Assume here that h is one to one on G, and Lipschitz on G, Lemma 20.0.2 implies one can
define a measure ν, on the σ− algebra of Lebesgue measurable subsets of G as
follows.
ν(E) ≡ ℋn (h (E ∩A )) .
Recall that A is the set in G on which Dh
(x )
exists. This is all except a set of measure zero
and so one could actually replace the right side with ℋ^{n}
(h (E ))
because the new material has
ℋ^{n} measure zero. By Lemma 20.0.2, this is a measure and ν ≪ m_{n}. If m_{n}
Therefore by the corollary to the Radon Nikodym theorem, Corollary 25.1.3 on Page 1885,
or Theorem 18.5.2, the Radon Nikodym theorem for Radon measures, there exists
f ≥ 0,f
(x)
= 0 if x
∈∕
A, and
∫ ∫
ν (E ) = E fdmn = A∩E fdmn.
In fact,
f ∈ L1loc(ℝn )
Indeed, for a ball B,
∫
∞ > Knm (B ) ≥ ℋn (h(B ∩ A)) ≡ ν (B ) = fdm
n B n
What is f? I will show that
f (x ) = J∗(x) ≡ det(U (x))
a.e. Here U
(x)
≡
(Dh (x)∗Dh (x))
^{1∕2}. Theorem 20.0.10 and the fundamental theorem of
calculus implies that for a.e. x ∈ G,
Note there are no measurability questions in the above formula because h^{−1}
(F)
is a Borel set
due to the continuity of h. The Borel measurability of J_{∗}
(x)
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_{∗}
(x )
is just a continuous function of these partial
derivatives. However, things are not so clear if F is only assumed ℋ^{n} measurable. Is there a
similar formula for F only ℋ^{n} measurable?
First consider the case where E is only ℋ^{n} measurable but
ℋn (E ∩ h (A )) = 0.
By Theorem 19.1.5 on Page 1247, there exists a Borel set F ⊇ E ∩ h
∫ ∫
= XA ∩h−1(F)(x)J∗ (x )dmn = XA ∩h−1(E)(x)J∗(x)dmn, (20.21)
(20.21)
which shows 20.19 holds in this case where E is ℋ^{n} measurable and
ℋn (E ∩ h (A )) = 0.
Now let A_{R}≡ A ∩ B
(0,R)
for large R and let E be ℋ^{n} measurable. By Theorem 19.1.5,
there exists F ⊇ E ∩ h
(A )
R
such that F is Borel and
ℋn (F ∖ (E ∩ h (AR ))) = 0. (20.22)
(20.22)
Then
(E ∩ h (AR ))∪ (F ∖ (E ∩ h (AR ))∩ h(AR )) = F ∩ h (AR )
and so
XAR∩h−1(F)J∗ = XAR∩h−1(E)J∗ + XAR ∩h−1(F∖(E ∩h(AR)))J∗
where from 20.22 and 20.21, 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,
∫ ∫ ∫
X (y )dℋn = X (y)dℋn = X (h(x))J (x)dm
E∩h(AR ) F ∩h(AR) AR F ∗
∫ ∫
= X −1 (x )J (x)dm = X −1 (x)J (x) dm . (20.23)
AR∩h (F ) ∗ n AR∩h (E) ∗ n
(20.23)
Since this holds for any R, it holds for 20.23 with A replacing A_{R} and the function
x → X −1 (x)J (x )
A∩h (E) ∗
is Lebesgue measurable.Writing this in a more familiar form yields
From this, it follows that if s is a nonnegative, ℋ^{n} measurable simple function, 20.24
continues to be valid with s in place of X_{E}. Then approximating an arbitrary nonnegative ℋ^{n}
measurable function g by an increasing sequence of simple functions, it follows that
20.24 holds with g in place of X_{E} and there are no measurability problems because
x → g
(h (x))
J_{∗}
(x)
is Lebesgue measurable. This proves the following theorem which is the
area formula.
Theorem 20.0.11Let h : ℝ^{n}→ ℝ^{m}be Lipschitz continuous. Also let hbe one to one on a measurable set G ⊆ ℝ^{n}and let m ≥ n. Let A ⊆ G be the setof x ∈ G on which Dh
(x)
exists, and let g : h
(A )
→ [0,∞] be ℋ^{n}measurable.Then
x → (g∘ h)(x)J∗(x)
is Lebesgue measurable and
∫ ∫
g(y)dℋn = g(h(x))J∗(x)dmn
h(A) A
where J_{∗}
(x)
= det
(U (x))
= det
(Dh (x)∗Dh (x))
^{1∕2}.
Since ℋ^{n} = m_{n} on ℝ^{n}, 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 C^{1} so this is much better and
in addition it is not limited to h having values in ℝ^{n}. 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 ℝ^{n}.
The conclusion has to do with integrals over G. It is not really necessary to have h be
Lipschitz continuous on ℝ^{n}, 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.
Definition 20.0.12Let h : ℝ^{n}→ ℝ^{m}. This function is said to be locally Lipschitz iffor every x ∈ ℝ^{n}, there exists a ball B_{x}containing x and a constant K_{x}such that for ally,z∈ B_{x},
|h (z)− h (y)| ≤ Kx |z− y |
The proof uses a little generalization of Lemma 20.0.1.
Lemma 20.0.13If h is locally Lipschitz and m_{n}
(T)
= 0, then
ℋn (h(T)) = 0
Proof:Let
Tk ≡ {x ∈ T : h has Lipschitz constant k near x}.
Thus T = ∪_{k}T_{k}. I will show h
(Tk)
has ℋ^{n} 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}
(V )
< 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 18.1.4, there are N_{n} sets of these balls
{G1,⋅⋅⋅,GNn}
such that
the balls in G_{k} are disjoint and the union of all balls in the N_{n} sets covers T_{k}. Then
Then an easy generalization of the above formula is the following.
Theorem 20.0.14Let h : ℝ^{n}→ ℝ^{m}be locally Lipschitz. Also suppose that h is oneto one on G, a measurable subset of ℝ^{n}. Then let g : h
(G)
→ [0,∞] be ℋ^{n}measurable. Itfollows that
x → (g∘ h)(x)J∗(x)
is Lebesgue measurable and
∫ ∫
g(y)dℋn = g(h(x))J∗(x)dmn
h(G) G
where J_{∗}
(x)
= det
(U (x))
= det
( )
Dh (x)∗Dh (x)
^{1∕2}.
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 18.3.2,
there exists a sequence of these balls
{Bi}
such that they are disjoint and G ∖∪_{i}B_{i}
has measure zero. Then, using Theorem 20.0.11 and the above Lemma 20.0.13,
∫ ∫ ∫
g (y )dℋn = g(y)dℋn = ∑ g(y)dℋn
h(G) h(∪iBi) i h(Bi)
∑ ∫
= g(h (x))J∗(x)dmn
∫i Bi ∫
= g(h(x))J (x)dm = g(h(x))J (x)dm ■
∪iBi ∗ n G ∗ n