28.1. THE DEFINITION 995

If∞

∑k=1

µ(S∩ (Kk \Kk+1))< ∞, (28.1.4)

then µ(S∩ (Kn \K))→ 0 because it is dominated by the tail of a convergent series so itsuffices to show 28.1.4.

M

∑k=1

µ(S∩ (Kk \Kk+1)) =

∑k even, k≤M

µ(S∩ (Kk \Kk+1))+ ∑k odd, k≤M

µ(S∩ (Kk \Kk+1)). (28.1.5)

By the construction, the distance between any pair of sets, S∩(Kk \Kk+1) for different evenvalues of k is positive and the distance between any pair of sets, S∩(Kk \Kk+1) for differentodd values of k is positive. Therefore,

∑k even, k≤M

µ(S∩ (Kk \Kk+1))+ ∑k odd, k≤M

µ(S∩ (Kk \Kk+1))≤

µ(⋃

k even

S∩ (Kk \Kk+1))+µ(⋃

k odd

S∩ (Kk \Kk+1))≤ 2µ (S)< ∞

and so for all M, ∑Mk=1 µ(S∩ (Kk \Kk+1))≤ 2µ (S) showing 28.1.4 .

With the above theorem, the following theorem is easy to obtain.

Theorem 28.1.5 The σ algebra of H s measurable sets contains the Borel sets and H s

has the property that for all E ⊆ Rn, there exists a Borel set F ⊇ E such that H s(F) =H s(E).

Proof: Let dist(A,B) = 2δ 0 > 0. Is it the case that

H s(A)+H s(B) = H s(A∪B)?

This is what is needed to use Caratheodory’s criterion.Let {C j}∞

j=1be a covering of A∪B such that diam(C j)≤ δ < δ 0 for each j and

H sδ(A∪B)+ ε >

∞

∑j=1

β (s)(r (C j))s.

ThusH s

δ(A∪B)̇+ ε > ∑

j∈J1

β (s)(r (C j))s + ∑

j∈J2

β (s)(r (C j))s

whereJ1 = { j : C j ∩A ̸= /0}, J2 = { j : C j ∩B ̸= /0}.

Recall dist(A,B) = 2δ 0, J1∩ J2 = /0. It follows

H sδ(A∪B)+ ε > H s

δ(A)+H s

δ(B).