Next I will consider the case where h is not necessarily one to one. Recall the major
theorem presented earlier on which the proof of the area formula depended, Theorem
27.0.10 on Page 3302. Here it is.
Theorem 35.2.1Let h : U → ℝ^{m}where U is an open set in ℝ^{n}for n ≤ m and supposeh is locally Lipschitz at every point of a Lebesgue measurable subset, A of U. Also supposethat for every x ∈ A, Dh
Lemma 35.2.2Let h : U → ℝ^{m}where U is an open set in ℝ^{n}for n ≤ m and supposeh is locally Lipschitz at every point of a Lebesgue measurable subset, A of U.Let
N ≡ {x ∈ A : Dh (x) does not exist} (35.2.7)
(35.2.7)
and let
S ≡ {x ∈ A0 ≡ A ∖N : J (x) = 0} (35.2.8)
(35.2.8)
Then ℋ^{n}
(h(S ∪ N))
= 0.
Proof: By Rademacher’s theorem, N has measure 0. Therefore, ℋ^{n}
∩S for k a positive integer large
enough that S_{k}≠∅. By Theorem 27.0.10 on Page 3302 stated above, if x ∈ S_{k}, there exists
r_{x} such that 5r_{x}< min
(Rx,1)
and if r ≤ 5r_{x},
n
ℋ--(h(B-(x,r))) < -εn-n,B (x,r) ⊆ B (0,k)∩ U (35.2.9)
mn (B(x,r)) 5 k
(35.2.9)
Then by the Vitali covering theorem, there exists a sequence of disjoint balls of this sort,
{B }
i
_{i=1}^{∞} such that the balls having 5 times the radius but the same center,
n ∞∑ n ( (^ )) ∞∑ n n
ℋ (h(Sk)) ≤ ℋ h Bi ≤ 5 ℋ (h (Bi))
i=∞1 i=1
≤ ∑ 5n--ε-m (B ) ≤ ε-m (B (0,k)) = εα (n)
i=1 5nkn n i kn n
Since ε > 0 is arbitrary, it follows ℋ^{n}
(h (S ))
k
= 0 and now letting k → 0, it follows
ℋ^{n}
(h (S ))
= 0. This proves the lemma.
The following very technical lemma provides the necessary theory to generalize to
functions which are not one to one.
Lemma 35.2.3Let h : U → ℝ^{m}where U is an open set in ℝ^{n}for n ≤ m and supposeh is locally Lipschitz at every point of a Lebesgue measurable subset, A of U.Let
N ≡ {x ∈ A : Dh (x) does not exist}
and let
S ≡ {x ∈ A0 ≡ A ∖N : J (x) = 0}
Let B = A ∖
(S ∪N )
. Then there exist measurable disjoint sets,
{Ei }
_{i=1}^{∞}such thatA = ∪_{i=1}^{∞}E_{i}and h is one to one on E_{i}. Furthermore, h^{−1}is Lipschitz onh
(Ei)
.
Proof: Let C be a dense countable subset of B and let ℱ be a countable dense
subset of the invertible elements of ℒ
(ℝn,ℝn)
. For i a positive integer and
T ∈ℱ,c ∈C
{ ( 1) }
E (c,T,i) ≡ b ∈ B c,i ∩B : (a),(b) both hold
cover B and that they are measurable sets. To
begin with consider the measurability question. Inequality
(a)
is the same as
saying
2
3 |T v| ≤ |Dh (b)v| for all v
which is the same as saying
2 | − 1 |
3 |v| ≤ |Dh (b)T v| for all v.
Let
{v }
i
denote a dense countable subset of ℝ^{n}. Letting
{ }
2 || −1 ||
Si ≡ b :3 |vi| ≤ Dh (b )T vi
it follows easily that S_{i} is measurable because the component functions of the matrix of
Dh
(b)
are limits of difference quotients of continuous functions so they are Borel
measurable. (Note that if B were Borel, then S_{i} would also be Borel.) Now by
continuity,
{ }
∪ ∞ Si = b : 2|v| ≤ ||Dh (b)T− 1v || for all v
i=1 3
and so this set is measurable also. Inequality
(b)
also determines a measurable set by
similar reasoning. It is the same as saying that for all
|v|
< 2∕i,
1
|h (b+ v) − h(b)− Dh (b)(v)| ≤- |T (v)|
2
Use
{v }
i
a countable dense subset of B
(0,2∕i)
in a similar fashion to
(a)
.
Next I need to show these sets cover B. Let x ∈ B. Then pick c_{i}∈ B
is an equivalent norm, there exists some r > 0 such that
|Tv|
≥ r
|v |
for
all v. Therefore,
|a − b| ≤ 6 |h (a)− h(b)|.
r
In other words,
||h−1(h (a))− h− 1(h (b))|| = |a − b| ≤ 6 |h (a) − h (b)|.
r
which completes the proof.
With these lemmas, here is the main theorem which is a generalization of
Theorem 35.1.3. First remember that from Lemma 27.0.2 on Page 3278 a locally
Lipschitz function maps Lebesgue measurable sets to Hausdorff measurable
sets.
Theorem 35.2.4Let U be an open set in ℝ^{n}and h : U → ℝ^{m}. Let h be locallyLipschitz on A, a Lebesgue measurable subset of U and let g : h
(A)
→ ℝ be a nonnegativeℋ^{n}measurable function. Also let
# (y) ≡ Number of elements of h−1(y)
Then # is ℋ^{n}measurable,
x → (g ∘h)(x)J (x)
is Lebesgue measurable, and
∫ n ∫
# (y)g(y)dℋ = g(h(x))J (x )dmn
h(A) A
where J
(x)
= det
(U (x))
= det
( ∗ )
Dh (x) Dh (x)
^{1∕2}.
Proof: Let B = A ∖
(S ∪ N )
where S is the set of points where J
(x)
= 0 and N
is the set of points, x of A where Dh
(x)
does not exist. Also from Lemma
35.2.3 there exists
{E }
i
_{i=1}^{∞}, a sequence of disjoint measurable sets whose
union equals B such that h is one to one on each E_{i}. Then from Theorem 35.1.3
∫
A g(h (x))J (x)dmn
∫ ∑∞ ∫
= g(h (x ))J (x)dmn = g (h (x))J (x)dmn
B i=1 Ei
∑∞ ∫ ∫ ( ∞∑ )
= g(y)dℋn = Xh(Ei)(y) g(y)dℋn. (35.2.11)
i=1 h(Ei) h(B) i=1
Now #
(y)
=
(∑ )
∞i=1 Xh(Ei)(y)
on h
(B )
and # differs from this ℋ^{n} measurable function
only on h
(S ∪ N)
, which by Lemma 35.2.2 is a set of ℋ^{n} measure zero. Therefore, # is
ℋ^{n} measurable and the last term of 35.2.11 equals
( )
∫ ∞∑ n ∫ n
h(A) Xh(Ei)(y) g(y)dℋ = h(A)# (y)g(y)dℋ .
i=1