where F = RU with R preserving distances and U a symmetric matrix having all positive
eigenvalues. The area formula gives a generalization of this simple relationship to the case
where F is replaced by a nonlinear mapping h. It contains as a special case the earlier change
of variables formula. There are two parts to this development. The first part is to generalize
Lemma 19.4.5 to the case of nonlinear maps. When this is done, the area formula can be
presented.
In the first version of the area formula h will be a Lipschitz function,
|h (x)− h(y)| ≤ K |x− y|
defined on ℝ^{n} which is one to one on G, a measurable subset of ℝ^{n}. This is no loss of
generality because of Theorem 17.3.4.
The following lemma states that Lipschitz maps take sets of measure zero to sets of
measure zero. It also gives a convenient estimate. It involves the consideration of ℋ^{n} as an
outer measure. Thus it is not necessary to know the set B is measurable.
Lemma 20.0.1If h is Lipschitz with Lipschitz constant K then
ℋn (h (B)) ≤ Kn ℋn (B)
Also, if T is a set in ℝ^{n}, m_{n}
(T )
= 0, then ℋ^{n}
(h(T ))
= 0. It is not necessary that h be one toone.
Proof: Let
{Ci}
_{i=1}^{∞} cover B with each having diameter less than δ and let this cover
be such that
∑ 1
β(n) -diam (Ci)n < ℋnδ (B )+ ε
i 2
Then
{h(Ci)}
covers h
(B )
and each set has diameter no more than Kδ. Then
( )n
ℋn (h(B )) ≤ ∑ β (n) 1 diam (h (C ))
K δ i 2 i
∑ ( )n
≤ Kn β (n) 1diam (Ci) ≤ Kn (ℋnδ (B)+ ε)
i 2
Since ε is arbitrary, this shows that
ℋnKδ(h(B )) ≤ Kn ℋnδ (B )
Now take a limit as δ → 0. The second claim follows from m_{n} = ℋ^{n} on Lebesgue measurable
sets of ℝ^{n}. ■
Lemma 20.0.2If S is a Lebesgue measurable set and h is Lipschitz then h
(S )
is ℋ^{n}measurable. Also, if h is Lipschitz with constant K,
ℋn (h(S)) ≤ Knmn (S)
It is not necessary that h be one to one.
Proof:The estimate follows from Lemma 20.0.1 and the observation that, as shown
before, Theorem 19.2.3, if S is Lebesgue measurable in ℝ^{n}, then ℋ^{n}
(S)
= m_{n}
(S)
. The
estimate also shows that h maps sets of Lebesgue measure zero to sets of ℋ^{n} measure
zero. Why is h
(S)
ℋ^{n} measurable if S is Lebesgue measurable? This follows from
completeness of ℋ^{n}. Indeed, let F be F_{σ} and contained in S with m_{n}
(S ∖F )
= 0.
Then
h (S) = h (S ∖F )∪ h(F )
The second set is Borel and the first has ℋ^{n} measure zero. By completeness of ℋ^{n}, h
if δ is small enough. The other inequality is shown exactly similar. ■
The following is a simplified version of an argument in [10]. Assume the following:
Dh (x) exists at a.e.x ∈ G say at all x ∈ A ⊆ G (20.2)
(20.2)
By regularity, we can and will assume A is a Borel set. Of course this is automatic if h is
Lipschitz which is the case of interest here but the condition 20.2 could likely be obtained in
other situations also.
For x ∈ A, let Dh
(x)
≡ R
(x)
U
(x)
where R
(x )
preserves lengths and
U
(x)
≡
( ∗ )
Dh (x )Dh (x)
^{1∕2}. Let A^{+} denote those points of A for which U
(x)
^{−1} exists.
Thus this is a measurable subset of A.
Let B be a Lebesgue measurable subset of A^{+} and let b ∈ B. Let S be a countable dense
subset of the space of symmetric invertible matrices and let C be a countable dense subset of
B.
||T (h−1(x))− T (h−1(y))|| ≤---1---|x− y | (20.8)
(1− 2ε)
(20.8)
Here the variables are in the appropriate sets, T
(E(T,c,i))
in the first inequality and
h
(E(T,c,i))
in the second and h^{−1} refers to the restriction to h
(E (T,c,i))
of the inverse
image of h. Thus, on this set, h^{−1} is actually a function.
Now let
(Ek,Tk)
result form a disjoint union of measurable subsets of the countably many
E
(T,c,i)
such that B = ∪_{k}E_{k}. Thus the above Lipschitz conditions hold for T_{k} in place of T.
This proves most of the following lemma. It is not necessary to assume h is one to one in
this lemma. h^{−1} refers to the inverse image of h restricted to h
(Ek)
as discussed
above.
Lemma 20.0.8There are disjoint measurable sets E_{k}whose union equals B andsymmetric linear transformations T_{k}such that
||h (T−1 (x))− h (T−1 (y))|| ≤ (1+ 2ε)|x− y| (20.9)
k k
(20.9)
( )
||T (h−1 (x )) − T (h−1(y))|| ≤ --1--- |x − y| (20.10)
k k 1− 2ε