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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?
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?
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