The area formula was discussed above. This formula implies that for E a measurable
set
∫
ℋn (f (E)) = X (x )J (x)dm
E ∗
where f : ℝ^{n}→ ℝ^{m} for f a Lipschitz mapping and m ≥ n. It is a version of the change of
variables formula for multiple integrals. The coarea formula is a statement about the
Hausdorff measure of a set which involves the inverse image of f. It is somewhat reminiscent
of Fubini’s theorem. Recall that if n > m and ℝ^{n} = ℝ^{m}× ℝ^{n−m}, we may take a
product measurable set, E ⊆ ℝ^{n}, and obtain its Lebesgue measure by the formula
∫ ∫
mn (E) = XE (y,x) dmn−mdmm
∫ℝm ℝn−m ∫
y n−m y
= ℝm mn −m (E )dmm = ℝm ℋ (E )dmm.
Thus, the notion of product measure yields a formula for the measure of a set in terms of the
inverse image of one of the projection maps onto a smaller dimensional subspace. The coarea
formula gives a generalization of 20.41 in the case where π_{1} is replaced by an arbitrary
Lipschitz function mapping ℝ^{n} to ℝ^{m}.
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. I found this formula in
[10]. This is a good place to obtain a slightly different proof. This argument follows [19] which
came from [10].
To begin with we give the linear algebra identity which will be used. Recall that for a real
matrix A^{∗} is just the transpose of A. Thus AA^{∗} and A^{∗}A are symmetric.
Theorem 20.3.1Let A be an m × n matrix and let B be an n × m matrix form ≤ n. Then for I an appropriate size identity matrix,
det(I + AB ) = det(I + BA )
Proof:Use block multiplication to write
( ) ( ) ( )
I + AB 0 I A = I + AB A +ABA
B I 0 I B BA + I
( )( ) ( )
I A I 0 = I + AB A + ABA
0 I B I + BA B I + BA
Hence
( I + AB 0 )( I A ) ( I A ) ( I 0 )
B I 0 I = 0 I B I + BA
so
( ) ( ) ( ) ( )
I A −1 I + AB 0 I A I 0
0 I B I 0 I = B I + BA
which shows that the two matrices
( ) ( )
I + AB 0 I 0
B I , B I + BA
are similar and so they have the same determinant. Thus
det(I + AB ) = det(I + BA )
Note that the two matrices are different sizes. ■
Corollary 20.3.2Let A be an m × n real matrix. Then
det(I + AA ∗) = det(I + A∗A ).
We need to use a version of the chain rule described in the next theorem.
Theorem 20.3.3Let f and g be Lipschitz mappings from ℝ^{n}to ℝ^{n}withf
(g (x ))
= x on A, a measurable set. Then for a.e. x∈ A, Dg
(f (x ))
, Df
(x)
, andD
(f ∘ g)
(x)
all exist and
I = D (g∘f)(x) = Dg (f (x))Df (x).
The proof of this theorem is based on the following lemma.
Lemma 20.3.4If h : ℝ^{n}→ ℝ^{n}is Lipschitz, then if h
(x)
= 0 for all x∈ A, then
det
(Dh (x))
= 0 a.e.
Proof:By the Area formula, 0 = ∫_{}
{0}
#
(y )
dy = ∫_{A}
|det(Dh (x))|
dx and so
det
(Dh (x))
= 0 a.e. ■
Proof of the theorem: On A, g
(f (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∘ f)
(x)
−I = 0. Let M be the set of points
in f
(ℝn )
where g fails to be differentiable and let N_{2}≡ g
(M )
∩ A, also a set of measure
zero. Finally let N_{3} be the set of points where f fails to be differentiable. Then if
x
∕∈
N_{1}∪ N_{2}∪ N_{3}, the chain rule implies I = D
(g∘ f)
(x)
= Dg
(f (x))
Df
(x)
.
■
It is convenient to define the following [10] for a measure space
(Ω,S,μ)
and
f : Ω → [0,∞], an arbitrary function, maybe not measurable.
∫ ∗ ∫ ∗ ∫
fdμ ≡ fdμ ≡ inf{ gdμ : g ≥ f, and g measurable}
Ω Ω
Lemma 20.3.5Suppose f_{n}≥ 0 and
∫
lim sup ∗f dμ = 0.
n→∞ n
Then there is a subsequence f_{nk}such that f_{nk}
(ω)
→ 0 a.e. ω.
Proof:For n large enough, ∫^{∗}f_{n}dμ < ∞. Let n be this large and pick g_{n}≥ f_{n}, g_{n}
measurable, such that
∩_{n=1}^{∞}∪_{m≥n}[g_{km}≥ m^{−1}], a set of measure
zero, for all m large enough, [g_{km}< m^{−1}] and so g_{km}
(ω)
→ 0 a.e. ω. Since f_{km}
(ω)
≤ g_{km}
(ω)
,
this proves the lemma. ■
It might help a little before proceeding further to recall the concept of a level surface of a
function of n variables. If f : U ⊆ ℝ^{n}→ ℝ, such a level surface is of the form f^{−1}
(y)
and we
would expect it to be an n− 1 dimensional thing in some sense. In the next lemma, consider a
more general construction in which the function has values in ℝ^{m}. In this more general
case, one would expect f^{−1}
(y)
to be something which is an n − m dimensional
thing.
Lemma 20.3.6Let A ⊆ ℝ^{p }and let f : ℝ^{p}→ ℝ^{m}be Lipschitz. Then
∫ ∗
ℋs (A ∩ f−1(y))dℋm ≤ β(s)β-(m-)(Lip(f))m ℋs+m (A).
ℝm β (s+ m)
Proof:The formula is obvious if ℋ^{s+m}
(A)
= ∞ so assume
ℋs+m (A ) < ∞.
∞ j ( j) − 1 j
A ⊆ ∪i=1Bi,r B i ≤ j ,B i is closed,
and
s+m −1 ∑∞ ( ( j))s+m
ℋ j−1 (A )+ j ≥ β (s+ m ) r B i (20.42)
i=1
(20.42)
There is a subsequence of the
{ }
Bj
i
denoted as {B_{ik}^{j}}_{k=1}^{∞} which consists of those balls
which also contain f^{−1}
(y)
covering A ∩ f^{−1}
(y)
. Now define
j ( ( j))s
gi (y) ≡ β (s) r Bi Xf(Bji)(y).
This picks out the appropriate subsequence such that y ∈ f
( )
Bj
ik
. If f^{−1}
(y )
∕∈
B_{i}^{j}, this
indicator function X_{f(Bji)
} just gives 0. Thus
( ) ∑∞ ( ( ))s ∑∞
ℋsj−1 A ∩ f−1(y) ≤ β (s) r Bjik = gji (y) ,
k=1 i=1
a Borel measurable function. It follows,
∫ ∗ ∫ ∗
ℋs (A ∩f−1(y))dℋm = lim ℋsj−1 (A ∩f−1 (y ))dℋm
ℝm ℝmj→ ∞
∫ ∗ ∑∞ j m
≤ ℝm lim ji→nf∞ gi (y) dℋ .
i=1