36.2. INTEGRATION ON MANIFOLDS 1259

=∞

∑j=1

∫g′j(

Γ′j

)η j(h′j (v)

)f(h′j (v)

)J j (v)dv

the definition of Λ f using(Vi,g′i

).

This proves the lemma.This lemma and the Riesz representation theorem for positive linear functionals implies

the part of the following theorem which says the functional is well defined.

Theorem 36.2.4 Let Γ be a Cm,1 manifold. Then there exists a unique Radon measure, µ ,defined on Γ such that whenever f is a continuous function having compact support whichis defined on Γ and (Γi,gi) denotes an atlas and {ψ i} a partition of unity subordinate tothis atlas,

Λ f =∫

Γ

f dµ =∞

∑i=1

∫giΓi

ψ i f (hi (u))Ji (u)du. (36.2.4)

Also, a subset, A, of Γ is µ measurable if and only if for all r,gr (Γr ∩A) is νr measurablewhere νr is the measure defined by

νr (gr (Γr ∩A))≡∫

gr(Γr∩A)Jr (u)du

Proof: To begin, here is a claim.Claim : A set, S ⊆ Γi, has µ measure zero if and only if giS has measure zero in giΓi

with respect to the measure, ν i.Proof of the claim: Let ε > 0 be given. By outer regularity, there exists a set, V ⊆ Γi,

open3 in Γ such that µ (V ) < ε and S ⊆ V ⊆ Γi. Then giV is open in Rn and contains giS.Letting h≺ giV and h1 (x)≡ h(gi (x)) for x ∈ Γi it follows h1 ≺V . By Corollary 36.1.7 onPage 1256 there exists a partition of unity such that spt(h1)⊆ {x ∈ Rp : ψ i (x) = 1}. Thusψ jh1 (h j (u)) = 0 unless j = i when this reduces to h1 (hi (u)). It follows

ε ≥ µ (V )≥∫

Vh1dµ =

∫Γ

h1dµ

=∞

∑j=1

∫g jΓ j

ψ jh1 (h j (u))J j (u)du

=∫

giΓi

h1 (hi (u))Ji (u)du =∫

giΓi

h(u)Ji (u)du

=∫

giVh(u)Ji (u)du

Now this holds for all h≺ giV and so∫giV

Ji (u)du≤ ε.

3This means V is the intersection of an open set with Γ. Equivalently, it means that V is an open set in thetraditional way regarding Γ as a metric space with the metric it inherits from Rm.