying to find which sigma-algebras [itex]\mathcal{T}_1, \mathcal{T}_2[/itex] will make Lebesgue's definition and the abstract definition coincide. Obviously, we must take [itex]\mathcal{T}_2=\mathcal{T}(\{\{[-\infty,b]\}:b\in\mathbb{R}\})=\mathcal{B}_{\mathbb{R}}[/itex] (the borelian sigma-algebra). Now, if I were allowed to take [itex]\mathcal{T}_1=\mathcal{L}_{\mathbb{R}}[/itex] I would have succeeded, but [itex]\mathcal{L}_{\mathbb{R}}[/itex] is not a sigma-algebra on E. The next best thing is the trace of [itex]\mathcal{L}_{\mathbb{R}}[/itex] on E (aka maybe the induced sigma-algebra on E by [itex]\mathcal{L}_{\mathbb{R}}[/itex]) defined by [itex]\mathcal{L}_E=\{EM:M\in \mathcal{L}_{\mathbb{R}}\}[/itex].

    But this does not seem to work. I need to check now that we have the equivalence (for all b in R, the preimage of [itex]]-\infty, b][/itex] by f is in [itex]\mathcal{L}_{\mathbb{R}}[/itex]) <==>(for all b in R the preimage of [itex]]-\infty, b][/itex] by f is in [itex]\mathcal{L}_E[/itex]).

    The ==> part is trivial but I don'T know how to prive the <== part, and actually, I would think that it is not necessarily true, for instance if E is not a part of [itex]\mathcal{L}_{\mathbb{R}}[/itex].

    Any ideas???
     
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  3. Sep 26, 2007 #2

    quasar987

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    I just noticed that the trace of a sigma-algebra on a set E is only defined if E is itself an element of the sigma-algebra, such that the problem I rise concerning part <== does not exists.
     
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