The coarea formula involves a function, h which maps a subset of ℝ^{n} to ℝ^{m} where m ≤ n
instead of m ≥ n as in the area formula. The symbol, Lip
(h )
will denote the Lipschitz
constant for h.
It is possible to obtain the coarea formula as a computation involving the area
formula and some simple linear algebra and this is the approach taken here. To begin
with, here is the necessary linear algebra.
Theorem 35.3.1Let A be an m × n matrix and let B be an n × m matrix for m ≤ n.Then
n− m
pBA (t) = t pAB (t),
so the eigenvalues of BA and AB are the same including multiplicities except that BAhas n − m extra zero eigenvalues.
Proof:Use block multiplication to write
( ) ( ) ( )
AB 0 I A = AB ABA
B 0 0 I B BA
( ) ( ) ( )
I A 0 0 = AB ABA .
0 I B BA B BA
Therefore,
( )−1( ) ( ) ( )
I A AB 0 I A = 0 0
0 I B 0 0 I B BA
It follows that
( )
0 0
B BA
and
( )
AB 0
B 0
have the same characteristic polynomials
because the two matrices are simlar. Thus
( tI − AB 0 ) ( tI 0 )
det − B tI = det − B tI − BA
and so noting that BA is an n × n matrix and AB is an m × m matrix,
tm det (tI − BA ) = tn det (tI − AB )
and so det
(tI − BA )
= p_{BA}
(t)
= t^{n−m} det
(tI − AB )
= t^{n−m}p_{AB}
(t)
. This proves the
theorem.
The following corollary is what will be used to prove the coarea formula.
Corollary 35.3.2Let A be an m × n matrix. Then
det(I + AA ∗) = det(I + A ∗A ).
Proof: Assume m ≤ n. From Theorem 35.3.1AA^{∗} and A^{∗}A have the eigenvalues,
λ_{1},⋅⋅.λ_{m}, necessarily nonnegative, with the same multiplicities and some zero eigenvalues
which have differing multiplicities. The eigenvalues, λ_{1},⋅⋅ .λ_{m} are the zeros of p_{AA∗}
(t)
.
Thus there is an orthogonal transformation, P such that
( )
λ1
|| ... 0 ||
∗ || λm || ∗
A A = P || 0 || P .
|| . ||
( 0 .. )
0
The other main ingredient is the following version of the chain rule.
Theorem 35.3.3Let h and g be locally Lipschitz mappings from ℝ^{n}to ℝ^{n}withh
(g(x))
= x on A, a Lebesgue measurable set. Then for a.e. x∈ A, Dg
(h (x))
, Dh
(x)
,and D
(h∘ g)
(x)
all exist and
I = D (g ∘h) (x) = Dg (h(x))Dh (x).
The proof of this theorem is based on the following lemma.
Lemma 35.3.4If h : ℝ^{n}→ ℝ^{n}is locally Lipschitz, then if h
(x)
= 0 for all x∈A, then det
(Dh (x))
= 0 a.e.
Proof:By the case of the Area formula which involves mappings which are not
one to one, 0=∫_{}
{0}
#
(y)
dy= ∫_{A}
|det(Dh (x ))|
dx and so det
(Dh (x))
= 0
a.e.
Proof of the theorem: On A, g
(h (x))
−x = 0 and so by the lemma, there exists a
set of measure zero, N_{1} such that if x
∕∈
N_{1}, D
(g ∘h)
(x)
− I = 0. Let M be
the set of measure zero of points in h
n
(ℝ )
where g fails to be differentiable
and let N_{2}≡ g
(M )
∩ A, also a set of measure zero because locally Lipschitz
maps take sets of measure zero to sets of measure zero. Finally let N_{3} be the
set of points where h fails to be differentiable. Then if x
∈∕
N_{1}∪ N_{2}∪ N_{3},
the chain rule implies I = D
(g ∘h )
(x)
= Dg
(h (x))
Dh
(x)
. This proves the
theorem.
Lemma 35.3.5Let h : ℝ^{p}→ ℝ^{m}be Lipschitz continuous and δ > 0. Then if A ⊆ ℝ^{p}iseither open or compact,
y → ℋs (A ∩h− 1(y))
δ
is Borel measurable.
Proof: Suppose first that A is compact and suppose for δ > 0,
( )
ℋsδ A ∩ h−1(y) < t
Then there exist sets S_{i}, satisfying
diam (S ) < δ,A ∩h −1(y) ⊆ ∪∞ S ,
i i=1 i
and
∑∞
α (s)(r (Si))s < t.
i=1
I claim these sets can be taken to be open sets. Choose λ > 1 but close enough to 1
that
∞
∑ α (s)(λr(Si))s < t
i=1
Replace S_{i} with S_{i} + B
(0,ηi)
where η_{i} is small enough that
diam (Si)+ 2ηi < λdiam (Si).
Then
diam (Si + B (0,ηi)) ≤ λ diam (Si)
and so r
(Si + B (0,ηi))
≤ λr
(Si)
. Thus
∑∞ s
α(s)r(Si + B (0,ηi)) < t.
i=1
Hence you could replace S_{i} with S_{i} + B
(0,ηi)
and so one can assume the sets S_{i} are
open.
Claim:If z is close enough to y, then A ∩ h^{−1}
(z)
⊆∪_{i=1}^{∞}S_{i}.
Proof:If not, then there exists a sequence {z_{k}} such that
zk → y,
and
x ∈ (A ∩ h−1(z ))∖∪∞ S .
k k i=1 i
By compactness of A, there exists a subsequence still denoted by k such that
zk → y,xk → x ∈ A ∖ ∪∞i=1Si.
Hence
h(x) = lim h(xk) = lim zk = y.
k→ ∞ k→ ∞
But x
∕∈
∪_{i=1}^{∞}S_{i} contrary to the assumption that A ∩ h^{−1}
(y)
⊆∪_{i=1}^{∞}S_{i}.
It follows from this claim that whenever z is close enough to y,
ℋsδ(A ∩h −1(z)) < t.
This shows
{z ∈ ℝp : ℋs (A∩ h− 1(z)) < t}
δ
is an open set and so y →ℋ_{δ}^{s}
( −1 )
A ∩ h (y)
is Borel measurable whenever A is compact.
Now let V be an open set and let
is Borel measurable for each
δ > 0. Without loss of generality, ℋ^{s+m}
(A)
< ∞. Now let B_{i} be closed sets with
diam
(Bi)
< δ,A ⊆∪_{i=1}^{∞}B_{i}, and
s+m ∑∞ s+m
ℋ δ (A )+ ε > α (s + m)r (Bi) .
i=1
Note each B_{i} is compact so y →ℋ_{δ}^{s}
( )
Bi ∩ h−1(y)
is Borel measurable. Thus
∫ s( − 1 )
ℝm ℋ δ A ∩h (y) dy
∫ ∑ ( )
≤ ℋsδ Bi ∩ h−1 (y) dy
ℝm∫ i
= ∑ ℋs (B ∩ h−1 (y))dy
i ℝm δ i
∑ ∫
≤ ℋsδ (Bi)dy
∑i h(Bi)
= mm (h(Bi))ℋsδ(Bi)
i
∑ m m m s
≤ (Lip(h)) 2 α (m )r(Bi) α(s)r(Bi)
i ∑
= (Lip(h))m α(m-)α(s)2m α(s+ m )r(Bi)m+s
α (m + s) i
m α(m )α(s) m ( s+m )
≤ (Lip(h)) α-(m-+-s)-2 ℋ δ (A )+ ε
Since ε is arbitrary,
∫ s( − 1 ) m α(m )α(s) m s+m
ℝm ℋ δ A ∩h (y) dy ≤ (Lip(h)) α-(m--+s)-2 ℋ δ (A )
Taking a limit as δ → 0 this proves the lemma.
Next I will show that whenever A is Lebesgue measurable,
( )
y → ℋn−m A ∩ h−1(y)
is m_{m} measurable and the above estimate holds.
Lemma 35.3.7Let A be a Lebesgue measurable subset of ℝ^{n}and let h : ℝ^{n}→ ℝ^{m}beLipschitz. Then
n−m ( −1 )
y → ℋ A ∩ h (y)
is Lebesgue measurable. Furthermore, for all A Lebesgue measurable,
∫ n−m ( −1 ) m m α(n−-m-)α-(m-)-
ℝm ℋ A ∩ h (y) dy ≤ 2 (Lip(h)) α (n ) mn (A)
Proof:Let A be a bounded Lebesgue measurable set in ℝ^{n}. Then by inner and
outer regularity of Lebesgue measure there exists an increasing sequence of
compact sets,
{Kk }
contained in A and a decreasing sequence of open sets,