Manifolds are things which locally appear to be ℝ^{n} for some n. The extent to which they
have such a local appearance varies according to various analytical characteristics which
the manifold possesses.
Definition 36.2.1Let U ⊆ ℝ^{n}be an open set and let h : U → ℝ^{m}. Then for r ∈ [0,1),h ∈ C^{k,r}
(U )
for k a nonnegative integer means that D^{α}h exists for all
|α |
≤ k and eachD^{α}h is Holder continuous with exponent r. That is
|D αh(x)− D αh (y )| ≤ K |x − y |r.
Also h ∈ C^{k,r}
(-)
U
if it is the restriction of a function of C^{k,r}
n
(ℝ )
to U.
Definition 36.2.2Let Γ be a closed subset of ℝ^{p}where p ≥ n. Suppose Γ =
∪_{i=1}^{∞}Γ_{i}where Γ_{i} = Γ∩W_{i}for W_{i}a bounded open set. Suppose also
{Wi }
_{i=1}^{∞}islocally finite. This means every bounded open set intersects only finitely many. Alsosuppose there are open bounded sets, U_{i}having Lipschitz boundaries and functionsh_{i} : U_{i}→ Γ_{i}which are one to one, onto, and in C^{m,1}
(Ui)
. Suppose also there existfunctions, g_{i} : W_{i}→ U_{i}such that g_{i}is C^{m,1}
(Wi)
, and g_{i}∘ h_{i} = idon U_{i}whileh_{i}∘g_{i} = idon Γ_{i}. The collection of sets, Γ_{j}and mappings, g_{j},
{( )}
Γ j,gj
is calledan atlas and an individual entry in the atlas is called a chart. Thus
( )
Γ j,gj
is achart. Then Γ as just described is called a C^{m,1}manifold. The number, m is justa nonnegative integer. When m = 0 this would be called a Lipschitz manifold, theleast smooth of the manifoldsdiscussed here.
For example, take p = n + 1 and let
hi(u) = (u1,⋅⋅⋅,ui,ϕi(u ),ui+1,⋅⋅⋅,un)T
for u =
(u1,⋅⋅⋅,ui,ui+1,⋅⋅⋅,un)
^{T}∈ U_{i} for ϕ_{i}∈ C^{m,1}
(Ui)
and g_{i} : U_{i}× ℝ → U_{i} given
by
gi(u1,⋅⋅⋅,ui,y,ui+1,⋅⋅⋅,un) ≡ u
for i = 1,2,
⋅⋅⋅
,p. Then for u ∈ U_{i}, the definition gives
gi ∘hi(u) = gi(u1,⋅⋅⋅,ui,ϕi(u),ui+1,⋅⋅⋅,un) = u
and for Γ_{i}≡ h_{i}
(Ui)
and
(u1,⋅⋅⋅,ui,ϕi(u),ui+1,⋅⋅⋅,un)
^{T}∈ Γ_{i},
hi ∘ gi(u1,⋅⋅⋅,ui,ϕi(u),ui+1,⋅⋅⋅,un)
= hi(u) = (u1,⋅⋅⋅,ui,ϕi(u),ui+1,⋅⋅⋅,un)T .
This example can be used to describe the boundary of a bounded open set and since
ϕ_{i}∈ C^{m,1}
(U )
i
, such an open set is said to have a C^{m,1} boundary. Note also
that in this example, U_{i} could be taken to be ℝ^{n} or if U_{i} is given, both h_{i} and
and g_{i} can be taken as restrictions of functions defined on all of ℝ^{n} and ℝ^{p}
respectively.
The symbol, I will refer to an increasing list of n indices taken from
{1,⋅⋅⋅,p}
.
Denote by Λ
(p,n)
the set of all such increasing lists of n indices.
∞ ∫
= ∑ η (h ′(v )) f (h′(v))J (v)dv
j=1 g′j(Γ ′j) j j j j
′
the definition of Λf using (Vi,gi).
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 36.2.4Let Γ be a C^{m,1}manifold. Then there exists a unique Radonmeasure, μ, defined on Γ such that whenever f is a continuous function having compactsupport which is defined on Γ and
(Γ i,gi)
denotes an atlas and
{ψi}
a partition of unitysubordinate to this atlas,
∫ ∑∞ ∫
Λf = f dμ = ψif (hi(u))Ji(u)du. (36.2.4)
Γ i=1 giΓ i
(36.2.4)
Also, a subset, A, of Γ is μ measurable if and only if for all r,g_{r}
(Γ r ∩ A)
is ν_{r}measurable where ν_{r}is the measure defined by
∫
ν (g (Γ ∩ A)) ≡ J (u )du
r r r gr(Γr∩A) r
Proof: To begin, here is a claim.
Claim : A set, S ⊆ Γ_{i}, has μ measure zero if and only if g_{i}S has measure zero in g_{i}Γ_{i}
with respect to the measure, ν_{i}.
Proof of the claim: Let ε > 0 be given. By outer regularity, there exists a set, V ⊆ Γ_{i},
open^{3}
in Γ such that μ
(V)
< ε and S ⊆ V ⊆ Γ_{i}. Then g_{i}V is open in ℝ^{n} and contains g_{i}S.
Letting h ≺ g_{i}V and h_{1}
(x)
≡ h
(gi(x))
for x ∈ Γ_{i} it follows h_{1}≺ V . By Corollary 36.1.7
on Page 4437 there exists a partition of unity such that spt
(h1)
⊆
p
{x ∈ ℝ : ψi(x) = 1}
.
Thus ψ_{j}h_{1}
(hj (u))
= 0 unless j = i when this reduces to h_{1}
(hi(u ))
. It follows
∫ ∫
ε ≥ μ(V) ≥ h dμ = hdμ
V 1 Γ 1
∞∑ ∫
= ψjh1(hj(u))Jj(u)du
j∫=1 gjΓ j ∫
= giΓ i h1(hi(u))Ji(u)du = giΓ i h (u )Ji(u )du
∫
= h(u)Ji(u)du
giV
Now this holds for all h ≺ g_{i}V and so
∫
Ji(u )du ≤ ε.
giV
Since ε is arbitrary, this shows g_{i}V has measure no more than ε with respect to the
measure, ν_{i}. Since ε is arbitrary, g_{i}S has measure zero.
Consider the converse. Suppose g_{i}S has ν_{i} measure zero. Then there exists an open
set, O ⊆ g_{i}Γ_{i} such that O ⊇ g_{i}S and
∫
J (u)du < ε.
O i
Thus h_{i}
(O)
is open in Γ and contains S. Let h ≺ h_{i}
(O )
be such that
∫
Γ hdμ + ε > μ (hi(O )) ≥ μ(S) (36.2.5)
(36.2.5)
As in the first part, Corollary 36.1.7 on Page 4437 implies there exists a partition of unity
such that h
(x)
= 0 off the set,
{x ∈ ℝp : ψ (x ) = 1}
i
and so as in this part of the argument,
∫ ∞ ∫
hd μ ≡ ∑ ψjh(hj(u))Jj(u)du
Γ j=1 gjUj
∫
= h (hi(u )) Ji(u) du
∫giΓ i
= h(hi(u))Ji(u )du
O∩giΓ i
∫
≤ O Ji(u )du < ε (36.2.6)
≤ 2ε. Since ε is arbitrary, this proves the
claim.
For the last part of the theorem, it suffices to let A ⊆ Γ_{r} because otherwise, the
above argument would apply to A ∩ Γ_{r}. Thus let A ⊆ Γ_{r} be μ measurable.
By the regularity of the measure, there exists an F_{σ} set, F and a G_{δ} set, G
such that Γ_{r}⊇ G ⊇ A ⊇ F and μ
(G ∖F )
= 0.(Recall a G_{δ} set is a countable
intersection of open sets and an F_{σ} set is a countable union of closed sets.) Then
since Γ_{r} is compact, it follows each of the closed sets whose union equals F is
a compact set. Thus if F = ∪_{k=1}^{∞}F_{k}, g_{r}
(F )
k
is also a compact set and so
g_{r}
(F )
= ∪_{k=1}^{∞}g_{r}
(F )
k
is a Borel set. Similarly, g_{r}
(G)
is also a Borel set. Now by the
claim,
∫
Jr(u)du = 0.
gr(G∖F)
Since g_{r} is one to one,
grG ∖grF = gr (G ∖ F)
and so
g (F) ⊆ g (A ) ⊆ g (G )
r r r
where g_{r}
(G)
∖ g_{r}
(F)
has measure zero. By completeness of the measure, ν_{r}, g_{r}
(A)
is
measurable. It follows that if A ⊆ Γ is μ measurable, then g_{r}
(Γ r ∩A )
is ν_{r} measurable
for all r. The converse is entirely similar. This proves the theorem.
Corollary 36.2.5Let f ∈ L^{1}
(Γ ;μ)
and suppose f
(x)
= 0 for all x
∈∕
Γ_{r}where
(Γ r,gr)
is a chart. Then
∫ ∫ ∫
Γ fdμ = Γ r fdμ = grΓ r f (hr (u )) Jr(u )du. (36.2.7)
(36.2.7)
Furthermore, if
{(Γ i,gi)}
is an atlas and
{ψi}
is a partition of unity as described earlier,then for any f ∈ L^{1}
(Γ ,μ)
,
∫ ∞ ∫
f dμ = ∑ ψ f (h (u))J (u)du. (36.2.8)
Γ r=1 grΓ r r r r
(36.2.8)
Proof:Let f ∈ L^{1}
(Γ ,μ )
with f = 0 off Γ_{r}. Without loss of generality assume f ≥ 0
because if the formulas can be established for this case, the same formulas are obtained
for an arbitrary complex valued function by splitting it up into positive and negative
parts of the real and imaginary parts in the usual way. Also, let K ⊆ Γ_{r} a compact set.
Since μ is a Radon measure there exists a sequence of continuous functions,
{fk}
,f_{k}∈ C_{c}
(Γ r)
, which converges to f in L^{1}
(Γ ,μ)
and for μ a.e. x. Take the partition
of unity,
{ψi}
to be such that
K ⊆ {x : ψr(x) = 1}.
Therefore, the sequence
{fk(hr(⋅))}
is a Cauchy sequence in the sense that
∫
kl,lim→∞ |fk(hr (u ))− fl(hr(u))|Jr(u)du = 0
gr(K )
It follows there exists g such that
∫
|f (h (u ))− g (u )|J (u)du → 0,
gr(K ) k r r
and
g ∈ 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. h_{r}
(u)
∈ K. Therefore,
∫ ∫ ∫
f dμ = lim f dμ = lim f (h (u))J (u)du
K k→∞ K k k→ ∞ gr(K )k r r
∫ ∫
= g(u)Jr(u)du = f (hr(u))Jr(u)du. (36.2.9)
gr(K ) gr(K )