27.3 Integration And The Degree
There is a very interesting application of the degree to integration [?]. Recall Lemma ??.
I want to generalize this to the case where h :ℝn → ℝn is only Lipschitz continuous,
vanishing outside a bounded set. In the following proposition, let ϕε be a symmetric
Ω will be a bounded open set. By Theorem 24.6.7, h satisfies
For any p > n,
where ψm is a mollifier.
Proposition 27.3.1 Let S ⊆ h(∂Ω)C such that
where Ω is a bounded open set and also let h be Lipschitz continuous, vanishing outside
some bounded set. Then whenever ε > 0 is small enough,
for all y ∈ S.
Proof: Let ε0 > 0 be small enough that for all y ∈ S,
Now let ψm be a mollifier as m →∞ with support in B
Thus hm ∈ C∞
and for any
p > n,
as m →∞. The first claim above is obvious and the second follows by 27.3.42. Choose M
such that for m ≥ M,
Thus hm ∈Uy ∩ C2
y ∈ S.
For y ∈ S, let z ∈ B
ε < ε0
and suppose x ∈ ∂
Ω, and k,m ≥ M
. Then for
showing that for each y ∈ S, B
. By Lemma ??
all y ∈ S,
for all k,m ≥ M. By this lemma again, which says that for small enough ε the integral is
constant and the definition of the degree in Definition 21.2.3,
for all ε small enough. For x ∈ ∂Ω, y ∈ S, and t ∈
and so by Theorem 21.3.4
, the part about homotopy, for each y ∈ S,
whenever ε is small enough. Fix such an ε < ε0 and use 27.3.45 to conclude the right side
of the above equation is independent of m > M.
By 27.3.43, there exists a subsequence still denoted by m such that Dhm
p > n,
is bounded in
1 and so the integrands in
the following are uniformly integrable. By the Vitali convergence theorem, one can pass
to the limit as follows.
This proves the proposition.
Next is an interesting change of variables theorem. Let Ω be a bounded open set with
the property that ∂Ω has measure zero and let h be Lipschitz continuous on ℝn. Then
from Lemma 27.0.1, h
also has measure zero.
Now suppose f ∈ Cc
There are finitely many components of
which have nonempty intersection with spt
. From the Proposition
Actually, there exists an ε small enough that for all y ∈ spt
This is because spt
is at a positive distance from the compact set
for all ε
Using the uniform continuity of f,
you can now pass to a limit and obtain using the fact
This has proved the following interesting lemma.
Lemma 27.3.2 Let f ∈ Cc
Ω a bounded open set and let h be Lipschitz
on ℝn. Say ∂
Ω has measure zero so that h
has measure zero. Then everything is
measurable which needs to be and
Note that h is not necessarily one to one. Next is a simple corollary which replaces
in the case that
is one to one. Also another assumption is
made on there being finitely many components.
Corollary 27.3.3 Let f ∈ Lloc1
and let h be one to one and satisfy 27.3.41
- 27.3.42, ∂
Ω has measure zero for
Ω a bounded open set and h
finitely many components. Then everything is measurable which needs to be
Proof: Since d
= 0 for all
large enough due to
is no loss of generality in assuming f
is in L1
. For all
, a set of
is bounded by some constant, depending on the maximum
of the degree on the various components of
. Then from Proposition
This time, use the area formula to write
and so using the dominated convergence theorem in 27.3.47, it equals
which needs to converge to 0 as ε → 0. However, from the area formula, Theorem 27.0.11
applied to the inside integral, the above equals
which converges to 0 by continuity of translation in L1
. Thus as in the lemma,
and this proves the corollary.
Note that in this corollary h is one to one.