≡ 0. Thus it is uniformly continuous
on this compact set and so there exists δ > 0 such that if
∥v∥
< δ, then
|ϕ(x,v)− ϕ (x,0 )| = |ϕ(x,v)| < ε, (8.5)
(8.5)
this for all x ∈U.
h (x + v)− h (x) = Dh (x )v(+ o (v) )
= Dh (x )v + Dh −1(h(x))o (v )
Let f : V → ℝ be a bounded, uniformly continuous function.
Let ℬ_{m} be a collection of disjoint half open rectangles as in Lemma 8.1.4 such that each has diameter
no more than 2^{−m} and each rectangle of ℬ_{m+1} is either a subset of a rectangle of ℬ_{m} or is equal to a
rectangle of ℬ_{m} such that ∪ℬ_{m} = U. Let m be large enough that the diameters of all these half open
rectangles are less than δ. Denote the rectangles of ℬ_{m} as
{Rmi }
_{i=1}^{∞} and let the center of these be
denoted by x_{i}^{m}. Also let m be large enough that
m m m
|f (h(xi ))|det(Dh (xi ))|− f (h(x))|det(Dh (x))|| < ε for all x ∈ Ri
A basic version of the theorems to be presented is the following.
Lemma 8.3.1Let U and V be bounded open sets in ℝ^{p}and leth,h^{−1}be C^{1}defined respectively onÛ⊇Uand
ˆV
⊇V such that h
(U )
= V and let f be a bounded uniformly continuous function defined on U.Then
∫ ∫
V f (y)dmp = U f (h (x))|det(Dh (x))|dmp
Proof:Let x ∈ U. By the assumption that h and h^{−1} are C^{1},
h (x + v)− h (x) = Dh (x )v+ o (v)
= Dh (x )(v + Dh −1(h(x))o (v ))
Let an upper bound for
∥∥ −1 ∥∥
Dh (h(x))
be C. It exists because V is compact and h^{−1} is C^{1} on an open set
containing this compact set. Therefore, since all the boxes in ℬ_{m} are in diameter less than
δ,
to within ε. Let r_{i}^{m} be half the diameter of R_{i}^{m}. Thus ∑_{i}m_{p}
(B (0,rim))
= m_{p}
(U)
. This is by translation
invariance of the measure which shows that
mp (Rmi ) = mp (B (xmi ,rmi )) = mp (B (0,rmi ))
Then
∫ ∑ ∞ ∫
V f (y)dmp = i=1 h(Rmi )f (y )dmp
∑ ∞ ∫
≤ εmp (V )+ i=1 h(Rmi )f (h (xmi ))dmp
∑ ∞
≤ εmp (V )+ i=1 f (h(xmi ))mp (h (Rmi ))
∑ ∞ m m
≤ εmp (V )+ i=1f (h (xi ))mp (Dh (xi )(B (0,(1+ C ε)ri)))
p∑ ∞ ∫ m m
= εmp (V) +(1 +C ε) i=1 Rmi f (h(xi ))|det(Dh (x i ))|dmp
p∑ ∞ (∫ m )
≤ εmp (V )+ (1+ C ε) i=1 Rmi f (h (x))|det(Dh (x))|dmp + 2εmp (Ri )
p∑ ∞ ∫ p
≤ εmp (V )+ (1 + Cε) i=1 Rmi f (h(x))|det(Dh (x))|dmp + (1 + Cε) 2εmp (U )
Since ε > 0 is arbitrary, this shows
∫ ∫
f (y)dm ≤ f (h(x))|det(Dh (x))|dm (8.7)
V p U p
(8.7)
whenever f is uniformly continuous and bounded on V. Now x → f
(h(x))
|det(Dh (x))|
has the same
properties as f and so, using the same argument with U and V switching roles and replacing h with
h^{−1},
∫
U f (h (x)) |det(Dh (x))|dmp
∫ ∫
≤ f (h (h −1(y)))||det(Dh (h− 1(y)))||||det(Dh −1(y))||dm = f (y )dm
V p V p
by the chain rule. This with 8.7 proves the lemma. ■
The Lebesgue integral is defined for nonnegative functions and then you break up an arbitrary function
into positive and negative parts. Thus the most convenient theorems involve nonnegative functions which
do not involve assumptions of uniform continuity and such things. The next corollary gives such a result.
Tlhis will remove assumptions that U,V are bounded and the need for larger open sets on which h,h^{−1} are
defined and C^{1}.
Corollary 8.3.2Let U be an open set in ℝ^{p}and let h be a one to one C^{1}function such thath
(U)
= V and
|detDh (x)|
≠0 for all x. Let f be continuous and nonnegative defined on V.Then
∫ ∫
f (y)dmp = f (h (x))|det(Dh (x))|dmp
V U
Proof:Let U_{k}≡
(− k,k)
^{p}∩
{ ( ) }
x ∈ U : dist x,UC > 1k
. Thus U_{k}⊆ U_{k+1} for all k and U_{k} is
closed and bounded, hence compact. The inverse function theorem implies V_{k}≡ h
(Uk )
is open
and h^{−1} is C^{1} on V_{k}. Also f is uniformly continuous on U_{k} hence on U_{k} as well. It follows
that
∫ ∫
Xh(Uk)(y )f (y)dmp = XUk (x)f (h(x))|det(Dh (x))|dmp
V U
Now use the monotone convergence theorem and let k →∞. ■
Lemma 8.3.3Let Z ⊆
{x ∈ U : det(Dh (x)) = 0}
be a compact subset of U an open set and leth : U → ℝ^{p}be C^{1}. Then m_{p}
is the usual Euclidean norm. Then ϕ is uniformly continuous on Z ×B
(0,1)
. Therefore, there
exists δ > 0 such that if
|v|
< δ, it follows that
|h (x+ v)− (h (x) + Dh (x )v)| < ε|v | (8.8)
(8.8)
for all x ∈ Z. Also let max
{∥Dh (x)∥ : x ∈ Z}
= C and all ε will be less than 1.
Let V be an open set containing Z such that m_{p}
(V)
< m_{p}
(Z )
+ ε and let ℬ_{m} be a countable set of
disjoint half open rectangles whose union is V as in Lemma 8.1.4. Let m be large enough that all of these
rectangles have diameters less than δ, all distances being measured with respect to the Euclidean norm.
Consider only those rectangles which have nonempty intersection with Z. Denote this set by
ˆℬ
_{m} If R is one
of these, then for x ∈ Z ∩ R,
h (R)− h (x ) ⊆ Dh (x)B (0,d)+ B (0,εd)
where d is the diameter of R. Thus Dh
(x)
B
(0,d)
is in a subspace of dimension k < p. It has diameter no
more than Cd. By Proposition 8.2.5 and translation invariance,
p p−1
mp (h(R)− h (x)) = mp(h (R )) ≤ 2 (2Cd + εd) εd
≤ C(p)dpε
Now the diameter d of R is
√p
r where r is the length of a side. Therefore, modifying C
(p)
,
mp (h (R)) ≤ C (p)rpε
Then
∑ ∑
mp (h (Z)) ≤ mp (h (Z)) ≤ C (p) mp (R)ε
R∈ˆℬm R∈ˆℬm
≤ C (p)m (V )ε < C (p)(m (Z) + ε)ε
p p
Since ε is arbitrary, it follows that m_{p}
(h (Z ))
= 0. ■
This implies the following major lemma.
Lemma 8.3.4Let h : U → ℝ^{p}where U is open. Let Z =
Corollary 8.3.5Let U be an open set in ℝ^{p}and let h be a one to one C^{1}function such that h
(U )
= V .Let f be continuous and nonnegative defined on V. Then
∫ ∫
f (y)dm = f (h (x))|det(Dh (x))|dm
V p U p
Proof:Let Û≡
{x ∈ U : |det(Dh (x))| > 0}
. Then
∫ ∫
f (h (x))|det(Dh (x))|dmp = ˆ f (h (x))|det(Dh (x ))|dmp
U ∫U ∫
= f (y )dmp = f (y)dmp
h(ˆU) h(U )
because h
({x ∈ U : |det(Dh (x))| = 0})
has measure zero. ■
It is easy to generalize this corollary to the case where f is nonnegative and only Borel meaurable. Let
R = ∏_{i=1}^{p}
(ai,bi)
, a open rectangle. Then let g^{k}
(x)
≡∏_{i=1}^{p}g_{i}^{k}
(xi)
where g_{i}^{k}
(t)
≥ 0, is continuous,
piecewise linear, equals 0 off
(ai,bi)
and 1 on
[ ]
ai + 1k,bi − 1k
. Thus lim_{k→∞}g^{k}
(x)
= X_{R}
(x )
and
g^{k}
(x)
≤ g^{k+1}
(x)
for all k. Therefore, if h and U are as in Corollary 8.3.5, we can apply the monotone
convergence theorem and obtain
∫ ∫
XR (h(x))|det(Dh (x))|dmp = lim gk (h (x))|det(Dh (x))|dmp
U k→ ∞ U
∫ ∫
= lim gk(y)dm = X (y)dm
k→∞ h(U) p h(U) R p
Now let K be the pi system of open rectangles. Thus σ
(K)
= ℬ
(ℝp)
. Let R_{k}≡∏_{i=1}^{p}
(− k,k)
{ ∫ ∫ }
p
G ≡ E ∈ ℬ(ℝ ) : U XE ∩Rk (h(x))|det (Dh (x))|dmp = h(U )XE∩Rk (y) dmp
Then it is routine to verify that G is closed with respect to disjoint unions and complements. The
assertion about disjoint unions is obvious. Consider the one about complements. Say E ∈G.
Then
. Now let k →∞ and apply the monotone convergence
theorem. The following theorem is now almost obvious because it was just shown that the change of
variables formula holds for indicator functions of Borel sets and hence for every nonnegative simple
function.
Theorem 8.3.6Let f
(y)
≥ 0 and let it be Borel measurable. Also let h be a one to one C^{1}function onthe open set U such that h
(U)
= V. Then
∫ ∫
f (y)dm = f (h (x))|det(Dh (x))|dm
V p U p
Proof:By Theorem 5.1.9, there exists an increasing sequence of Borel measurable simple
functions
{sk}
which converges pointwise to f
(y)
. Then by the monotone convergence theorem,
∫ ∫ ∫
f (y)dm = lim s (y)dm = lim s (h(x))|det (Dh (x))|dm
V p k∫→∞ V k p k→∞ U k p
= U f (h(x))|det (Dh (x))|dmp■
In fact, it is only necessary to assume that f is ℱ_{p} measurable. This is fairly routine to show from the
regularity result of Proposition 8.1.3 and the fact that the Lebesgue measure is complete on ℱ_{p}.
The difficulty is that x →f
(h (x ))
is not known to be measurable. You end up showing that
x → f