## Examples 7.3.7(e): Measurable Sets

We need to show that for every set*we have that*

**A**becausem^{*}() mA^{*}((a, )) + mA^{*}((-, a])A

*comp(a, ) = (-, a]*. If

*m*is infinite, there's nothing to prove. Therefore we can assume that

^{*}(*)***A***m*is finite. Then, because of the definition of outer measure as an infimum, there exists a countable collection of open intervals

^{*}(*)***A***that cover*

**I**_{n}*and*

**A**for anyl(I_{n}) m^{*}() +A

*> 0*. Define sets

*and*

**J**_{n}*as*

**K**_{n}Then we have the following properties:J_{n}=I_{n}(a, )

K_{n}=I_{n}(-, a])

But then we have that

andJ_{n}are intervals (or empty) so thatK_{n}mand^{*}(J_{n}) = l(J_{n})mand^{*}(K_{n}) = l(K_{n})

l(J_{n}) + l(K_{n}) = l(I_{n})andJ_{n}I_{n}so thatK_{n}I_{n}In particular, all sums are absolutely convergent because the measure ofl(andJ_{n}) l(I_{n})l(K_{n}) l(I_{n})is finite.A(and(a, ))AJ_{n}(so that(-, a])AK_{n}because of subadditivity and (1).mand^{*}((a, )) mA^{*}(J_{n}) l(J_{n})

m^{*}((-, a]) mA^{*}(K_{n}) l(K_{n})

Since this inequality holds for everym^{*}((a, )) + mA^{*}((-, a]) l(AJ_{n}) + l(K_{n}) =

l(J_{n}) + l(K_{n}) = l(I_{n}) m^{*}() +A

*> 0*, it implies what we wanted to prove.