# Equivalent Definition of $n$-dimensional Lebesgue measure

This is from Folland Real Analysis section 2.6 where he gave a definition of $$n$$-dimensional Lebesgue measure

Lebesgue measure $$m^n$$ on $$\mathbb{R}^n$$ is the completion of the n-fold product of Lebesgue measure on $$\mathbb{R}$$ with itself, that is, the completion of $$m \times \cdots \times m$$ on $$\mathcal{B}_{\mathbb{R}} \bigotimes \cdots \bigotimes \mathcal{B}_{\mathbb{R}} = \mathcal{B}_{\mathbb{R}^n}$$ or equivalently the completion of $$m \times \cdots \times m$$ on $$\mathcal{L} \bigotimes \cdots \bigotimes \mathcal{L}$$, where $$\mathcal{L}$$ is the Lebesgue measuarble set on $$\mathbb{R}$$.

Why "or equivalently" holds, why these two product measures' completion is the same one?

EDIT: This is indeed a gap here, which Folland failed to explain. And I found a proof somewhere else. I will post the proof later when I have time.

The definitions are indeed equivalent. Recall that Given a (possibly incomplete) measure space $$(X, \mathcal A, \mu)$$, there is an extension $$(X, \mathcal A_1, \mu_1)$$ of this measure space that is complete. The smallest such extension (i.e., the smallest $$\sigma$$-algebra $$\mathcal A_1$$) is called the completion of the measure space [1].
Let $$(\mathbb{R}^n, \mathcal A_1, \mu_1)$$ be the completion of $$m \times \cdots \times m$$ on $$\mathcal{B}_{\mathbb{R}} \otimes \cdots \otimes \mathcal{B}_{\mathbb{R}} = \mathcal{B}_{\mathbb{R}^n}$$.
Now the product $$\sigma$$-algebra $$\mathcal{L}^{\otimes n} :=\mathcal{L}\otimes \cdots \otimes \mathcal{L}$$ is sandwiched between $$\mathcal{B}_{\mathbb{R}^n}$$ and $$\mathcal A_1$$, because $$m$$ on $$\mathcal{L}$$ is the completion of $$m$$ on $$\mathcal{B}_{\mathbb{R}}$$. Thus $$\mathcal A_1$$ is a complete $$\sigma$$-algebra (for $$m^n$$) which contains $$\mathcal{L}^{\otimes n}$$. It must be the smallest such $$\sigma$$-algebra, since it is the smallest complete $$\sigma$$-algebra that contains $$\mathcal{B}_{\mathbb{R}^n}$$.
• This looks good to me. But I think If you prove $\mathcal{L}^{\otimes n}$ is not a complete measure(There are many obvious examples), then this answer would be perfect. Aug 6 at 17:11