First is an easy result about approximation of continuous functions with smooth
ones.
Theorem 21.1.1Let Ω be a bounded open set in ℝnand let f ∈ Cc
n
(ℝ )
. Thenthere exists g ∈ Cc∞
n
(ℝ )
with
||g− f||
∞< ε.
Proof:Form g ≡ f ∗ψn for a mollifier ψn. This will approximate f uniformly on Ω,
and will be in Cc∞
(ℝn )
. ■
Using the Weierstrass approximation theorem, you could also get g to equal a
function from G described in the development of the Fourier transform for all x ∈ Ω.
Simply apply Theorem 7.2.9 to the functions in G.
Applying this result to the components of a vector valued function yields the following
corollary.
Corollary 21.1.2If f ∈ C
(Ω;ℝn)
for Ω a bounded subset of ℝn, then for all ε > 0,there exists g ∈ C∞
(Ω;ℝn )
such that
||g − f||∞ < ε.
The following lemma is a wonderful application of the Vitali covering theorem.
Lemma 21.1.3Let h be differentiable on U. If T ⊆ U and mp
(T)
= 0, thenmp
(h(T ))
= 0.
Proof:For k ∈ ℕ
T ≡ {x ∈ T : ||Dh (x)|| < k}
k
and let ε > 0 be given. Since Tk is a subset of a set of measure zero, it is measurable, but
we don’t need to pay much attention to this fact. Now by outer regularity, there exists an
open set V , containing Tk which is contained in U such that mp
(V)
< ε. Let x ∈ Tk.
Then by differentiability,
h(x + v) = h (x )+ Dh (x)v+ o (v)
and so there exist arbitrarily small rx< 1 such that B
(x,5rx)
⊆ V and whenever
||v||
≤ 5rx,
||o(v )||
<
1
5
||v||
. Thus
h (B(x,5rx)) ⊆ Dh (x)(B(0,5rx))+ h(x)+ B (0,rx) ⊆ B(0,k5rx)+
+B (0,rx)+ h (x) ⊆ B (h (x),(5k + 1)rx) ⊆ B (h (x),6krx)
From the Vitali covering theorem, there exists a countable disjoint sequence of these
balls,
{B (xi,ri)}
i=1∞ such that
{B(xi,5ri)}
i=1∞ =
{ }
^Bi
i=1∞ covers Tk Then letting
mp denote the outer measure determined by mp,
∑∞ ---( (^ )) ∞∑
≤ mp h Bi ≤ mp (B (h (xi),6krxi))
i=1 i=1
∞ ∞
= ∑ m (B (x ,6kr )) = (6k)p∑ m (B (x,r ))
i=1 p i xi i=1 p i xi
≤ (6k)pm (V ) ≤ (6k)pε.
p
Since ε > 0 is arbitrary, this shows mp
(h (Tk))
=mp
(h (Tk))
= 0. Now
mp (h (T)) = kli→m∞ mp (h(Tk)) = 0.■
The following is Sard’s lemma. It is important in defining the degree.
Lemma 21.1.4(Sard) Let U be an open set in ℝpand let h : U → ℝpbe differentiable.Let
Z ≡ {x ∈ U : detDh (x) = 0} .
Then mp
(h(Z ))
= 0.
Proof: For convenience, assume the balls in the following argument come from
||⋅||
∞.
First note that Z is a Borel set because h is continuous and so the component functions
of the Jacobian matrix are each Borel measurable. Hence the determinant is also Borel
measurable.
Suppose that U is a bounded open set. Let ε > 0 be given. Also let V ⊇ Z with
V ⊆ U open, and
mp (Z)+ ε > mp(V ).
Now let x ∈ Z. Then since h is differentiable at x, there exists δx> 0 such that if r < δx,
then B
(x,r)
⊆ V and also o
(v)
< η
∥v∥
for
∥v∥
< r. Thus
h (x+B (0,r)) = h(B (x,r)) ⊆ h (x )+ Dh (x)(B (0,r)) +B (0,rη), η < 1.
Regard Dh
(x)
as an n×n matrix, the matrix of the linear transformation Dh
(x)
with
respect to the usual coordinates. Since x ∈ Z, it follows that there exists an invertible
matrix A such that ADh
(x)
is in row reduced echelon form with a row of zeros on the
bottom. Therefore,
mp (A (h (B (x,r)))) ≤ mp (ADh (x )(B (0,r))+ AB (0,rη)) (21.1.1)
(21.1.1)
The diameter of ADh
(x)
(B (0,r))
is no larger than
||A||
||Dh (x)||
2r and it lies in
ℝp−1×
{0}
. The diameter of AB
(0,rη)
is no more than
||A||
(2rη)
.Therefore, the
measure of the right side in 21.1.1 is no more than
[(||A||||Dh (x)||2r+ ||A ||(2η))r]p−1(rη)
p
≤ C (||A ||,||Dh (x)||)(2r) η
That is,
m (A (h(B (x,r)))) ≤ C (||A ||,||Dh (x)||)(2r)p η
p
Hence from the change of variables formula for linear maps,
C-(||A||,||Dh--(x)||)
mp (h(B (x,r))) ≤ η |det(A)| mp (B (x,r))
Then letting δx be still smaller if necessary, corresponding to sufficiently small
η,
m (h (B (x,r))) ≤ εm (B (x,r))
p p
The balls of this form constitute a Vitali cover of Z. Hence, by the Vitali covering
theorem, Corollary 11.4.6, there exists
{Bi}
i=1∞,Bi = Bi
(xi,ri)
, a collection of disjoint
balls, each of which is contained in V, such that mp