## 7.3. Measures

### Theorem 7.3.9: Properties of Lebesgue measure

- All intervals are measurable and the measure of an interval is its length
- All open and closed sets are measurable
- The union and intersection of a finite or countable number of measurable sets is again measurable
- If
is measurable and**A**is the union of countable number of measurable sets**A**, then**A**_{n}*m(*) m(**A****A**_{n}) - If
is measurable and**A**is the union of countable number of disjoint measurable sets**A**, then**A**_{n}*m(*) = m(**A****A**_{n})

### Proof:

First let's review a few facts we have shown before:- The outer measure of an interval is its length
- Intervals of the form
*(a, )*are measurable - Complements of measurable sets are measurable
- Union and intersection of two measurable sets are measurable

**Lemma 1:**Suppose

*is a collection of sets such that the union of two elements and the complement of every element from*

**O***is again part of*

**O***(such a collection is called an*

**O***Algebra of Sets*). If

*{*is any countable collection of elements from

**E**_{n}}*then there is another countable collection*

**O***{*of

**F**_{n}}*disjoint*elements from

*such that*

**O***.*

**E**_{n}=**F**_{n}

**Lemma 2:**If

*are finitely many disjoint measurable sets, then*

**E**_{1},**E**_{2}, ...,**E**_{n}m((AE_{1}...E_{n})) = m(AE_{1}) + ... + m(AE_{n})

We will first prove the proposition and give the proofs of these lemmas at the end.

**Proof of statement 3:** We have already shown that the union and
intersection of two measurable sets is again measurable, so we need to
prove the statement for countable unions and intersections. Because of facts
(3) and (4) measurable sets form an algebra of sets so that it is sufficient
to show statement 3 for countable unions of *disjoint* measurable sets.

Let *{ E_{n} }* be a countable collection of
disjoint measurable sets and

*be their union. Define*

**E**Because of lemma 2 we have thatF_{n}=E_{1}E_{2}...E_{n}

Becausem^{*}(AF_{n}) = m(AE_{1}) + ... + m(AE_{n})

*comp(*we know

*) comp(***E****F**_{n})Thus, for any setm^{*}(comp()) mE^{*}(comp(F_{n}))

*we have:*

**A**because the setm(AE_{1}) + ... + m(AE_{n}) + m^{*}(comp(A)) =E

= m^{*}(AF_{n}) + m^{*}(comp(A))E

m^{*}(AF_{n}) + m^{*}(comp(AF_{n})) =

= m^{*}()A

*is measurable. But this is true for any integer*

**F**_{n}*n*so that

because of subadditivity. But that proves that the countable unionm^{*}() m(AAE_{n}) + m^{*}(comp(A))E

m(A) + mE^{*}(comp(A))E

*of the*

**E***is measurable. That countable intersections of measurable sets are measurable follows from de Morgan laws and because complements of measurable sets are measurable.*

**E**_{n}
**Proof of statement 1:** Let's focus on an open interval
*(a, b)*. We know that
*(a, )* and
*(-, b] =
comp(b, ))* are both
measurable. But

Since each set on the right is measurable and countable unions of measurable sets are measurable, intervals of the form(-, b) = (-, b-1/n]

*(-, b)*are also measurable. But

so that(a, b) = (a, ) (-, b)

*(a, b)*is measurable. Similarly,

*[a, b]*is measurable. Since the outer measure of an interval is its length, and intervals are now measurable, their (Lebesgue) measure must also be their length.

**Proof of statement 2:** We have
shown before that an open set
*U R*
can be written as a countable union of open intervals. By (1) intervals are
measurable and by (3) countable unions of measurable sets are measurable.
Therefore open sets are measurable. But closed sets are the complements of
open sets, and complements of measurable sets are measurable. Therefore
closed sets are measurable.

**Proof of statement 4:** This follows immediately from the subadditivity
property of outer measure so there is nothing to prove.

**Proof of statement 5:**
In lemma 2 we can set *A = R* to get
that if

*are finitely many disjoint measurable sets, then*

**E**_{1},**E**_{2}, ...,**E**_{n}We now need to make the step to countably many sets. Ifm(E_{1}...E_{n}) = m(E_{1}) + ... + m(E_{n})

*{*is a countable collection of disjoint measurable sets, then

**E**_{j}}so thatBut for finitely many sets we know that

so that for allBecause that statement holds for alln

*n*we conclude that

The reverse inequality follows immediately from subadditivity (statement 4), so that we have proved equality, and hence statement 5.

It remains to prove the lemmas we have used.

**Proof of Lemma 1:**
Because
*
A B =
comp( comp(A)
comp(B))
* we know that intersections of two sets from

*must also be part of*

**O***. The same is true (by induction) for finite unions, intersections, or complements of sets in*

**O***.*

**O**
Now let *{ E_{n} }* be a countable collection
of sets in

*and recursively define sets*

**O***as follows:*

**F**_{n}forF_{1}=E_{1}

F_{n}=E_{n}- (E_{1}...E_{n-1}

*n > 1*. Because

*all*

*-***A***=***B***comp(***A***)***B***are part of*

**F**_{n}*. It is left as an exercise to show that (i) the*

**O***are disjoint and (ii) the union of the*

**F**_{n}*is the same as the union of the*

**F**_{n}*.*

**E**_{n}

**Proof of Lemma 2:**
First let's show that for two disjoint measurable
sets * E* and

*we have*

**F**We know thatm((AE)) = m(FA) + m(EA)F

*is measurable, hence*

**F**Butm^{*}((AE)) = mF^{*}((AE)F) + mF^{*}((AE) comp(F))F

*and*

**E***are disjoint so that*

**F**Therefore(AE)F=FAF

(AE) comp(F) =FAE

It is now easy (and left as an exercise) to use induction to finish the proof.m^{*}((AE)) = mF^{*}(A) + mF^{*}(A)E