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

[quasar987]

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

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

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 \mathfrak{L_{\mathbb{R}^2}}=(\mathfrak{L_{\mathbb{R}}}\times \mathfrak{L_{\mathbb{R}}})_{\lambda \times \lambda}.

To justify this assertion, my text proceeds to show that

\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}}

How does the conclusion follow?

 

[morphism]

\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}}

How does the conclusion follow?

Take completions.

 

[quasar987]

I see. We'll have the double inclusion

\mathfrak{L_{\mathbb{R}^2}} \subset (\mathfrak{L_{\mathbb{ R}}}\times \mathfrak{L_{\mathbb{R}}})_{\lambda \times \lambda}

and

(\mathfrak{L_{\mathbb{ R}}}\times \mathfrak{L_{\mathbb{R}}})_{\lambda \times \lambda}\subset \mathfrak{L_{\mathbb{R}^2}}