With Rademacher’s theorem, one can give a general change of variables formula involving
Lipschitz maps. First here is an elementary estimate.
Lemma 34.6.1Suppose V is an n− 1 dimensional subspace of ℝ^{n}and K is a compactsubset of V . Then letting
K ≡ ∪ B (x,ε) = K + B (0,ε),
ε x∈K
it follows that
m (K ) ≤ 2nε(diam (K)+ ε)n−1.
n ε
Proof:Let an orthonormal basis for V be
{v1,⋅⋅⋅,vn−1}
and let
{v ,⋅⋅⋅,v ,v }
1 n−1 n
be an orthonormal basis for ℝ^{n}. Now define a linear transformation, Q by Qv_{i} = e_{i}. Thus
QQ^{∗} = Q^{∗}Q = I and Q preserves all distances because
| | | | | |
|| ∑ ||2 ||∑ ||2 ∑ 2 ||∑ ||2
||Q aiei|| = || aivi|| = |ai| = || aiei|| .
i i i i
where B^{n−1} refers to the ball taken with respect to the usual norm in ℝ^{n−1}. Every point
of K_{ε} is within ε of some point of K and so it follows that every point of QK_{ε} is within ε
of some point of QK. Therefore,
n−1
QK ε ⊆ B (Qk0,diam (QK )+ ε)× (− ε,ε),
To see this, let x ∈ QK_{ε}. Then there exists k ∈ QK such that
|k − x|
< ε. Therefore,
|(x1,⋅⋅⋅,xn−1)− (k1,⋅⋅⋅,kn−1)|
< ε and
|xn − kn|
< ε and so x is contained in the set
on the right in the above inclusion because k_{n} = 0. However, the measure of the set on
the right is smaller than
Next is the definition of a point of density. This is sort of like an interior point but not
as good.
Definition 34.6.2Let E be a Lebesgue measurable set. x ∈ E is a point of densityif
m(E ∩ B(x,r))
lrim→0 -------------= 1.
m (B(x,r))
You see that if x were an interior point of E, then this limit will equal 1. However, it
is sometimes the case that the limit equals 1 even when x is not an interior
point. In fact, these points of density make sense even for sets that have empty
interior.
Lemma 34.6.3Let E be a Lebesgue measurable set. Then there exists a set ofmeasure zero, N, such that if x ∈ E ∖ N, then x is a point of density of E.
Proof: Consider the function, f
(x)
= X_{E}
(x)
. This function is in L_{loc}^{1}
(ℝn)
. Let
N^{C} denote the Lebesgue points of f. Then for x ∈ E ∖ N,
In this section, Ω will be a Lebesgue measurable set in ℝ^{n} and h : Ω → ℝ^{n} will be
Lipschitz. Recall the following definition and theorems. See Page 11.4.2 for the proofs and
more discussion.
Definition 34.6.4Let ℱ be a collection of balls that cover a set, E, which havethe property that if x ∈ E and ε > 0, then there exists B ∈ℱ, diameter of B < εand x ∈ B. Such a collection covers E in the sense of Vitali.
Theorem 34.6.5Let E ⊆ ℝ^{n}and suppose m_{n}(E) < ∞ where m_{n}is the outermeasuredetermined by m_{n}, n dimensional Lebesgue measure, and letℱ, be acollection of closed balls of bounded radii such that ℱ covers E in the sense of Vitali.Then there exists a countable collection of disjoint balls from ℱ, {B_{j}}_{j=1}^{∞}, suchthat m_{n}(E ∖∪_{j=1}^{∞}B_{j}) = 0.
Now this theorem implies a simple lemma which is what will be used.
Lemma 34.6.6Let V be an open set in ℝ^{r},m_{r}
(V )
< ∞. Then there exists a sequenceof disjoint open balls {B_{i}} having radii less than δ and a set of measure 0, T, suchthat
V = (∪ ∞i=1Bi)∪ T.
As in the proof of the change of variables theorem given earlier, the first step is to
show that h maps Lebesgue measurable sets to Lebesgue measurable sets. In showing this
the key result is the next lemma which states that h maps sets of measure zero to sets of
measure zero.
Lemma 34.6.7If m_{n}
(T )
= 0 then m_{n}
(h(T))
= 0.
Proof:Let V be an open set containing T whose measure is less than ε. Now using
the Vitali covering theorem, there exists a sequence of disjoint balls
{Bi}
, B_{i} = B
(xi,ri)
which are contained in V such that the sequence of enlarged balls,
{ }
^Bi
, having the
same center but 5 times the radius, covers T. Then
( ( ))
mn (h(T)) ≤ mn h ∪∞i=1^Bi
∑∞ ( ( ))
≤ mn h ^Bi
i=1
∑∞ n n n n n ∞∑
≤ α (n )(Lip(h)) 5 ri = 5 (Lip (h)) mn (Bi )
i=1 n n n n i=1
≤ (Lip(h)) 5 mn (V ) ≤ ε(Lip(h)) 5 .
Since ε is arbitrary, this proves the lemma.
With the conclusion of this lemma, the next lemma is fairly easy to obtain.
Lemma 34.6.8If A is Lebesgue measurable, then h
(A)
is m_{n}measurable.Furthermore,
mn (h(A)) ≤ (Lip(h))nmn (A). (34.6.9)
(34.6.9)
Proof:Let A_{k} = A ∩ B
(0,k )
,k ∈ ℕ. Let V ⊇ A_{k} and let m_{n}
(V)
< ∞. By Lemma
34.6.6, there is a sequence of disjoint balls
∞∑ n n∑∞ n
≤ α(n)(Lip(h)ri) = Lip(h) mn (Bi) = Lip(h) mn (V ).
i=1 i=1
Therefore,
m--(h(A )) ≤ Lip(h)n m (V ).
n k n
Since V is an arbitrary open set containing A_{k}, it follows from regularity of Lebesgue
measure that
m--(h (A )) ≤ Lip(h)nm (A ). (34.6.10)
n k n k
(34.6.10)
Now let k →∞ to obtain 34.6.9. This proves the formula. It remains to show h
(A )
is
measurable.
By inner regularity of Lebesgue measure, there exists a set, F, which is the countable
union of compact sets and a set T with m_{n}
(T )
= 0 such that
F ∪ T = A .
k
Then h
(F )
⊆ h
(Ak)
⊆ h
(F )
∪ h
(T)
. By continuity of h, h
(F)
is a countable union of
compact sets and so it is Borel. By 34.6.10 with T in place of A_{k},
---
mn (h(T)) = 0
and so h
(T)
is m_{n} measurable. Therefore, h
(Ak )
is m_{n} measurable because m_{n} is a
complete measure and this exhibits h
(Ak)
between two m_{n} measurable sets whose
difference has measure 0. Now
∞
h(A) = ∪k=1h(Ak )
so h
(A)
is also m_{n} measurable and this proves the lemma.
The following lemma, depending on the Brouwer fixed point theorem and
found in Rudin [?], will be important for the following arguments. The idea is
that if a continuous function mapping a ball in ℝ^{k} to ℝ^{k} doesn’t move any
point very much, then the image of the ball must contain a slightly smaller
ball.
Lemma 34.6.9Let B = B
(0,r)
, a ball in ℝ^{k}and let F :B→ ℝ^{k}be continuous andsuppose for some ε < 1,
|F (v)− v| < εr
for allv ∈B. Then
F (B) ⊇ B-(0,r(1-− ε)).
Proof:Suppose a ∈B
(0,r(1− ε))
∖ F
(-)
B
and let
r-(a-−-F(v))
G (v) ≡ |a− F (v)| .
Then by the Brouwer fixed point theorem, G
(v)
= v for some v ∈B. Using the formula
for G, it follows
|v|
= r. Taking the inner product with v,
2 2 ---r-----
(G (v),v) = |v| = r = |a − F (v)| (a− F (v ),v)
----r----
= |a − F (v)| (a− v + v− F (v),v)
----r----
= |a − F (v)| [(a − v,v)+ (v − F (v ),v )]
----r----[ 2 ]
= |a − F (v)| (a,v )− |v| + (v− F (v),v)
----r----[ 2 2 2 ]
≤ |a − F (v)| r (1− ε)− r +r ε = 0,
a contradiction. Therefore, B
(0,r(1− ε))
∖ F
(B)
= ∅ and this proves the lemma.
Now let Ω be a Lebesgue measurable set and suppose h : ℝ^{n}→ ℝ^{n} is Lipschitz
continuous and one to one on Ω. Let
N ≡ {x ∈ Ω : Dh (x) does not exist} (34.6.11)
(34.6.11)
{ −1 }
S ≡ x ∈ Ω ∖N : Dh (x) does not exist (34.6.12)
(34.6.12)
Lemma 34.6.10Let x ∈ Ω ∖
(S ∪ N )
. Then if ε ∈
(0,1)
the following hold for all rsmall enough.
( (------))
mn h B (x,r) ≥ mn (Dh (x)B (0,r(1− ε))) , (34.6.13)
(34.6.13)
h (B (x,r)) ⊆ h (x )+ Dh (x)B (0,r (1 + ε)), (34.6.14)
(34.6.14)
( (------))
mn h B (x,r) ≤ mn (Dh (x)B (0,r(1 + ε))) (34.6.15)