Hello Mathematics Community. I was hoping someone could assist me in solving the following problem from Terrence Tao's Introduction to Measure Theory book. I am using the free online version of the text, and this problem can be found in section 1.7, on page 196; this is part (vi) of the exercise.

Show that the Lebesgue $\sigma-$algebra on $\mathbb{R}^{d+d'}$ is the completion of the product of the Lebesgue $\sigma-$algebras of $\mathbb{R}^d$ and $\mathbb{R}^{d'}$ with respect to $d+d'-$dimensional Lebesgue measure.

What I do know is that a $\sigma-$algebra is complete if it contains all the sets with outer measure 0. So if our measurable space is, say, $(X, \mathcal{B})$, then for any $E \subset X$ with $\mu^*(E)-0,$ and any $F$ $\in \mathcal{P(x)}$, we have

$\mu^*(F \cap E) + \mu^*(F \cap E^c)-\mu^*(F \cap E^c) \leq \mu^*(F)$. So $E$ is in the $\sigma-$algebra.

Furthermore, the book defines a product $\sigma-$algebra $\mathcal{B}_X \times \mathcal{B}_Y$ to be the $\sigma-$algebra generated by the union of the two $\sigma-$algebras:

$\mathcal{B}_X \times \mathcal{B}_Y := \langle \pi^*_X(\mathcal{B}_X) \cup \pi^*_Y(\mathcal{B}_Y) \rangle$ where $\pi_X$ and $\pi_Y$ are coordinate projection maps, respectively. How would I begin to prove this? Is Tonelli's Theorem applicable here since we are working with product measures?

As always, any help is greatly appreciated. Thanks in advance.

A link to the free version is provided below:



1 Answer 1


Given measurable spaces $(X,\mathcal{M})$ and $(Y,\mathcal{N})$, we can consider the product measure space $(X\times Y,\mathcal{M}\times\mathcal{N})$, where $\mathcal{M}\times\mathcal{N}$ is the $\sigma$-algebra generated by products $M\times N$, where $M\in\mathcal{M}$ and $N\in\mathcal{N}$ (this coincides with how you defined the product $\sigma$-algebra, and I think it is easier to see it this way).

If we have measure spaces $(X,\mathcal{M},\mu)$ and $(Y,\mathcal{N},\eta)$, then we can induce a measure $\mu\times\eta$, called the product measure, on $\mathcal{M}\times\mathcal{N}$ with the property that $(\mu\times\eta)(M\times N)=\mu(M)\cdot\eta(N)$ for $M\in\mathcal{M}$ and $N\in\mathcal{N}$. If the spaces in question are $\sigma$-finite, then the measure $\mu\times\eta$ with this property is unique.

Clearly, all this is true if we consider products of a finite number of measure spaces.

Given a topological space $X$, let $\mathscr{B}_X$ denote the Borel $\sigma$-algebra on $X$ (which is generated by open sets). If $X$ is second countable (i.e., if it has a countable topological basis), then $\mathscr{B}_{X^n}=\mathscr{B}_X\times\cdots\times\mathscr{B}_X=:(\mathscr{B}_X)^n$ (you can easily verify this). In particular, $\mathscr{B}_{\mathbb{R}^n}=(\mathscr{B}_\mathbb{R})^n$.

The usual definition of Lebesgue measure on $\mathbb{R}^n$ is the following: There exists a unique measure $\lambda$ on $\mathscr{B}_{\mathbb{R}^n}$ such that $\lambda_n(I_1\times\cdots I_n)=\operatorname{length}(I_1)\cdots\operatorname{length}(I_n)$ for all intervals $I_1,\ldots,I_n$. The Lebesgue measure on $\mathbb{R}^n$ is the completion of this measure, and it is also denoted by $\lambda$, and it is then defined in some other $\sigma$-algebra $\mathscr{L}_{\mathbb{R}^n}$ which contains $\mathscr{B}_{\mathbb{R}^n}$. When restricted to Borel sets, we have $\lambda_n=\lambda_1\times\cdots\times\lambda_1$.

However, it is not true that $\mathscr{L}_{\mathbb{R}^{n+m}}=\mathscr{L}_{\mathbb{R}^n}\times\mathscr{L}_{\mathbb{R}^n}$ (to be precise, this is not true if we assume the Axiom of Choice).

Finally, to solve the question, you should first show that $\mathscr{L}_{\mathbb{R}^n}\times\mathscr{L}_{\mathbb{R}^m}\subseteq\mathscr{L}_{\mathbb{R}^{n+m}}$. You can do this using the definition of completion of measures, and using the fact that $\lambda_{n+m}=\lambda_n\times\lambda_m$ on Borel sets. Since $\mathscr{L}_{\mathbb{R}^{n+m}}$ is a complete $\sigma$-algebra (with respect to $\lambda_n$), then it contains the completion of $\mathscr{L}_{\mathbb{R}^n}\times\mathscr{L}_{\mathbb{R}^m}$.

On the other hand, it is clear that $\mathscr{L}_{\mathbb{R}^n}\times\mathscr{L}_{\mathbb{R}^m}$ contains $\mathscr{B}_{\mathbb{R}^n}\times\mathscr{B}_{\mathbb{R}^m}=\mathscr{B}_{\mathbb{R}^{n+m}}$, so the completion of $\mathscr{L}_{\mathbb{R}^n}\times\mathscr{L}_{\mathbb{R}^m}$ contains the completion of $\mathscr{B}_{\mathbb{R}^{n+m}}$, which is by definition $\mathscr{L}_{\mathbb{R}^{n+m}}$ (here we again use the fact that $\lambda_{n+m}=\lambda_n\times\lambda_m$).

  • $\begingroup$ So by the definition of completion of measures, which means that it contains all subsets of outer measure zero, and the fact that $\mathscr{L}_{\mathbb{R}^{n+m}}$ is a complete $\sigma-$algebra, then we obtain the desired result? $\endgroup$
    – Jamil_V
    Nov 4, 2014 at 15:09
  • $\begingroup$ The first part (previous to last paragraph), yes. $\endgroup$ Nov 4, 2014 at 18:34
  • $\begingroup$ Ok, I have been reading this over a bit. The first part, as mentioned, is done by definition of completion of measures and that $\mathscr{L}_{\mathbb{R}^{n+m}}$ is a complete $\sigma-$algebra. But the second part of this problem is to show that this is the completion over $d+d'$ dimensional Lebesgue measure. Is this correct? Or is the $d$ and $d'$ equal to the $m$ and $n$ in the answer? $\endgroup$
    – Jamil_V
    Nov 6, 2014 at 4:21
  • $\begingroup$ Oh, yes. You got the idea right. This gets kind of confusin because everything is called "Lebesgue", even though we sometimes restrict to Borel sets, etc... and I was using $n$ and $m$ instead of $d$ and $d'$, which maybe was not a great idea. $\endgroup$ Nov 7, 2014 at 22:02

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