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$M^*(a,b)=b-a$ we know that this fact but how we can prove closed intervals are Lebesgue measurable. I tried to prove by using $\cap ((a-\frac1n),(b+\frac1n))$ But ı totaly stucked :( please help me guys

user1008646
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Manzara
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2 Answers2

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Let $[a,b]$ be the closed interval in consideration. Since $[a,b]$ is a cover over itself, $M^*([a,b])\leq b-a$. it suffices to show for a countable arbitrary covering (of closed cubes) $\{Q_j\}$ of $[a,b]$, $b-a\leq \sum_{j} |Q_j|$.

Fix $\epsilon>0$, and for each $Q_j=[a_j,b_j]$, let $R_j=(a_j+\epsilon a_j, b_j+\epsilon b_j)$. Then this generates an open cover for $[a,b]$, so by compactness, there is a finite subcover of $\{R_j\}$ that covers $[a,b]$. Thus we have $$ b-a\leq \sum_{j=1}^n M^*(R_j) \leq \sum_{j=1}^\infty M^*(Q_j) $$ Our conclusion follows. (See Shakarchi's Real Analysis for a general proof).

The key part here is you need to show what I have stated for arbitrary countable closed cube coverings.

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Let $[a,b]$ be the closed interval in question. Define $I_n = ]n,a-\frac1n[$, $J_n=]b+\frac1n,n[$. Since $I_n$ and $J_n$ er open intervals they are lebesgue measurable, and therefore so is the union $O_n = I_n \cup J_n$. Now it follows that the countable union $O = \cup_{n\in \mathbb N}O_n$ is also lebesgue measurable. Finally we now know that $O^C = \mathbb R \setminus O$ is measurable. But $O^C = [a,b]$, and we're done.

Martin
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  • This proof works as well, but I believe it uses more technology in comparison to the proof I gave. –  Oct 24 '17 at 05:31
  • Your proof, as I understand it, shows that if $[a,b]$ is lebesgue-measurable, then it's measure is $b-a$. But you miss (again as I read it) that $[a,b]$ is lebesgue measurable in the first place - but this in turn may depend on your definition of the lebesgue measure. I've only used the fact that the lebesgue measure is a borel measure, so I do not require any other results. – Martin Oct 24 '17 at 06:14
  • I am using exterior measure. Nothing about Borel/Lebesgue measurable sets, but frankly, this is probably not totally true...too tired to discuss basic measure theory, we can go both ways anyways haha. –  Oct 24 '17 at 06:15
  • So true - it's a silly discussion :) – Martin Oct 24 '17 at 07:05