# Is the n-dim lebesgue measure the product of the lebesgue measure?

Let $\mathfrak{M}$ be the sigma-algebra of the n-dimensional Lebesgue measurable sets (Completion of translation invariant measure having 1 for $[0,1]^n$)

Let $\Sigma$ be the sigma-algebra of the Lebesgue-measurable sets in $\mathbb{R}$.

Is $\mathfrak{M}$ the sigma-algebra generated by $\{\prod_{i=1}^n p(i) : p\in \Sigma^n\}$?

• Almost. It's the completion of that $\sigma$-algebra with respect to the Lebesgue measure. – Daniel Fischer Feb 17 '14 at 19:24
• @Daniel What do you mean by Almost? The sigma algebra generated by $\otimes_{i=1}^n \Sigma$ is bigger than the Borel algebra of $\mathbb{R}^n$ and i don't know whether it is smaller than the sigma-algenra of n-dim Lebesgue measurable sets. Nonetheless, of course its completion is the Sigma-algenra of n-dim Lebesgue measurable sets. I'm not asking that. Is it strictly smaller or the same? – John. p Feb 17 '14 at 19:44
• @Daniel Note that $\otimes_{i=1}^n \Sigma$ is strictly bigger than the borel algebra of $\mathbb{R}^n$. Of course its completion is the sigma-algebra of n-dim Lebesgue measure. But I don't think it's easy to answer whether $\otimes_{i=1}^n \Sigma$ is exactly the sigma-algebra of n-dim Lebesgue measurable sets. And i believe it is. – John. p Feb 17 '14 at 19:51

The product $\sigma$-algebra $\bigotimes^n\Sigma$ lies strictly between the Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R}^n)$ and the $\sigma$-algebra $\mathfrak{M}$ of Lebesgue-measurable subsets of $\mathbb{R}^n$ (for $n > 1$). Since the latter is already the completion of the Borel $\sigma$-algebra with respect to the Lebesgue measure, it is also the completion of $\bigotimes^n\Sigma$ with respect to the Lebesgue measure.
For the other, let $N\subset \mathbb{R}$ a non-measurable set. Then $N\times \{0\}^{n-1}$ is a Lebesgue null-set in $\mathbb{R}^n$, but does not belong to $\bigotimes^n\Sigma$. To see the latter, note that for $M\in \bigotimes^n\Sigma$, every section $M_y = \{x \in \mathbb{R} : (x,y) \in M\}, \; y \in \mathbb{R}^{n-1}$ arbitrary, lies in $\Sigma$.