# 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}. 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}. 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}. 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}. 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}. But $b_{k}>b$ and $a_{1} 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

where by $\overline{I}$ we the topological closure of $I$. Thus for each $\varepsilon>0$, we have $l(I)-\varepsilon, 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 ProofThatTheOuterLebesgueMeasureOfAnIntervalIsItsLength 2013-03-22 14:47:04 2013-03-22 14:47:04 Simone (5904) Simone (5904) 6 Simone (5904) Proof msc 28A12 LebesgueOuterMeasure