# Interactive Real Analysis

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## 7.3. Measures

### 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 + 2
Since was arbitrary we have
(*)      m*([a, b]) b - a
To show the other inequality, we must show that for any collection In of intervals whose union covers [a, b] we have
l(In) b - a
Take 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).
Continue in this fashion until either one of the bj > b or until all intervals are numbered. Since there are only finitely many open intervals, the process must terminate in finitely many steps, say in k steps, and since the intervals cover [a, b] we must have that bk > b.

Subintervals with k = 6

Then

l(In) (b1 - a1) + (b2 - a2) + ... + (bk - ak) =
= -a1 + (b1 - a2) + (b2 - a3) + ... + (bk-1 - ak) + bk
bk - a1 b - a
Therefore, for any finite collection of open intervals In we have:
(**)      l(In) b - a
Now 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.

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