proof that the outer (Lebesgue) measure of an interval is its length

We begin with the case in which we have a bounded interval, say $[a,b]$ . Since the open interval $(a-\varepsilon,b+\varepsilon)$ contains $[a,b]$ for each positive number $\varepsilon$ , we have $m^{*}[a,b]\leq b-a+2\varepsilon$ . But since this is true for each positive $\varepsilon$ , we must have $m^{*}[a,b]\leq b-a$ . Thus we only have to show that $m^{*}[a,b]\geq b-a$ ; for this it suffices to show that if $\{I_{n}\}$ is a countable open cover by intervals of $[a,b]$ , then

\sum l(I_{n})\geq b-a.

By the Heine-Borel theorem, any collection of open intervals $[a,b]$ contains a finite subcollection that also cover $[a,b]$ and since the sum of the lengths of the finite subcollection is no greater than the sum of the original one, it suffices to prove the inequality for finite collections $\{I_{n}\}$ that cover $[a,b]$ . Since $a$ is contained in $\bigcup I_{n}$ , there must be one of the $I_{n}$ ’s that contains $a$ . Let this be the interval $(a_{1},b_{1})$ . We then have $a_{1}<a<b_{1}$ . If $b_{1}\leq b$ , then $b_{1}\in[a,b]$ , and since $b_{1}\not\in(a_{1},b_{1})$ , there must be an interval $(a_{2},b_{2})$ in the collection $\{I_{n}\}$ such that $b_{1}\in(a_{2},b_{2})$ , that is $a_{2}<b_{1}<b_{2}$ . Continuing in this fashion, we obtain a sequence $(a_{1},b_{1}),\dots,(a_{k},b_{k})$ from the collection $\{I_{n}\}$ such that $a_{i}<b_{i-1}<b_{i}$ . Since $\{I_{n}\}$ is a finite collection our process must terminate with some interval $(a_{k},b_{k})$ . But it terminates only if $b\in(a_{k},b_{k})$ , that is if $a_{k}<b<b_{k}$ . Thus

	$\displaystyle\sum l(I_{n})$	$\displaystyle\geq\sum l(a_{i},b_{i})$
		$\displaystyle=(b_{k}-a_{k})+(b_{k-1}-a_{k-1})+\dots+(b_{1}-a_{1})$
		$\displaystyle=b_{k}-(a_{k}-b_{k-1})-(a_{k-1}-b_{k-2})-\dots-(a_{2}-b_{1})-a_{1}$
		$\displaystyle>b_{k}-a_{1},$

since $a_{i}<b_{i-1}$ . But $b_{k}>b$ and $a_{1}<a$ and so we have $b_{k}-a_{1}>b-a$ , whence $\sum l(I_{n})>b-a$ . This shows that $m^{*}[a,b]=b-a$ .

If $I$ is any finite interval, then given $\varepsilon>0$ , there is a closed interval $J\subset I$ such that $l(J)>l(I)-\varepsilon$ . Hence

l(I)-\varepsilon<l(J)=m^{*}J\leq m^{*}I\leq m^{*}\overline{I}=l(\overline{I})=% l(I),

where by $\overline{I}$ we the topological closure of $I$ . Thus for each $\varepsilon>0$ , we have $l(I)-\varepsilon<m^{*}I\leq l(I)$ , and so $m^{*}I=l(I)$ .

If now $I$ is an unbounded interval, then given any real number $\Delta$ , there is a closed interval $J\subset I$ with $l(J)=\Delta$ . Hence $m^{*}I\geq m^{*}J=l(J)=\Delta$ . Since $m^{*}I\geq\Delta$ for each $\Delta$ , it follows $m^{*}I=\infty=l(I)$ .

References

Royden, H. L. Real analysis. Third edition. Macmillan Publishing Company, New York, 1988.

Title	proof that the outer (Lebesgue) measure of an interval is its length
Canonical name	ProofThatTheOuterLebesgueMeasureOfAnIntervalIsItsLength
Date of creation	2013-03-22 14:47:04
Last modified on	2013-03-22 14:47:04
Owner	Simone (5904)
Last modified by	Simone (5904)
Numerical id	6
Author	Simone (5904)
Entry type	Proof
Classification	msc 28A12
Related topic	LebesgueOuterMeasure