Topics In Analysis

27.0.2 The Area Formula

Assume here that h is one to one on G, and Lipschitz on G, Lemma 27.0.2 implies one can define a measure ν, on the σ− algebra of Lebesgue measurable subsets of G as follows.

n ν(E ) ≡ ℋ (h(E ∩ A)).

Recall that A is the set in G on which Dh

exists. This is all except a set of measure zero and so one could actually replace the right side with ℋⁿ because the new material has ℋⁿ measure zero. By Lemma 27.0.2, this is a measure and ν ≪ m. If m_n = 0, then by Lemma 27.0.2, ℋⁿ = 0. In fact, by this lemma, ℋⁿ ≤ Kⁿm_n so it is also clear that ν is σ finite.

Therefore by the corollary to the Radon Nikodym theorem, Corollary 18.1.3 on Page 1825, there exists f ≥ 0,f

= 0 if xA, and

∫ ∫ ν (E ) = fdm = fdm . E A∩E n

In fact,

f ∈ L1 (ℝn ) loc

Indeed, for a ball B,

∫ ∞ > Knmn (B) ≥ ℋn (h (B ∩A )) ≡ ν(B) = f dmn B

What is f? I will show that f

= J_∗ ≡ det a.e. Here U ≡^1∕2. Theorem 27.0.10 and the fundamental theorem of calculus implies that for a.e. x ∈ G,

Therefore, f = J_∗ a.e.

Note that one could define a measure ν_R as ℋⁿ

and it would still be the case that ν_R ≪ m_n and the Radon Nikodym derivative would still be J_∗ for x ∈ A ∩ B. You would just apply the above Theorem 27.0.10 for G = A ∩ B and the new A would be A ∩ B.

Now let F be a Borel set in ℝ^m. Recall this implies F is ℋⁿ measurable. Then

∫ = A XF (h(x))J∗(x)dm. (27.0.19)

(27.0.19)

Similarly, if

AR ≡ A ∩ B(0,R ),

∫ ∫ ∫ X (y )dℋn = X (y )dℋn = X (h (x)) J (x )dm. F∩h(AR) h(AR) F AR F ∗

Note there are no measurability questions in the above formula because h⁻¹

is a Borel set due to the continuity of h. The Borel measurability of J_∗ also follows from the observation that h is continuous and therefore, the partial derivatives are Borel measurable, being the limit of continuous functions. Then J_∗ is just a continuous function of these partial derivatives. However, things are not so clear if F is only assumed ℋⁿ measurable. Is there a similar formula for F only ℋⁿ measurable?

First consider the case where E is only ℋⁿ measurable but

ℋn (E ∩h (A)) = 0.

By Theorem 26.1.5 on Page 3224, there exists a Borel set F ⊇ E ∩ h

such that

ℋn (F ) = ℋn (E ∩ h(A)) = 0.

Then from 27.0.19,

XA ∩h−1(F)(x)J∗ (x) = 0 a.e.

But

0 ≤ XA∩h−1(E)(x)J∗(x) ≤ XA ∩h−1(F)(x)J∗ (x ) (27.0.20)

(27.0.20)

which shows the two functions in 27.0.20 are equal a.e. Therefore by completeness of Lebesgue measure, X_A∩h⁻¹

J_∗ is Lebesgue measurable and so from 27.0.19,

∫ ∫ n n 0 = XE∩h(A)(y)dℋ = XF ∩h(A)(y )dℋ

∫ ∫ = XA∩h−1(F)(x)J∗(x)dmn = XA∩h−1(E )(x )J∗(x)dmn, (27.0.21)

(27.0.21)

which shows 27.0.19 holds in this case where E is ℋⁿ measurable and

ℋn (E ∩h (A)) = 0.

Now let A_R ≡ A ∩ B

for large R and let E be ℋⁿ measurable. By Theorem 26.1.5, there exists F ⊇ E ∩ h such that F is Borel and

n ℋ (F ∖(E ∩h (AR))) = 0. (27.0.22)

(27.0.22)

Then

(E ∩ h(A )) ∪(F ∖(E ∩ h(A ))∩ h (A )) = F ∩ h(A ) R R R R

and so

XA ∩h−1(F)J∗ = XA ∩h−1(E )J∗ + XA ∩h−1(F∖(E∩h(A )))J∗ R R R R

where from 27.0.22 and 27.0.21, the second function on the right of the equal sign is Lebesgue measurable and equals zero a.e. Therefore, the first function on the right of the equal sign is also Lebesgue measurable and equals the function on the left a.e. Thus,

∫ n ∫ n ∫ XE ∩h(AR)(y)dℋ = XF ∩h(AR )(y) dℋ = A XF (h (x ))J∗ (x) dm R

∫ ∫ = XAR ∩h−1(F)(x)J∗(x)dmn = XAR∩h−1(E)(x)J∗(x)dmn. (27.0.23)

(27.0.23)

Since this holds for any R, it holds for 27.0.23 with A replacing A_R and the function

x → XA∩h−1(E)(x )J∗(x)

is Lebesgue measurable.Writing this in a more familiar form yields

∫ ∫ X (y)dℋn = X (h (x))J (x)dm . (27.0.24) h(A) E A E ∗ n

(27.0.24)

From this, it follows that if s is a nonnegative, ℋⁿ measurable simple function, 27.0.24 continues to be valid with s in place of X_E. Then approximating an arbitrary nonnegative ℋⁿ measurable function g by an increasing sequence of simple functions, it follows that 27.0.24 holds with g in place of X_E and there are no measurability problems because x → g

J_∗ is Lebesgue measurable. This proves the following theorem which is the area formula.

Since ℋⁿ = m_n on ℝⁿ, this is just a generalization of the usual change of variables formula except that here, one does not even need to know that h is C¹ so this is much better and in addition it is not limited to h having values in ℝⁿ. Also note that you could replace A with G since they differ by a set of measure zero thanks to Rademacher’s theorem.

Note that if you assume that h is Lipschitz on G then it has a Lipschitz extension to ℝⁿ. The conclusion has to do with integrals over G. It is not really necessary to have h be Lipschitz continuous on ℝⁿ, but you might as well assume this because of the existence of the Lipschitz extension. However, it can all be generalized to a situation in which h is not known to be Lipschitz on G. An assumption of locally Lipschitz is sufficient. The definition follows.

The proof uses a little generalization of Lemma 27.0.1.

Proof: Let

Tk ≡ {x ∈ T : h has Lipschitz constant k near x}.

Thus T = ∪_kT_k. I will show h

has ℋⁿ measure zero and then it will follow that

h (T) = ∪ ∞k=1h (Tk), the h(Tk) increasing in k,

must also have measure zero.

Let ε > 0 be given. By outer regularity, there exists an open set V containing T_k such that m_n

< ε. For x ∈ T_k it follows there exists r_x < 1 such that the ball centered at x with radius r_x is contained in V and in this ball, h has Lipschitz constant k. By the Besicovitch covering theorem, Theorem 29.1.4, there are N_n sets of these balls such that the balls in G_k are disjoint and the union of all balls in the N_n sets covers T_k. Then

Since ε is arbitrary, this shows that ℋⁿ = 0. Hence ℋⁿ = 0 also, since it is the sum of the ℋⁿ. ■

Then an easy generalization of the above formula is the following.

Proof: Let C consist of balls of radius less than 1 covering G such that for B ∈C, h is Lipschitz continuous on B. By the Besicovitch covering theorem, Theorem 29.4.2, there exists a sequence of these balls

such that they are disjoint and G ∖∪_iB_i has measure zero. Then, using Theorem 27.0.11 and the above Lemma 27.0.13,