1260 CHAPTER 36. INTEGRATION ON MANIFOLDS

Since ε is arbitrary, this shows giV has measure no more than ε with respect to the measure,ν i. Since ε is arbitrary, giS has measure zero.

Consider the converse. Suppose giS has ν i measure zero. Then there exists an open set,O⊆ giΓi such that O⊇ giS and ∫

OJi (u)du < ε.

Thus hi (O) is open in Γ and contains S. Let h≺ hi (O) be such that∫Γ

hdµ + ε > µ (hi (O))≥ µ (S) (36.2.5)

As in the first part, Corollary 36.1.7 on Page 1256 implies there exists a partition of unitysuch that h(x) = 0 off the set,

{x ∈ Rp : ψ i (x) = 1}

and so as in this part of the argument,∫Γ

hdµ ≡∞

∑j=1

∫g jU j

ψ jh(h j (u))J j (u)du

=∫

giΓi

h(hi (u))Ji (u)du

=∫

O∩giΓi

h(hi (u))Ji (u)du

≤∫

OJi (u)du < ε (36.2.6)

and so from 36.2.5 and 36.2.6 µ (S)≤ 2ε . Since ε is arbitrary, this proves the claim.For the last part of the theorem, it suffices to let A ⊆ Γr because otherwise, the above

argument would apply to A∩Γr. Thus let A ⊆ Γr be µ measurable. By the regularity ofthe measure, there exists an Fσ set, F and a Gδ set, G such that Γr ⊇ G ⊇ A ⊇ F andµ (G\F) = 0.(Recall a Gδ set is a countable intersection of open sets and an Fσ set is acountable union of closed sets.) Then since Γr is compact, it follows each of the closed setswhose union equals F is a compact set. Thus if F = ∪∞

k=1Fk, gr (Fk) is also a compact setand so gr (F) = ∪∞

k=1gr (Fk) is a Borel set. Similarly, gr (G) is also a Borel set. Now by theclaim, ∫

gr(G\F)Jr (u)du = 0.

Since gr is one to one,grG\grF = gr (G\F)

and sogr (F)⊆ gr (A)⊆ gr (G)

where gr (G) \ gr (F) has measure zero. By completeness of the measure, νr, gr (A) ismeasurable. It follows that if A⊆ Γ is µ measurable, then gr (Γr ∩A) is νr measurable forall r. The converse is entirely similar. This proves the theorem.