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

1. Nov 25, 2007

### 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?

Last edited: Nov 25, 2007
2. Nov 25, 2007

### morphism

Take completions.

3. Nov 25, 2007

### 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

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