In this section theorems are proved which yield change of variables formulas for C^{1} functions.
More general versions can be seen in Kuttler [20], Kuttler [19], and Rudin [28].
You can obtain more by exploiting the Radon Nikodym theorem and the Lebesgue
fundamental theorem of calculus, two topics which are best studied in a more advanced
course. Instead, I will present some good theorems using the Vitali covering theorem
directly.
A basic version of the theorems to be presented is the following. If you like, let the balls be
defined in terms of the norm
||x|| ≡ max{|x | : k = 1,⋅⋅⋅,p}
k
Recall that C_{c}
(U )
refers to continuous functions which are zero off a compact subset of
U.
Lemma 9.7.1Let U and V be bounded open sets in ℝ^{p}and leth,h^{−1}be C^{1}functionssuch that h
(U)
= V . Also let f ∈ C_{c}
(V )
. Then
∫ ∫
f (y)dmp = f (h(x))|det (Dh (x))|dmp
V U
Proof:First note h^{−1}
(spt(f))
is a closed subset of the bounded set U and so it is
compact. Thus x → f
(h(x))
|det(Dh (x))|
is bounded and continuous.
Let x ∈ U. By the assumption that h and h^{−1} are C^{1},
h (x+B (0,r))− h (x ) ⊆ Dh (x)(B (0,(1+ ε)r)). (9.10)
(9.10)
Making r still smaller if necessary, one can also obtain
|f (y)− f (h (x))| < ε (9.11)
(9.11)
for any y ∈ h
(B(x,r))
and also
|f (h(x1))|det(Dh (x1))|− f (h (x))|det(Dh (x))|| < ε (9.12)
(9.12)
whenever x_{1}∈ B
(x,r)
. The collection of such balls is a Vitali cover of U. By Corollary 9.3.6
there is a sequence of disjoint closed balls
{Bi}
such that U = ∪_{i=1}^{∞}B_{i}∪ N where
m_{p}
(N )
= 0. Denote by x_{i} the center of B_{i} and r_{i} the radius. Then by Lemma 9.6.1, the
monotone convergence theorem, 9.10 - 9.12, and the change of variables formula for linear
maps Theorem 9.5.3,
∫ ∑ ∫
V f (y)dmp = ∞i=1 h(Bi) f (y)dmp
∑ ∞ ∫
≤ εmp (V )+ i=1 h(Bi)f (h(xi))dmp
≤ εm (V)+ ∑ ∞ f (h (x))m (h(B ))
∑ p i=1 i p i
≤ εmp (V)+ ∞i=1f (h (xi))mp (Dh (xi)(B (0,(1+ ε)ri)))
p∑ ∞ ∫
= εmp (V)+ (1+ ε) i=1 Bi f (h (xi))|det(Dh (xi))|dmp
p ∑ ∞ (∫ )
≤ εmp(V )+ (1+ ε) i=1 Bi f (h(x))|det(Dh (x))|dmp + εmp(Bi)
≤ εm (V )+ (1+ ε)p ∑ ∞ ∫ f (h(x))|det (Dh (x))|dm + (1 + ε)pεm (U)
p i=∫1 Bi p p
= εmp (V) + (1 + ε)p U f (h (x ))|det(Dh (x ))|dmp + (1+ ε)p εmp(U )
Since ε > 0 is arbitrary, this shows
∫ ∫
f (y)dmp ≤ f (h(x))|det(Dh (x))|dmp (9.13)
V U
(9.13)
whenever f ∈ C_{c}
(V )
. Now x →f
(h (x))
|det(Dh (x))|
is in C_{c}
(U )
and so using the same
argument with U and V switching roles and replacing h with h^{−1},
∫
f (h (x ))|det(Dh (x))|dm
U p
∫ ( ( −1 )) | ( ( − 1 ))|| ( − 1 )| ∫
≤ f h h (y) |det Dh h (y ) ||det Dh (y )|dmp = f (y)dmp
V V
by the chain rule. This with 9.13 proves the lemma. ■
The next task is to relax the assumption that f is continuous. This will make use of the
following simple lemma.
Lemma 9.7.2Let S be a nonempty set in ℝ^{p}and let
dist(x,S ) ≡ inf {|x − y| : y ∈ S }
Then this function of x is continuous. If K is a compact subset of an open set G, there existsa function f : G →
[0,1]
such thatspt
(f )
≡
{x : f (x) ⁄= 0}
is a compact subset of G andf
(x )
= 1 for all x ∈ K. This situation is written as K ≺ f ≺ G.
Proof: First consider the claim about the function. Let x_{1},x_{2} be two points. Say
dist
Now consider the second claim about f. Since K is compact, there exists a positive
distance, 2δ between K and G^{C}. (Why?) Now let D_{1},
⋅⋅⋅
,D_{m} be closed balls of radius δ
which cover K. Hence K ∪∪_{i=1}^{m}D_{i}≡ H is a compact set which contains K and is contained
in G. Now consider the open sets B
(0,k)
∩
{x : dist(x,GC ) > 1}
k
. These open sets cover H
and are increasing so there exists one of them W which contains H. Note that the closure of
this set is compact because it is closed and bounded. Hence K ⊆ H ⊆ W ⊆ G and W is
compact. Let
( )
-----dist-x,W-C--------
f (x) = dist(x,W C)+ dist(x,H )
Then since H is compact, it has positive distance to W^{C} and so the denominator is nonzero.
Hence the function given has the desired properties. ■
Note that the lemma proves a little more than needed by having the function equal 1 on
an open set containing K.
Corollary 9.7.3Let U and V be boundedopen sets in ℝ^{p}and leth,h^{−1}be C^{1}functionssuch that h
(U)
= V . Also let E ⊆ V be measurable. Then
∫ ∫
X (y )dm = X (h (x))|det(Dh (x ))|dm .
V E p U E p
Proof:Suppose E ⊆ H ⊆ V where H is compact. By regularity, there exist compact sets
K_{k} and a decreasing sequence of open sets G_{k}⊆ V such that
Kk ⊆ E ⊆ Gk
and m_{p}
(Gk ∖ Kk)
< 2^{−k}. By Lemma 9.7.2, there exist f_{k} such that K_{k}≺ f_{k}≺ G_{k}. Then
f_{k}
(y )
→X_{E}
(y )
a.e. because if y is such that convergence fails, it must be the case
that y is in G_{k}∖ K_{k} for infinitely many k and ∑_{k}m_{p}
(Gk ∖Kk )
< ∞. This set
equals
N = ∩∞m=1 ∪∞k=m Gk ∖Kk
and so for each m ∈ ℕ
m (N ) ≤ m (∪∞ G ∖K )
p ∞p k=m k k ∞
≤ ∑ mp(Gk ∖Kk ) < ∑ 2−k = 2−(m −1)
k=m k=m
showing m_{p}
(N )
= 0. (Borel Cantelli lemma.)
Then f_{k}
(h (x ))
must converge to X_{E}
(h (x))
for all x
∕∈
h^{−1}
(N)
, a set of measure zero by
Lemma 9.6.1. Thus X_{E}
(h(x))
= lim_{k→∞}f_{k}
(h (x))
off h^{−1}
(N)
and so by completeness of
Lebesgue measure, x →X_{E}
(h (x))
is measurable. Then
∫ ∫
V fk (y) dmp = U fk(h (x ))|det(Dh (x))|dmp.
Since V is bounded, G_{1} is compact. Therefore,
|det(Dh (x))|
is bounded independent of k
and so, by the dominated convergence theorem, using a dominating function, X_{V } in the
integral on the left and X_{G1}
|det(Dh )|
on the right, it follows
∫ ∫
XE (y) dmp = XE (h (x))|det(Dh (x ))|dmp.
V U
For an arbitrary measurable E, let V = ∪_{k=1}^{∞}H_{k} where H_{k} is compact and
H_{k}⊆ H_{k+1},∪_{k}H_{k} = V. Let E_{k} = H_{k}∩ E replace E in the above with E_{k} and use the
monotone convergence theorem letting k →∞. ■
You don’t need to assume the open sets are bounded.
Corollary 9.7.4Let U and V be open sets in ℝ^{p}and leth,h^{−1}be C^{1}functions such thath
(U)
= V . Also let E ⊆ V be measurable. Then
∫ ∫
X (y )dm = X (h (x))|det(Dh (x ))|dm .
V E p U E p
Proof:Since both h,h^{−1} are continuous, h maps open sets to open sets. Let
Û_{n} = B
(0,n)
∩ U where n is large enough that this intersection is nonempty. Let
V_{n} = h
(ˆ )
Un
∩B
(0,n)
, an open bounded set, and let U_{n} = h^{−1}
(Vn)
, also an open bounded
set. Then if E is a Lebesgue measurable set, the above implies
∫ ∫
XE∩Vn (y )dmp = XE ∩Vn (h(x))|detDh (x )|dmp
V U
Now let n →∞ and use the monotone convergence theorem.■
With this corollary, the main theorem follows.
Theorem 9.7.5Let U and V be open sets in ℝ^{p}and leth,h^{−1}be C^{1}functions suchthat h
(U )
= V. Then if g is a nonnegative Lebesgue measurable function,
∫ ∫
g (y )dmp = g (h (x))|det(Dh (x))|dmp. (9.14)
V U
(9.14)
Proof:From Corollary 9.7.4, 9.14 holds for any nonnegative simple function
in place of g. In general, let
{s }
k
be an increasing sequence of simple functions
which converges to g pointwise. Then from the monotone convergence theorem
∫ ∫ ∫
g (y) dmp = lim skdmp = lim sk(h(x))|det(Dh (x))|dmp
V k∫→ ∞ V k→ ∞ U
= U g(h (x))|det(Dh (x ))|dmp. ■
Of course this theorem implies the following corollary by splitting up the function into the
positive and negative parts of the real and imaginary parts.
Corollary 9.7.6Let U and V be open sets in ℝ^{p}and leth,h^{−1}be C^{1}functions such thath
(U)
= V. Let g ∈ L^{1}
(V)
. Then
∫ ∫
g (y )dmp = g (h (x))|det(Dh (x))|dmp.
V U
This is a pretty good theorem but it isn’t too hard to generalize it. In particular, it is not
necessary to assume h^{−1} is C^{1}.
In what follows, it may be convenient to take
||x|| = max {|x|,i = 1,⋅⋅⋅,p} ≡ ||x||
i ∞
and use the operator norm for A ∈ℒ
p p
(ℝ ,ℝ )
.
class=”left” align=”middle”(Y )9.8. THE MAPPING IS ONLY ONE TO ONE