14.4. REGULARITY 329

In particular, the above lemma shows that all intervals are both Gδ and Fσ as are finiteunions of intervals. Indeed, R is both open and closed so in the lemma, you could takeA = R. Also note that /0 is both open and closed. In what follows, µ is a measure whichis defined on B (R) such that µ is finite on all bounded sets. We don’t really need toworry about where it comes from. However, this is a good time to review the properties ofmeasures in Lemma 14.1.4. I will use these as needed.

A measure satisfying the conclusions of the following theorem is called a regular mea-sure.

Theorem 14.4.3 Let µ be a measure defined on B (R) which is finite on boundedsets. Then for all E a Borel set, there is an Fσ set F and a Gδ set G such that

F ⊆ E ⊆ G, and µ (G\F) = 0 (14.1)

Also for all E Borel,

µ (E) = sup{µ (K) ,K ⊆ E and K compact}µ (E) = inf{µ (V ) ,V ⊇ E and V open} (14.2)

Proof: Letting I denote the open intervals, recall that σ (I ) = B (R). Let

Ak = [−(k+1) ,−k)∪ [k,k+1), k = 0,1, ...

These Ak are disjoint and partition R, each being both Gδ and Fσ . The following is adefinition of Borel sets such that the set intersected with each of these Ak is perfectly ap-proximated from the inside and outside by a Fσ and Gδ set respectively. Thus

G ≡{

E ∈B (R) : Fk ⊆ E ∩Ak ⊆ Gk,µ (Gk \Fk) = 0,Gk ⊆ Ak

},

for some Gk a Gδ set and Fk an Fσ set, this holding for all k ∈ 0,1,2, · · · . From Lemma14.4.2, I ⊆ G .

I want to show that G is closed with respect to countable disjoint unions and comple-ments. Consider complements first. Say E is Borel and

Fk ⊆ E ∩Ak ⊆ Gk, µ (Gk \Fk) = 0, Gk ⊆ Ak

From Lemma 14.4.2, each Ak is both Gδ and Fσ . Thus

Ak ∩GCk ⊆ EC ∩Ak ⊆ Ak ∩FC

k

From Lemma 14.4.2, the left end is Fσ and the right end is Gδ . This is because the com-plement of a Gδ is an Fσ and the complement of an Fσ is a Gδ . Now

µ(Ak ∩FC

k \(Ak ∩GC

k))

= µ((

Ak ∩FCk)∩(AC

k ∪Gk))

= µ(Ak ∩FC

k ∩Gk)= µ (Gk \Fk) = 0

since Gk ⊆ Ak. Thus G is closed with respect to complements.Next let Ei ∈ G and let the Ei be disjoint. Is ∪∞

i=1Ei ≡ E ∈ G ? Say for each k

Fk ⊆ Ei∩Ak ⊆ Gk, µ (Gk \Fk) = 0, Gk ⊆ Ak