8.8. CONSTRUCTING MEASURES FROM FUNCTIONALS 195

the definition, µ is outer regular and so it follows from Theorem 8.7.4 that µ is regularbecause it is finite on compact sets and X is the union of countably many compact sets soµ is σ finite. Thus µ (F)≤ µ (E)≤ µ (G) = µ (F)+µ (G\F) = µ (F).

The measure µ is uniquely determined. If µ,ν are two, then if K is compact, thereis a sequence of open sets Vn decreasing to K such that µ (K) = limn→∞ µ (Vn) ,ν (K) =limn→∞ ν (Vn) . Now let K ≺ fn ≺Vn. By Claim 2, L fn→ µ (K) and L fn→ ν (K) so µ = ν

on all compact sets. Therefore, µ = ν on all Fσ sets. Now every open set V is the countableunion of a sequence of increasing compact sets from the assumptions on X so µ = ν onevery open set. It follows that the two σ algebras are the same and the measures areequal. To see this, µ

(F̂)≤ µ (E) ≤ µ

(Ĝ),ν(F̃)≤ ν (E) ≤ ν

(G̃)

where these G and Fare respectively Gδ and Fσ with µ

(Ĝ\ F̂

)= 0 similar with the other pair. So let G = G̃∩

Ĝ,F = F̃∪ F̂ and then µ (G\F) = 0,ν (G\F) = 0. Now if the two σ algebras are Fµ ,Fν

then if E ∈Fµ , then E differs from F by a subset of a set of ν measure zero. Therefore, bycompleteness of ν it follows that E ∈Fν . Also µ (E) = µ (F) = ν (F) = ν (E) . The sameargument shows that Fν ⊆Fµ . ■

Definition 8.8.3 Let L f be given by Theorem 5.8.8. That is∫Rp

f dx =∫ b1

a1

· · ·∫ bp

ap

f (x1,x2, ...,xp)dxp · · ·dx1

whenever f vanishes outside of ∏pi=1 (ai,bi). The resulting measure defined in Theorem

8.8.2, denoted as mp is Lebesgue measure.

From the above theorem mp is a Borel measure meaning that the Borel sets are measur-able. Also it has the regularity properties. What does it do to boxes?

Theorem 8.8.4 Lebesgue mesure is translation invariant. This terminology meansthat mp (E) = mp (E +z). Also mp

(∏

pi=1 (ai,bi)

)= mp

(∏

pi=1 [ai,bi]

)= ∏

pi=1 (bi−ai) .

Proof: What is mp (R) where R = ∏pi=1 (ai,bi)? Let Rn = ∏

pi=1

(ai +

1n ,bi− 1

n

)and

let fn = 1 on Rn while vanishing off of R2n and piecewise linear in each variable. Thenfrom the definition, there is g ∈ Cc (R) such that mp (R) < Lg + ε where here g ≺ R.However, since the distance from the support of g to the boundary of R is positive, itfollows that for all n large enough, g ≤ fn and so mp (R) < L fn + ε . Now letting n→∞ and computing

∫ b1a1· · ·∫ bp

apfn (x1,x2, ...,xp)dxp · · ·dx1, (You could use Problem 17 on

Page 140.) it follows that mp (R) < ∏pi=1 (bi−ai)+ ε. Since ε is arbitrary, it follows that

mp (R) ≤ ∏pi=1 (bi−ai) . In fact these will be equal because for each n,L fn ≤ mp (R) and

as just observed, L fn→∏pi=1 (bi−ai). In the case of a closed box, ∏

pi=1 (ai +δ ,bi−δ )⊆

∏pi=1 [ai,bi]⊆∏

pi=1 (ai−δ ,bi +δ ) and so for every δ > 0 and sufficiently small,

mp

(p

∏i=1

[ai,bi]

)∈

[p

∏i=1

(bi−ai−2δ ) ,p

∏i=1

(bi−ai +2δ )

]

and so mp(∏

pi=1 [ai,bi]

)= mp

(∏

pi=1 (ai,bi)

)= ∏

pi=1 (bi−ai) .

Let K be all sets of the form ∏pi=1 (ai,bi) where −∞ ≤ ai < bi ≤ ∞. Clearly F is

closed with respect to finite intersections. Let G be the Borel sets F such that

mp (z+F ∩ (−m,m)p) = mp (F ∩ (−m,m)p) .