Kenneth Kuttler

36.2. INTEGRATION ON MANIFOLDS 1261

Corollary 36.2.5 Let f ∈ L1 (Γ; µ) and suppose f (x) = 0 for all x /∈ Γr where (Γr,gr) isa chart. Then ∫

f dµ =∫

Γr

f dµ =∫

grΓr

f (hr (u))Jr (u)du. (36.2.7)

Furthermore, if {(Γi,gi)} is an atlas and {ψ i} is a partition of unity as described earlier,then for any f ∈ L1 (Γ,µ),∫

f dµ =∞

∑r=1

∫grΓr

ψr f (hr (u))Jr (u)du. (36.2.8)

Proof: Let f ∈ L1 (Γ,µ) with f = 0 off Γr. Without loss of generality assume f ≥ 0because if the formulas can be established for this case, the same formulas are obtained foran arbitrary complex valued function by splitting it up into positive and negative parts ofthe real and imaginary parts in the usual way. Also, let K ⊆ Γr a compact set. Since µ is aRadon measure there exists a sequence of continuous functions, { fk} , fk ∈Cc (Γr), whichconverges to f in L1 (Γ,µ) and for µ a.e. x. Take the partition of unity, {ψ i} to be suchthat

K ⊆ {x : ψr (x) = 1} .

Therefore, the sequence { fk (hr (·))} is a Cauchy sequence in the sense that

limk,l→∞

∫gr(K)| fk (hr (u))− fl (hr (u))|Jr (u)du = 0

It follows there exists g such that∫gr(K)| fk (hr (u))−g(u)|Jr (u)du→ 0,

andg ∈ L1 (grK;νr) .

By the pointwise convergence and the claim used in the proof of Theorem 36.2.4,

g(u) = f (hr (u))

for µ a.e. hr (u) ∈ K. Therefore,∫K

f dµ = limk→∞

∫K

fkdµ = limk→∞

∫gr(K)

fk (hr (u))Jr (u)du

=∫

gr(K)g(u)Jr (u)du =

∫gr(K)

f (hr (u))Jr (u)du. (36.2.9)

Now let · · ·K j ⊆ K j+1 · · · and ∪∞j=1K j = Γr where K j is compact for all j. Replace K in

36.2.9 with K j and take a limit as j→ ∞. By the monotone convergence theorem,∫Γr

f dµ =∫

gr(Γr)f (hr (u))Jr (u)du.

36.2. INTEGRATION ON MANIFOLDS 1261Corollary 36.2.5 Let f € L'! (V3) and suppose f (x) =0 for all x ¢ 1, where (,,g,) isa chart. Then[fau=[ fau= ff (r(w))J,(u)du, (36.2.7)T T; g/lFurthermore, if {(T;,g;)} is an atlas and {y;} is a partition of unity as described earlier,then for any f € L' (T,w),[ fan = y Ls. Wf (hy (u)) J; (1) du (36.2.8)Proof: Let f € L' (I, yw) with f = 0 off I’. Without loss of generality assume f > 0because if the formulas can be established for this case, the same formulas are obtained foran arbitrary complex valued function by splitting it up into positive and negative parts ofthe real and imaginary parts in the usual way. Also, let K CI a compact set. Since U is aRadon measure there exists a sequence of continuous functions, {f,}, fi; € Ce (I), whichconverges to f in L'(I’,w) and for p a.e. x. Take the partition of unity, {y;} to be suchthatKC {x: w,(x) = 1}.Therefore, the sequence { f; (h, (-))} is a Cauchy sequence in the sense thatlim eh r(u)) — fi (hy (u))| J, (u) du = 0k,l,It follows there exists g such thatIf eZee (o)) ~ a (u)| J(u) du 0,andgE L! (g-K;V,).By the pointwise convergence and the claim used in the proof of Theorem 36.2.4,g(u) = f(h, (u))for Lt a.e. h,(u) € K. Therefore,I fdu = lim | fd it = lim i op filtte(w)) Jr (a) duk-00 J K k-00 2,[8m (u)du= [fit (u)) J, (u) du. (36.2.9)Now let ---Kj Kj41--- and U5_,K; =I’, where Kj is compact for all j. Replace K in36.2.9 with K; and take a limit as j —+ oo. By the monotone convergence theorem,[ fdau=[ — f(h,(u))J-(u)du.JT; J¢g-(T;)