Example 7.3.3(b): Outer Measure of Intervals
Find the outer measure of a closed interval [a, b]Since outer measure should be related to 'length' we expect that the outer measure of [a, b] is b - a. Indeed, that is the case.
Take the interval (a - , b + ). That interval covers [a, b] so that
m*([a, b]) b - a + 2Since was arbitrary we have
(*) m*([a, b]) b - aTo show the other inequality, we must show that for any collection In of intervals whose union covers [a, b] we have
l(In) b - aTake such a collection and assume, for the moment, that it is finite. Order that collection as follows:
- The first interval is the one that contains a. Call it (a1, b1).
- If b1 < b, then pick as the second interval that which contains b1. Call it (a2, b2).
- If b2 < b, then pick as the third interval that which contains b2. Call it (a3, b3).
Subintervals with k = 6
l(In) (b1 - a1) + (b2 - a2) + ... + (bk - ak) =Therefore, for any finite collection of open intervals In we have:
= -a1 + (b1 - a2) + (b2 - a3) + ... + (bk-1 - ak) + bk
bk - a1 b - a
(**) l(In) b - aNow take an arbitrary collection of intervals In that cover [a, b]. By the Heine-Borel theorem we can extract a finite subcover, i.e. a finite number of intervals that still cover [a, b]. We already know that for the finite subcover inequality (**) holds, and the sum for all intervals in the cover is even greater than the left side of (**). Therefore (**) holds for any collection of open sets covering [a, b], and thus
m*([a, b]) b - a
That together with (*) show that m*([a, b]) = b - a, as required.