Recall: $$ m^*(E)=\inf\left\{\sum_{k=1}^\infty\ell(I_k)\,:\,E\subseteq\bigcup_{k=1}^\infty I_k\right\} $$ where $$ I_k = \text{ an open, bounded interval on }\mathbb{R}\,\,\text{ and }\,\,\ell(I_k)=\text{ the length of }I_k $$

Here is the text of the proof as published:

$\textbf{Proposition 8}\,\,\textit{ Every interval is measurable.}$

$\textbf{Proof}\,\,$ As we observed above, the measurable sets are a $\sigma$-algebra. Therefore to show that every interval is measurable it suffices to show that every interval of the form $(a,\infty)$ is measurable (see Problem 11). Consider such an interval. Let $A$ be any set. We assume that $a$ does not belong to $A$. Otherwise, replace $A$ by $A\sim\{a\}$, leaving the outer measure unchanged. We must show that $$ m^*(A_1)+m^*(A_2)\le m^*(A) $$ where $$ A_1=A\cap (-\infty,a)\,\,\text{and}\,\,A_2=A\cap (a,\infty).\qquad(5) $$ By the definition of $m^*(A)$ as an infimum, to verify (5) it is necessary and sufficient to show that for any countable collection $\{I_k\}_{k=1}^\infty$ of open, bounded intervals that covers $A$, $$ m^*(A_1)+m^*(A_2)\le \sum_{k=1}^\infty\ell(I_k).\qquad(6) $$ Indeed, for such a covering, for each index $k$, define $$ I'_k=I_k\cap(-\infty,a)\,\,\text{and}\,\,I''_k=I_k\cap (a,\infty). $$ Then $I'_k$ and $I''_k$ are intervals and $$ \ell(I_k)=\ell(I'_k)+\ell(I''_k). $$ Since $\{I'_k\}_{k=1}^\infty$ and $\{I''_k\}_{k=1}^\infty$ are countable collections of open, bounded intervals that cover $A_1$ and $A_2$, respectively, by the definition of outer measure, $$ m^*(A_1)\le\sum_{k=1}^\infty \ell(I'_k)\,\,\text{and}\,\,m^*(A_2)\le\sum_{k=1}^\infty \ell(I''_k). $$ Therefore $$ \begin{align*} m^*(A_1)+m^*(A_2)&\le\sum_{k=1}^\infty \ell(I'_k) + \sum_{k=1}^\infty \ell(I''_k)\\ &=\sum_{k=1}^\infty [\ell(I'_k)+\ell(I''_k)]\\ &=\sum_{k=1}^\infty \ell(I_k). \end{align*} $$ Thus (6) holds and the proof is complete.

Overall, this proof is very clear and makes sense. I am having a problem with one of the assumptions in the beginning of the proof: $E$ is a measurable set if for $every$ set $A$ we have $$ m^*(A)=m^*(A\cap E)+m^*(A\cap E^C), $$ but this proof seems to ignore all sets containing $\{a\}$. Is this a flaw in the proof? I am not even sure why ignoring $\{a\}$ is necessary. Can someone clarify this for me?

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    $\begingroup$ A single pointy has measure $0$, so no issues there. $\endgroup$ – voldemort May 5 '14 at 2:00
  • $\begingroup$ I understand that the outer measure of $A$ and $A\sim\{a\}$ are the same. But why is it even necessary to remove $a$ in the first place? $\endgroup$ – Laars Helenius May 5 '14 at 2:01
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    $\begingroup$ Open covers and stuff... just to ensure that you have an open interval. Makes the proof easier. $\endgroup$ – voldemort May 5 '14 at 2:02

Removing $a$ allows you to use $(-\infty,a)$ instead of $(-\infty,a]$ in $(5)$. This in turn allows the $I_k'$ to cover $A_1$.

Removing $a$ is not an issue because, as the book says, $m^*(A\cap(-\infty,a])=m^*(A\cap(-\infty,a))$.

  • $\begingroup$ That is true. If $a\in A_1$, then none of $I'_k$ would be able to cover and hence $A_1\not\subseteq\bigcup_{k=1}^\infty I'_k$. So we could use intervals of the form $[a,\infty)$ and throw out the assumption, but it complicates the proof a bit, right? $\endgroup$ – Laars Helenius May 5 '14 at 2:22
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    $\begingroup$ Yes. To use the definition of the outer measure, you need a cover by open invervals. $\endgroup$ – Martin Argerami May 5 '14 at 2:25
  • $\begingroup$ Ah. So, actually, we have to proceed in the manner of the text. We have to have those open bounded covers otherwise the outer measure definition can't apply. Thanks! $\endgroup$ – Laars Helenius May 5 '14 at 3:11
  • $\begingroup$ You are very welcome. $\endgroup$ – Martin Argerami May 5 '14 at 3:15

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