We know $A \subseteq \mathbb{R}$ is null if given $\epsilon > 0$, there exists intervals $\{I_n\}_{n \geq 1}$ such that

$$ A \subseteq \bigcup_{n=1}^{\infty} I_n \text{ and } \sum|I_n| < \epsilon $$

My question is that if we change the words interval in the above definition with open intervals, closed intervals ,and intervals of the form $(a,b]$, $[a,b)$, then we get an equivalent notion?

  • 3
    $\begingroup$ Yes, because $(a,b)\subseteq (a,b],[a,b)$ and $[a,b]=\{a\}\cup (a,b]$ and the lengths are the same. $\endgroup$ – walcher Sep 3 '13 at 8:46

If you are dealing with Borel $\sigma$-field over $\mathbb{R}$ then for any property $P$ that depends only on the measure defined on that space, if $P$ holds for any one of the following class of sets, then it holds for any other class, and also for the entire Borel $\sigma$-field.

  • class of all intervals,
  • class of all open intervals,
  • class of all closed intervals,
  • class of all clopen intervals,
  • class of all half-lines etc.

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