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Simple question about the Lebesgue sigma-algebra on R^2

  1. Nov 25, 2007 #1

    quasar987

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    Denote the Lebesgue sigma-algebra on R by [tex]\mathfrak{L_{\mathbb{R}}}[/tex], the Borel sigma-algebra on R by [tex]\mathfrak{B_{\mathbb{R}}}[/tex], the Lebesgue measure on R by [tex]\lambda[/tex].

    Define the Lesbesgue sigma-algebra on R² [tex]\mathfrak{L_{\mathbb{R}^2}}[/tex] as the completion of the product Borel sigma-algebra with respect to the product Lebesgue measure on R². That is to say, [tex]\mathfrak{L_{\mathbb{R}^2}}=(\mathfrak{B_{\mathbb{R}}}\times \mathfrak{B_{\mathbb{R}}})_{\lambda \times \lambda}[/tex].

    Now according to my text, the Lebesgue sigma-algebra on R² is also the completion of the product Lebesgue sigma-algebra on R with respect to the product Lebesgue measure on R². That is to say, we also have [tex]\mathfrak{L_{\mathbb{R}^2}}=(\mathfrak{L_{\mathbb{R}}}\times \mathfrak{L_{\mathbb{R}}})_{\lambda \times \lambda}[/tex].

    To justify this assertion, my text proceeds to show that

    [tex]\mathfrak{B_{\mathbb{R}}}\times \mathfrak{B_{\mathbb{R}}}\subset \mathfrak{L_{\mathbb{R}}}\times \mathfrak{L_{\mathbb{R}}} \subset \mathfrak{L_{\mathbb{R}^2}}[/tex]

    How does the conclusion follow?
     
    Last edited: Nov 25, 2007
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  3. Nov 25, 2007 #2

    morphism

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    Take completions.
     
  4. Nov 25, 2007 #3

    quasar987

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    I see. We'll have the double inclusion

    [tex]\mathfrak{L_{\mathbb{R}^2}} \subset (\mathfrak{L_{\mathbb{ R}}}\times \mathfrak{L_{\mathbb{R}}})_{\lambda \times \lambda}[/tex]

    and

    [tex](\mathfrak{L_{\mathbb{ R}}}\times \mathfrak{L_{\mathbb{R}}})_{\lambda \times \lambda}\subset \mathfrak{L_{\mathbb{R}^2}}[/tex]
     
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