Lemma 20.6.4Let h : V → ℝ^{n}be a C^{1}function and suppose H is a compact subset of V . Then thereexists 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 20.6.5Let A be an open set. Write C^{k}
n
(A;ℝ )
to denote a C^{k}function whose domainis A and whose range is in ℝ^{n}. Let U be an open set in ℝ^{n}. Then h ∈ C^{k}
(-- n)
U ;ℝ
if there existsan open set V ⊇Uand a function g ∈ C^{1}
(V;ℝn )
such thatg = hon U. f ∈ C^{k}
(--)
U
means thesame thing except that 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 20.6.6Let U be a bounded open set such that ∂U has zero content and let h ∈ C
(-- )
U ;ℝn
be one to one and Dh
(x)
^{−1}exists for all x ∈ U. Then h
(∂U)
= ∂
(h (U ))
and ∂
(h (U ))
has zerocontent.
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 20.6.4 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 20.6.8Let U be a bounded open set and let fX_{U}∈ℛ
(ℝn)
. Then
∫ ∫
f (x+ p )XU−p (x )dx = f (x)XU (x)dx
A few more lemmas are needed.
Lemma 20.6.9Let 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
≤ |x − y|− |x − y|+ ε
1
≤ |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 20.6.10(Urysohn’s lemma for ℝ^{n}) Let H be a closed subset of an open set U. Thenthere exists a continuous function g : ℝ^{n}→
[0,1]
such that g
(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
( )
x,UC
> 0 for all
x ∈ ℝ^{n}. Now define a continuous function g as
( C)
g(x) ≡ -----dist-x,U---------.
dist(x,H )+ dist(x,UC )
It is easy to see this verifies the conclusions of the theorem and this proves the theorem. ■
Definition 20.6.11Define spt(f) (support of f)to be the closure of the set {x : f(x)≠0}. If V isan open set, C_{c}(V ) will be the set of continuous functions f, defined on ℝ^{n}having spt(f) ⊆ V .
Definition 20.6.12If K is a compact subset of an open set V , then K ≺ ϕ ≺ V if
ϕ ∈ Cc(V ),ϕ(K ) = {1},ϕ(ℝn ) ⊆ [0,1].
Also for ϕ ∈ C_{c}(ℝ^{n}), K ≺ ϕ if
ϕ(ℝn) ⊆ [0,1] and ϕ(K) = 1.
and ϕ ≺ V if
ϕ (ℝn ) ⊆ [0,1] and spt(ϕ) ⊆ V.
Theorem 20.6.13(Partition of unity)Let K be a compact subset of ℝ^{n}and suppose
n
K ⊆ V = ∪i=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 20.6.10 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 20.6.10, 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 [?] and [22] for similar proofs.
Lemma 20.6.14Let U be a bounded open set with ∂U having content 0. Also let h ∈ C^{1}
Proof: Let ε > 0 be given. Then by Theorem 20.6.7,
x → XU (x)f (h (x))|detDh (x)|
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 19.3.2 on Page 924 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
(O )
j
is an open set.
Finitely many of these open sets
{O }
j
_{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 20.6.13, there exists a partition of unity
dx. By Lemma 20.6.8 and Fubini’s theorem this
equals
∫ ∫
X (x)(ψ f)(h (p )+ F ∘⋅⋅⋅∘F ∘ G ∘⋅⋅⋅∘G (x))⋅
ℝn−1 ℝ Q−pj j i 1 n−1 n 1
|DF (Gn ∘⋅⋅⋅∘G1 (x))||DGn (Gn −1 ∘⋅⋅⋅∘G1 (x))|⋅
|DGn −1(Gn −2 ∘ ⋅⋅⋅∘ G1 (x))| (20.29)
⋅⋅⋅|DG2 (G1(x))||DG1 (x )|dx1dVn−1. (20.30)
The vertical lines in the above signify the absolute value of the determinant of the matrix on the
inside. Here dV_{n−1} is with respect to the variables x_{2},
⋅⋅⋅
,x_{n}. Also F denotes F_{1}∘
⋅⋅⋅
∘ F_{n−1}.
Now
T
G1 (x ) = (α(x),x2,⋅⋅⋅,xn)
and is one to one. Therefore, fixing x_{2},
⋅⋅⋅
,x_{n}, x_{1}→ α
(x)
is one to one. Also
|DG1 (x)| = |αx1 (x)|
Fixing x_{2},
⋅⋅⋅
,x_{n}, change the variable,
y1 = α (x1,x2,⋅⋅⋅,xn),dy1 = αx1 (x1,x2,⋅⋅⋅,xn)dx1
Thus
T − 1 −1 ′
x = (x1,x2,⋅⋅⋅,xn) = G1 (y1,x2,⋅⋅⋅,xn ) ≡ G 1 (x)
Then in (20.30) you can use Corollary 20.6.3 to write (20.30) as
∫ ∫ ( )
XQ−pj G −11(x′) (ψjf)
(ℝn−1 ℝ ( −1 ′))
h (pi)+ F1 ∘⋅⋅⋅∘Fn −1 ∘Gn ∘⋅⋅⋅∘G1 G 1 (x)
|| ( ( −1 ′ ))|||| ( ( −1 ′))||
⋅ DF Gn ∘ ⋅⋅⋅∘ G1 G 1 (x ) DGn Gn−1 ∘⋅⋅⋅∘G1 G 1 (x) ⋅