# 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^{} $(a-\epsilon ,b+\epsilon )$
contains $[a,b]$ for each positive number $\epsilon $, we have
${m}^{*}[a,b]\le b-a+2\epsilon $. But since this is true for each positive $\epsilon $, we must have ${m}^{*}[a,b]\le b-a$. Thus we only have to show that ${m}^{*}[a,b]\ge b-a$; for this it suffices to show that if $\{{I}_{n}\}$ is a countable^{} open cover by intervals of $[a,b]$, then

$$\sum l({I}_{n})\ge b-a.$$ |

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 $$. If ${b}_{1}\le b$, then ${b}_{1}\in [a,b]$, and since ${b}_{1}\notin ({a}_{1},{b}_{1})$, there must be an interval $({a}_{2},{b}_{2})$ in the collection $\{{I}_{n}\}$ such that ${b}_{1}\in ({a}_{2},{b}_{2})$, that is $$. Continuing in this fashion, we obtain a sequence $({a}_{1},{b}_{1}),\mathrm{\dots},({a}_{k},{b}_{k})$ from the collection $\{{I}_{n}\}$ such that $$. Since $\{{I}_{n}\}$ is a finite collection our process must terminate with some interval $({a}_{k},{b}_{k})$. But it terminates only if $b\in ({a}_{k},{b}_{k})$, that is if $$. Thus

$\sum l({I}_{n})$ | $\ge {\displaystyle \sum l({a}_{i},{b}_{i})}$ | ||

$=({b}_{k}-{a}_{k})+({b}_{k-1}-{a}_{k-1})+\mathrm{\dots}+({b}_{1}-{a}_{1})$ | |||

$={b}_{k}-({a}_{k}-{b}_{k-1})-({a}_{k-1}-{b}_{k-2})-\mathrm{\dots}-({a}_{2}-{b}_{1})-{a}_{1}$ | |||

$>{b}_{k}-{a}_{1},$ |

since $$. But ${b}_{k}>b$ and $$ 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 $\epsilon >0$, there is a closed interval$J\subset I$ such that $l(J)>l(I)-\epsilon $. Hence

$$ |

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

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

## 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 |