# Family of all elementary subsets of $\mathbb{R}^n$ is a ring, but not a $\sigma$-ring

I was going through the introduction to Lebesgue theory in Baby Rudin, where the property was given that:

$\mathcal{E}$ is a ring, but not a $\sigma$-ring.

$\mathcal{E}$ here represents the family of all elementary subsets of $\mathbb{R}^n$, $n$-dimensional Euclidean space.

An elementary set is the union of a finite number of intervals, where an interval is defined to be a set of points $(x_1,\dots,x_n)$ in $\mathbb{R}^n$ such that

$$a_i \leq x_i \leq b_i ~~~~~(i=1,\dots,n).$$

I am having trouble understanding why $\mathcal{E}$ does not satisfy the properties of a $\sigma$-ring.

I apologize if the question is very trivial, the notion of interval $I$ and $m(I)$ of $I$ is very new and abstract to me.

• You mean an elementary set is the union of finitely many compact cubes of $\Bbb{R}^{n}$?
– Yes
Dec 7, 2015 at 2:44
• @GudsonChou The book gives the definition of elementary set to be "the union of a finite number of intevals", where the interval is defined in $\mathbb{R}^n$. A closed condition is not given, I suppose it is not necessary. So yes, I am guessing it is the union of finitely many $k$-cells.
– user245273
Dec 7, 2015 at 2:46
• The conditions $a_{i} \leq x_{i} \leq b_{i}$ give compact cubes; I just wanted to double-check you copied things right. :)
– Yes
Dec 7, 2015 at 2:52