Lemma C.6.4Let h : V → ℝ^{n}be a C^{1}function and suppose H is a compact subsetof V . Then there exists a constant C independent of x ∈ H such that
|Dh (x)v| ≤ C |v|.
Proof: Consider the compact set H ×∂B
(0,1)
⊆ ℝ^{2n}. Let f : H ×∂B
(0,1)
→ ℝ be
given by f
(x,v)
=
|Dh (x)v |
. Then let C denote the maximum value of f. It follows
that for v ∈ ℝ^{n},
|| ||
||Dh (x)-v-|| ≤ C
|v|
and so the desired formula follows when you multiply both sides by
|v|
. ■
Definition C.6.5Let A be an open set. Write C^{k}
n
(A;ℝ )
to denote aC^{k}function whose domain is A and whose range is in ℝ^{n}. Let U be an open setin ℝ^{n}. Then h ∈ C^{k}
(-- n)
U;ℝ
if there exists an open set V ⊇Uand a functiong ∈ C^{1}
n
(V;ℝ )
such thatg = hon U. f ∈ C^{k}
(-)
U
means the same thing exceptthat f has values in ℝ. Also recall that x ∈ ∂U means that every open set whichcontains x contains points of U and points of U^{C}
Theorem C.6.6Let U be a bounded open set such that ∂U has zero contentand let h ∈ C
(-- )
U;ℝn
be one to one and Dh
(x)
^{−1}exists for all x ∈ U. Thenh
(∂U)
= ∂
(h(U))
and ∂
(h (U))
has zero content.
Proof: Let x ∈ ∂U and let g = h where g is a C^{1} function defined on an open set
containing U. By the inverse function theorem, g is locally one to one and an open
mapping near x. Thus g
(x)
= h
(x)
and is in an open set containing points of g
(U)
and
points of g
( C)
U
. These points of g
( C )
U
cannot equal any points of h
(U )
because g is
one to one locally. Thus h
(x)
∈ ∂
(h (U))
and so h
(∂U)
⊆ ∂
(h(U ))
. Now suppose
y ∈ ∂
(h(U ))
. By the inverse function theorem y cannot be in the open set h
(U)
. Since
y ∈ ∂
(h(U ))
, every ball centered at y contains points of h
(U )
and so y ∈h
(U)
∖h
(U )
.
Thus there exists a sequence,
{xn}
⊆ U such that h
(xn)
→y. But then, by the
continuity of h^{−1} which comes from the inverse function theorem, x_{n}→ h^{−1}
(y)
and so
h^{−1}
(y)
∈∕
U but is in U. Thus h^{−1}
(y )
∈ ∂U. (Why?) Therefore, y ∈ h
(∂U)
,
and this proves the two sets are equal. It remains to verify the claim about
content.
First let H denote a compact set whose interior contains U which is also in the
interior of the domain of g. Now since ∂U has content zero, it follows that for ε > 0
given, there exists a grid G such that if G^{′} are those boxes of G which have nonempty
intersection with ∂U, then
∑
′v (Q ) < ε
Q ∈G
and by refining the grid if necessary, no box of G has nonempty intersection with both U
and H^{C}. Refining this grid still more, you can also assume that for all boxes in
G^{′},
li-< 2
lj
where l_{i} is the length of the i^{th} side. (Thus the boxes are not too far from being
cubes.)
Let C be the constant of Lemma C.6.4 applied to g on H.
Now consider one of these boxes, Q ∈G^{′}. If x,y∈ Q, it follows from the chain rule
that
where U is a bounded open set with∂U having content 0. Then fX_{U}∈ℛ
(ℝn)
.
Proof: Let H be a compact set whose interior contains U which is also contained in
the domain of g where g is a continuous functions whose restriction to U equals
f. Consider gX_{U}, a function whose set of discontinuities has content 0. Then
gX_{U} = fX_{U}∈ℛ
(ℝn)
as claimed. This is by the big theorem which tells which functions
are Riemannn integrable. ■
The symbol U − p is defined as
{x − p : x ∈ U}
. It merely slides U by the vector p.
The following lemma is obvious from the definition of the integral.
Lemma C.6.8Let U be a bounded open set and let fX_{U}∈ℛ
n
(ℝ )
. Then
∫ ∫
f (x +p )XU−p (x)dx = f (x)XU (x)dx
A few more lemmas are needed.
Lemma C.6.9Let S be a nonempty subset of ℝ^{n}. Define
f (x) ≡ dist(x,S) ≡ inf{|x − y| : y ∈ S}.
Then f is continuous.
Proof: Consider
|f (x)− f (x )|
1
and suppose without loss of generality that
f
(x )
1
≥ f
(x)
. Then choose y ∈ S such that f
(x)
+ ε >
|x− y|
. Then
|f (x1)− f (x)| = f (x1)− f (x ) ≤ f (x1)− |x− y|+ ε
≤ |x1 − y|− |x − y|+ ε
≤ |x− x1|+ |x − y|− |x − y|+ ε
= |x− x1|+ ε.
Since ε is arbitrary, it follows that
|f (x1)− f (x)|
≤
|x − x1|
and this proves the lemma.
■
Theorem C.6.10(Urysohn’s lemma for ℝ^{n}) Let H be a closed subset ofan open set U. Then there exists a continuous function g : ℝ^{n}→
[0,1]
such thatg
(x)
= 1 for all x ∈ H and g
(x)
= 0 for all x
∕∈
U.
Proof: If x
∕∈
C, a closed set, then dist
(x,C)
> 0 because there exists δ > 0 such that
B
(x,δ)
∩ C = ∅. This is because, since C is closed, its complement is open. Therefore,
dist
(x,H)
+ dist
( C)
x,U
> 0 for all x ∈ ℝ^{n}. Now define a continuous function g
as
It is easy to see this verifies the conclusions of the theorem and this proves the theorem.
■
Definition C.6.11Define spt(f) (support of f)to be the closure of theset {x : f(x)≠0}. If V is an open set, C_{c}(V ) will be the set of continuous functionsf, defined on ℝ^{n}having spt(f) ⊆ V .
Definition C.6.12If K is a compact subset of an open set V , then K ≺ ϕ ≺ Vif
ϕ ∈ Cc(V),ϕ(K) = {1},ϕ(ℝn ) ⊆ [0,1].
Also for ϕ ∈ C_{c}(ℝ^{n}), K ≺ ϕ if
n
ϕ(ℝ ) ⊆ [0,1] and ϕ(K ) = 1.
and ϕ ≺ V if
n
ϕ (ℝ ) ⊆ [0,1] and spt(ϕ) ⊆ V.
Theorem C.6.13(Partition of unity)Let K be a compact subset of ℝ^{n}andsuppose
K ⊆ V = ∪ni=1Vi,Vi open and bounded.
Then there exist ψ_{i}≺ V_{i}with
∑n
ψi(x ) = 1
i=1
for all x ∈ K.
Proof: Let K_{1} = K ∖∪_{i=2}^{n}V_{i}. Thus K_{1} is compact because it is the intersection of a
closed set with a compact set and K_{1}⊆ V_{1}. Let K_{1}⊆ W_{1}⊆W_{1}⊆ V_{1}with
W_{1}compact. To obtain W_{1}, use Theorem C.6.10 to get f such that K_{1}≺ f ≺ V_{1}
and let W_{1}≡
{x : f (x) ⁄= 0}
.Thus W_{1},V_{2},
⋅⋅⋅
V_{n} covers K and W_{1}⊆ V_{1}.
Let K_{2} = K ∖ (∪_{i=3}^{n}V_{i}∪ W_{1}). Then K_{2} is compact and K_{2}⊆ V_{2}. Let
K_{2}⊆ W_{2}⊆W_{2}⊆ V_{2}W_{2} compact. Continue this way finally obtaining W_{1},
⋅⋅⋅
,W_{n},
K ⊆ W_{1}∪
⋅⋅⋅
∪ W_{n}, and W_{i}⊆ V_{i};W_{i} compact. Now let W_{i}⊆ U_{i}⊆U_{i}⊆ V_{i},U_{i}
compact.
PICT
By Theorem C.6.10, there exist functions ϕ_{i},γ such that U_{i}≺ ϕ_{i}≺ V_{i},∪_{i=1}^{n}W_{i}≺ γ ≺∪_{i=1}^{n}U_{i}.
Define
∪_{i=1}^{n}U_{i}. Consequently γ(y) = 0 for all y
near x and so ψ_{i}(y) = 0 for all y near x. Hence ψ_{i} is continuous at such x. If
∑_{j=1}^{n}ϕ_{j}(x)≠0, this situation persists near x and so ψ_{i} is continuous at such points.
Therefore ψ_{i} is continuous. If x ∈ K, then γ(x) = 1 and so ∑_{j=1}^{n}ψ_{j}(x) = 1. Clearly
0 ≤ ψ_{i}
(x)
≤ 1 and spt(ψ_{j}) ⊆ V_{j}. ■
The next lemma contains the main ideas. See [36] and [32] for similar proofs.
Lemma C.6.14Let U be a bounded open set with ∂U having content 0. Also leth ∈ C^{1}
is Riemannn integrable. Therefore, there exists a grid G such that, letting
g(x) = XU (x)f (h (x))|detDh (x)|,
ℒG(g)+ ε > UG(g).
Let K denote the union of the boxes Q of G which intersect U. Thus K is a compact
subset of V where V is a bounded open set containing U, and it is only the terms from
these boxes which contribute anything nonzero to the lower sum. By Theorem B.3.2 on
Page 1802 which is stated above and the inverse function theorem, it follows that for
p ∈ K, there exists an open set contained in U which contains p, denoted as O_{p} such
that for x ∈ O_{p}−p,
h (x+ p) − h(p) = F1 ∘⋅⋅⋅∘Fn −1 ∘Gn ∘⋅⋅⋅∘G1 (x)
where the G_{i} are primitive functions, and the F_{j} are flips. Also h
(Oj )
is an open
set.
Finitely many of these open sets
{Oj}
_{j=1}^{q} cover K. Let the distinguished point for
O_{j} be denoted by p_{j}. Now refine G if necessary, such that the diameter of every cell of
the new G which intersects U is smaller than a Lebesgue number for this open cover.
Denote by G^{′} those boxes of the new G which intersect U. Thus the union of these boxes
of G^{′} equals the set K and every box of G^{′} is contained in one of these O_{j}. By Theorem
C.6.13, there exists a partition of unity