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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.

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  • $\begingroup$ You mean an elementary set is the union of finitely many compact cubes of $\Bbb{R}^{n}$? $\endgroup$
    – Yes
    Dec 7, 2015 at 2:44
  • $\begingroup$ @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. $\endgroup$
    – user245273
    Dec 7, 2015 at 2:46
  • $\begingroup$ The conditions $a_{i} \leq x_{i} \leq b_{i}$ give compact cubes; I just wanted to double-check you copied things right. :) $\endgroup$
    – Yes
    Dec 7, 2015 at 2:52

1 Answer 1

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The reason is simple:

Finite union of elementary sets is always elementary set (union of a finite number of intervals), but countable union of elementary sets may not be elementary set anymore. It is easy to find examples.

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  • $\begingroup$ EDIT: Now that you point it out, the reason seems so obvious. Thanks for the help!! $\endgroup$
    – user245273
    Dec 7, 2015 at 3:03

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