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
(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 27.0.2, this is a measure
and ν ≪ m. If m_{n}
∫ ∫ ∫
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}
(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 26.1.5 on Page 3224, there exists a Borel set F ⊇ E ∩ h
which shows 27.0.19 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
26.1.5, there exists F ⊇ E ∩ h
(A )
R
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, ℋ^{n} measurable simple function, 27.0.24
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 27.0.24 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 27.0.11Let h : ℝ^{n}→ ℝ^{m}be Lipschitz continuous. Also let h be oneto one on a measurable set G ⊆ ℝ^{n}and let m ≥ n. Let A ⊆ G be the set ofx ∈ 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 27.0.12Let h : ℝ^{n}→ ℝ^{m}. This function is said to be locally Lipschitz if forevery 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 27.0.1.
Lemma 27.0.13If 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 29.1.4, there are N_{n} sets of these balls
{G ,⋅⋅⋅,G }
1 Nn
such that the balls in G_{k} are disjoint and the union of all balls in the N_{n}
sets covers T_{k}. Then
n N∑n ∑ n
ℋ (h (Tk)) ≤ {ℋ (h (B )) : B ∈ Gk}
k=1 n n
≤ Nnk mn (V) < Nnk ε
Since ε is arbitrary, this shows that ℋ^{n}
(h (Tk))
= 0. Hence ℋ^{n}
(h(T))
= 0 also, since it
is the sum of the ℋ^{n}
(h(Tk))
. ■
Then an easy generalization of the above formula is the following.
Theorem 27.0.14Let h : ℝ^{n}→ ℝ^{m}be locally Lipschitz. Also suppose that h is one toone 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 29.4.2, there
exists a sequence of these balls
{Bi}
such that they are disjoint and G ∖∪_{i}B_{i} has
measure zero. Then, using Theorem 27.0.11 and the above Lemma 27.0.13,
∫ n ∫ n ∑ ∫ n
g(y)dℋ = g(y)dℋ = g(y)dℋ
h(G) ∑h(∪∫iBi) i h(Bi)
= g (h (x)) J∗(x)dmn
i Bi
∫ ∫
= g(h(x))J∗(x)dmn = g(h (x ))J∗ (x )dmn ■
∪iBi G