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The definition of the outer measure of a set $A\subseteq\mathbb{R}$ is as follows:

$$ |A| = \inf \left\{ \Sigma_{k=1}^{\infty}\ \mathscr{l}(I_k): I_1, I_2,\dots\text{ are open intervals such that }A\subseteq\bigcup_{k=1}^{\infty} I_k \right\}. $$

My quick observation is that if two open intervals intersect, we can join them together to create another open interval. So starting from $I_1$, apply this process for every $I_k$ and we get a countable list of disjoint open intervals $O_1, O_2,\dots$ such that $A\subseteq\bigcup_{k=1}^{\infty} O_k$ and $\Sigma_{k=1}^{\infty}\ \mathscr{l}(I_k)\ge\Sigma_{k=1}^{\infty}\ \mathscr{l}(O_k)$. Since we're taking the $\inf$ of the length sum, we can replace $I_k$ by $O_k$.

So I wonder why the definition uses "open intervals" instead of "disjoint open intervals", or if using "disjoint open interval", we will get a different set function other than the outer measure?

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    The answer is yes. See https://math.stackexchange.com/questions/318299/any-open-subset-of-bbb-r-is-a-countable-union-of-disjoint-open-intervals – Akira Mar 12 '24 at 13:22
  • Ohm, it's just that simple, and yes thanks for your short answer! – Tran Khanh Mar 12 '24 at 13:26

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