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.


1 Answer 1


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 $\mathcal{B}_{\mathbb{R}^n}$.

[1] https://en.wikipedia.org/wiki/Complete_measure

  • $\begingroup$ 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. $\endgroup$
    – Beginner
    Aug 6 at 17:11

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