Analysis

20.0.1 Preliminary Results

It was shown in Lemma 19.4.5 that

n ℋ (FA ) = det(U )mn(A )

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 ℝⁿ which is one to one on G, a measurable subset of ℝⁿ. 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 ℋⁿ as an outer measure. Thus it is not necessary to know the set B is measurable.

Proof: Let

_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

covers h and each set has diameter no more than Kδ. Then

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 = ℋⁿ on Lebesgue measurable sets of ℝⁿ. ■

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 ℝⁿ, then ℋⁿ

= m_n. The estimate also shows that h maps sets of Lebesgue measure zero to sets of ℋⁿ measure zero. Why is h ℋⁿ measurable if S is Lebesgue measurable? This follows from completeness of ℋⁿ. Indeed, let F be F_σ and contained in S with m_n = 0. Then

h (S) = h (S ∖F )∪ h(F )

The second set is Borel and the first has ℋⁿ measure zero. By completeness of ℋⁿ, h

is ℋⁿ measurable. ■

By Theorem A.5.5 on Page 2102, when Dh

exists,

Dh (x) = R (x)U (x)

where

= , ≥ 0 and R^∗R = I so R preserves lengths. This convention will be used in what follows.

Proof: First note that

and so  

Then the following corollary follows from Lemma 20.0.3.

First is a simple lemma which is fairly interesting.

Proof: Suppose for some v≠0,

. Then replacing v with v∕, one can assume = 1. If ≥ 1 + δ, then

|| −1 || ∥∥T S−1∥∥ ≥ -TS--Sv--≥ 1 +δ |Sv|

Hence, since

= 1,

∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥ ∥T − S∥ ∥S−1∥ ≥ ∥(T − S)S −1∥ = ∥T S−1 − I∥ ≥ ∥TS −1∥− ∥I∥ ≥ δ

and so one would need to have

≥ δ∕.

If

≤ 1 − δ, then

∥ ∥ 1− ∥∥I − TS− 1∥∥ = 1− ∥I −-TS-−1∥|Sv| |Sv |

|Sv|− ||(I − TS −1)Sv|| |Sv − (S − T) v| |Tv| ≤ -------------------- ≤ --------------= ---- ≤ 1− δ |Sv| |Sv | |Sv|

Thus, as before, δ ≤

≤ and so again, you would need to have ≥ δ∕. Thus, if < δ∕, 20.1 must hold for all v≠0. Note that this shows that for small δ, T⁻¹ must exist because

|Tv| ≥ (1 − δ)|Sv| > 0

if v≠0. The last assertion follows by noting that if T⁻¹ is given to exist and S is close to T then

( ) |Sv-|∈ (1− δ,1+ δ) so |Tv|∈ -1--,--1-- ⊆ (1− ˆδ,1+ ˆδ) |T v| |Sv| 1+ δ 1 − δ

By choosing δ appropriately, one can achieve the last inclusion for given

. ■

In short, the above lemma says that if one of S,T is invertible and the other is close to it, then it is also invertible and the quotient of

and is close to 1. Then the following lemma is fairly obvious.

Proof: Consider the first claim. For

|o(Tv)| = |o-(T-v)||Tv-|≤ |o(Tv)|∥T∥ |v| |Tv| |v| |Tv |

Thus o

= o. It is similar for T replaced with S.

Consider the second claim. Pick δ small. Then by Lemma 20.0.5

|Lv| |Lv||Sv| |Lv| sup ---- = sup --------≤ (1 + δ) sup ----< b v⁄=0|Tv| v⁄=0|Sv||Tv| v⁄=0 |Sv|

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

≡ RU where R preserves lengths and U ≡^1∕2. Let A⁺ denote those points of A for which U⁻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.

Let ε be a small number. Since U

is invertible, Lemma 20.0.6 implies o = o and so

|h (a)− h(b)− Dh (b)(a − b)| < ε|U (b)(a− b )| (20.3)

(20.3)

provided that a ∈ B

for i sufficiently large. By Lemma 20.0.5,

|h(a)− h (b)− Dh (b)(a− b)| < ε|T (a − b)| (20.4)

(20.4)

where U

is replaced by another linear one to one and onto symmetric mapping T provided T is sufficiently close to U. Let T ∈S be such that 20.4 holds for all a ∈ B. Now let c ∈C be close enough to b that b ∈ B. Thus b ∈ E where for i ∈ ℕ,c ∈C, T ∈S, E consists of those b ∈ B such that for all a ∈ B, 20.4 holds and also

|Dh-(b)v-| |U-(b-)| ivn⁄=f0 |Tv| = ivn⁄=f0 |Tv| > 1− ε,

sup |Dh-(b-)v|= sup |U-(b)|< 1+ ε (20.5) v⁄=0 |Tv | v⁄=0 |T v|

(20.5)

Thus there are countably many of these sets E

. Some may be empty but as just shown, their union includes all of B. They are Borel measurable sets because b → Dh is Borel measurable. Indeed, thinking of Dh as a matrix, the entries are limits of difference quotients of a continuous function. Note that if a,b ∈ E then < 2∕i so you can switch a,b in 20.4. See the picture.

For a,b ∈ E

it follows from 20.4, 20.5

and also

|h (a)− h(b)| > (1− 2ε)|T (a− b)| (20.6)

(20.6)

and so

(1 + 2ε) |T (a − b)| > |h (a)− h(b)| > (1− 2ε)|T (a − b )|

Note that this proves that on E

the function h is one to one. This is interesting so is stated as the following lemma.

It follows from this that

| ( ) ( )| |h T−1(x) − h T−1(y) | ≤ (1+ 2ε)|x− y| (20.7)

(20.7)

||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

in the first inequality and h in the second and h⁻¹ refers to the restriction to h of the inverse image of h. Thus, on this set, h⁻¹ is actually a function.

Now let

result form a disjoint union of measurable subsets of the countably many E such that B = ∪_kE_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 as discussed above.

Proof: It only remains to verify the last claim. However, this follows right away from 20.11. This formula implies that

| −1 | (1 − ε)|v| < |U (b)Tk v| < (1 + ε)|v|

and so

B (0,1 − ε) ⊆ U (b) T−1B (0,1) ⊆ B (0,1+ ε) k

This implies

( ) α (n)(1− ε)n ≤ det U (b )T−k1 α (n) ≤ α (n) (1 + ε)n

and so

(1− ε)n|det(Tk)| ≤ det(U (b )) ≤ (1 + ε)n |det(Tk)| ■

This lemma, along with Lemma 20.0.1 about the relationship between Hausdorff measure and Lipschitz mappings, implies the following.

Summarizing,

( )n --1--- ℋn (h(Ek)) ≥ mn (Tk(Ek )) ≥---1--n-ℋn (h (Ek)) 1− 2ε (1+ 2ε)

Then the above inequality and 20.12 implies the following.

( )n ---1---nℋn (h(Ek)) ≤ mn (Tk(Ek)) ≤ --1--- |det(Tk)|mn (Ek) (1+ 2ε) 1− 2ε