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 . Since the open interval contains for each positive number , we have . But since this is true for each positive , we must have . Thus we only have to show that ; for this it suffices to show that if is a countable open cover by intervals of , then
By the Heine-Borel theorem, any collection of open intervals contains a finite subcollection that also cover 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 that cover . Since is contained in , there must be one of the ’s that contains . Let this be the interval . We then have . If , then , and since , there must be an interval in the collection such that , that is . Continuing in this fashion, we obtain a sequence from the collection such that . Since is a finite collection our process must terminate with some interval . But it terminates only if , that is if . Thus
since . But and and so we have , whence . This shows that .
If is any finite interval, then given , there is a closed interval such that . Hence
where by we the topological closure of . Thus for each , we have , and so .
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|
|Date of creation||2013-03-22 14:47:04|
|Last modified on||2013-03-22 14:47:04|
|Last modified by||Simone (5904)|