## 7.3. Measures

### Example 7.3.3(b): Outer Measure of Intervals

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

Since was arbitrary we havem^{*}([a, b]) b - a + 2

To show the other inequality, we must show that for any collection(*) m^{*}([a, b]) b - a

*I*of intervals whose union covers

_{n}*[a, b]*we have

Take such a collection and assume, for the moment, that it is finite. Order that collection as follows:l(I_{n}) b - a

- The first interval is the one that contains
*a*. Call it*(a*._{1}, b_{1}) - If
*b*, then pick as the second interval that which contains_{1}< b*b*. Call it_{1}*(a*._{2}, b_{2}) - If
*b*, then pick as the third interval that which contains_{2}< b*b*. Call it_{2}*(a*._{3}, b_{3})

*b*> 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

_{j}*k*steps, and since the intervals cover

*[a, b]*we must have that

*b*.

_{k}> b

Subintervals with

*k = 6*

Then

Therefore, for anyl(I_{n}) (b_{1}- a_{1}) + (b_{2}- a_{2}) + ... + (b_{k}- a_{k}) =

= -a_{1}+ (b_{1}- a_{2}) + (b_{2}- a_{3}) + ... + (b_{k-1}- a_{k}) + b_{k}

b_{k}- a_{1}b - a

*finite*collection of open intervals

*I*we have:

_{n}Now take an arbitrary collection of intervals(**) l(I_{n}) b - a

*I*that cover

_{n}*[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.