RIEMANN INTEGRATION ON GENERAL DOMAINS,
PARTITION OF UNITY
MATANIA BEN-ARTZI
March 2013
Ω ⊆ Rn is an open, connected, bounded domain.
Functions here are real, bounded , defined on Ω.
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∫
WE NEED TO DEFINE Ω f.
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Let Q = [a1 , b1 ] × [a2 , b2 ] × ... × [an , bn ] be a closed rectangular box, such that
Ω ⊆ Q.
Extend f to be zero in Q \ Ω :
{
f (x), x ∈ Ω,
fe(x) =
0, x ∈ Q \ Ω.
Definition: If fe(x) is integrable in Q we say that f is integrable in Ω and define
∫
∫
f=
fe.
Ω
Q
Remark. This definition is clearly independent of Q, namely, if Ω ⊆ Q1 for
some other box Q1 , and if fe(x) is integrable in Q then it is integrable in Q1 .
We need this definition to be valid for a large family of functions, for example,
all bounded continuous functions on Ω. This is a geometrical requirement on Ω :
Observation: Take f (x) ≡ 1 in Ω. Then the set of discontinuities of the extension fe(x) in Q is exactly the boundary ∂Ω.
Definition: Ω is called a Jordan Domain if its boundary ∂Ω has measure
zero.
THEOREM. Let Ω be a Jordan domain and let
Λ = {f,
Then
f is integrable in Ω} .
∫
• Λ is a linear space and Ω f is a linear functional on it.
• If f, g ∈ Λ then f · g, max(f, g) and min(f, g) are also in Λ.
• If f ∈ Λ then |f | ∈ Λ and if M = sup |f | and ϕ is a real continuous function
Ω
on an open interval containing [−M, M ], then also ϕ ◦ f ∈ Λ.
1
2
MATANIA BEN-ARTZI
Remark. In particular, for a Jordan domain the volume is defined by
∫
e
v(Ω) =
1.
Q
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GOAL: DEFINE THE INTEGRAL FOR ALL OPEN BOUNDED
DOMAINS.
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(In particular, to define the volume of all such domains).
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A) SOME TOPOLOGY IN Rn .
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Notation. C b Ω means that C is compact and contained in U.
Lemma 1. Let C b Ω. Then there is an open set V such that
C b V ⊂ V b Ω.
Proof. Cover C by finitely many boxes whose closures are contained in Ω. Then
take V =union of these boxes.
∞
Lemma 2. There exists an increasing sequence of open sets {Vm }m=1 , with the
following properties:
(1)
∞
∪
Vm = Ω.
m=1
(2)
Vm ⊂ Vm b Vm+1 ,
m = 1, 2....
∞
Proof. Let {Wj }j=1 be the countable set of all open boxes with rational centers and
rational sides, such that their closures are contained in Ω. Take inductively (with
V0 = ∅)
m
∪
Cm = Vm−1 Wj , m = 1, 2, ...
j=1
Then take Vm , according to the previous lemma
Cm b Vm ⊂ Vm b Ω.
Note. If K b Ω then K b Vm for some m.
∞
Definition.An increasing sequence of open sets {Vm }m=1 , having the properties
as in Lemma 2, is called an exhaustive sequence.
∞
Definition. Let {Wj }j=1 be a countable cover of Ω by open sets. We say that
the cover is locally finite if every compact set intersects only finitely many sets of
the cover.
Lemma 3. Ω has a locally finite cover of open sets with compact closures (can
be taken as open boxes).
RIEMANN INTEGRATION ON GENERAL DOMAINS, PARTITION OF UNITY
3
∞
Proof. Consider the increasing sequence {Vm }m=1 of Lemma 2. Define V0 = V−1 =
∅ and let
Am = Vm \ Vm−1 , m = 1, 2, ..
∞
∪
Then Am b Vm+1 \ Vm−2 and
Am = Ω.
m=1
Cover Am by finitely many open boxes whose closures are contained in Vm+1 \
Vm−2 .
∞
Lemma 4. Let {Wj }j=1 be a locally finite open cover of Ω. Assume that all
∞
the closures Wj are compact. Then there exists a locally finite open cover {Uj }j=1
such that
Uj b Wj , j = 1, 2, ...
Proof. The set C1 = W1 \
∞
∪
Wj = Ω \
j=2
∞
∪
Wj is compact. Use Lemma 1 to find an
j=2
open U1 such that
C1 b U1 ⊂ U1 b W1 .
Inductively, if U1 , . . . , Um−1 are constructed, consider
∞
m−1
[ ∪
∪ ]
C m = Wm \
Wj ∪
Uj ,
j=m+1
j=1
and find an open Um such that
Cm b Um ⊂ Um b Wm .
∞
{Uj }j=1
Clearly
is an open cover of Ω. It is locally finite because any compact set
intersects only finitely many sets Wj , hence only finitely many Uj .
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B) SMOOTH FUNCTIONS WITH COMPACT SUPPORT.
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Claim. Let (a, b) ⊆ R. Then there exists a function 0 ≤ ϕa,b (t) ∈ C ∞ (R) such
that
ϕa,b (t) > 0 ⇔ t ∈ (a, b).
Proof.
ϕa,b (t) = exp(−
1
1
) exp(−
),
(t − a)2
(t − b)2
t ∈ (a, b),
and ϕa,b (t) = 0 otherwise.
Lemma 5. Let ε > 0. There exists a monotone nondecreasing function 0 ≤
gε (t) ∈ C ∞ (R) such that
{
0, t ≤ 0,
gε (t) =
1, t ≥ ε.
Proof. With ϕ0,ε (t) as above define
∫t
gε (t) =
/ ∫∞
ϕ0,ε (τ )dτ
−∞
ϕ0,ε (τ )dτ.
−∞
4
MATANIA BEN-ARTZI
Claim. Let Q = (a1 , b1 )×(a2 , b2 )×...×(an , bn ) be an open rectangular box.Then
there exists a function 0 ≤ ϕQ (x) ∈ C ∞ (Rn ) such that
ϕQ (x) > 0 ⇔ x ∈ Q.
Proof. Take
ϕQ (x) = ϕa1 ,b1 (x1 )ϕa2 ,b2 (x2 ) . . . ϕan ,bn (xn ).
Notation. For any function f (x) (defined on Ω or on Rn ) we define its support
by
supp f = {x, f (x) ̸= 0}.
Definition. For any open set U ⊆ Rn , we define the space (of “test functions”)
as
C0∞ (U ) = {f ∈ C ∞ (U ), supp f b U } .
Lemma 6. Let C b U, where U ⊆ Rn is an open set. Then there exists ψC (x) ∈
C0∞ (U ) such that 0 ≤ ψC (x) ≤ 1 and in particular ψC (x) = 1 for x ∈ C.
m
Proof. Cover C by finitely many boxes {Qj }j=1 whose closures are contained in U.
m
∑
Let ϕQj (x) be as in the above Claim. Define λ(x) =
ϕQj (x).
j=1
Clearly λ(x) ∈ C0∞ (U ), and there exists ε > 0 such that λ(x) > ε for x ∈ C.
Therefore the function
ψC (x) = gε ◦ λ(x),
satisfies the requirements of the lemma, where gε is as in Lemma 5.
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C) PARTITION OF UNITY.
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∞
Definition. Let {Wj }j=1 be a locally finite cover of Ω, so that Wj b Ω for all
j.
∞
A sequence of nonnegative test functions {0 ≤ ϕj }j=1 ⊆ C0∞ (Ω) is a partition
of unity (subject to Wj ) if:
(1) supp ϕj b Wj , j = 1, 2, . . .
∞
∑
(2)
ϕj (x) = 1, x ∈ Ω.
j=1
Note. The sum above is finite for every x ∈ Ω.
∞
Lemma 7. Let {Wj }j=1 be a locally finite cover of Ω, so that Wj b Ω for all j.
∞
Then there exists a partition of unity {0 ≤ ϕj }j=1 ⊆ C0∞ (Ω) subject to this
cover.
∞
Proof. Take a locally finite open cover {Uj }j=1 such that (see Lemma 4)
Uj b Wj ,
Now take ψUj (x)
Define ψ(x) =
∈ C0∞ (Wj )
∞
∑
j=1
j = 1, 2, ...
as in Lemma 6.
ψUj (x) > 0, x ∈ Ω and then
ϕj (x) =
1
ψ (x).
ψ(x) Uj
RIEMANN INTEGRATION ON GENERAL DOMAINS, PARTITION OF UNITY
5
Here is a simple example of how the partition of unity can be used.
Corollary. Let f ∈ C0 (Ω) be a continuous function with compact support in
Ω. Then, for every ε > 0 there exists fε (x) ∈ C0∞ (Ω) such that
|fε (x) − f (x)| < ε,
x ∈ Ω.
Proof. Let δ > 0 be such that (by uniform continuity of f ),
|x − y| < δ ⇒ |f (x) − f (y)| < ε.
∞
{Uj }j=1
Let
be a locally finite open cover (say, by open boxes) such that diam(Uj ) <
∞
δ for all j, and let {ϕj }j=1 be a corresponding partition of unity.
∞
∑
Pick a point xj ∈ Uj and define fε (x) =
f (xj )ϕj (x). Note that the sum is
j=1
actually finite.
Now
|f (x) − fε (x)| = |
∞
∑
(f (x) − f (xj ))ϕj (x)| ≤ ε,
x ∈ Ω.
j=1
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D) DEFINITION OF THE INTEGRAL OVER GENERAL BOUNDED
DOMAINS.
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We take Q ⊇ Ω to be a box containing Ω. For the following definition the
integrals are taken over Q, since this is the only definition of integral we have so
far. A function of compact support in Ω is extended as zero to all of Q.
Definition. Let f : Ω → R be a bounded function. Assume that the set of
discontinuities of f is of measure zero.
∞
Let {0 ≤ ϕj }j=1 ⊆ C0∞ (Ω) be a partition of unity.
We define
∞ ∫
∑
J(f ) =
ϕj f.
j=1
Q
Remarks.
(1) Later we shall see that J(f ) coincides with the integral defined above in
the case of Jordan domains.
(2) The assumption on the discontinuities of f in Ω ensures that every ϕj f is
integrable.
(3) The sum in the definition of J(f ) is absolutely convergent since, for every
integer N,
∫
N
N ∫
∑
∑
|
ϕj f | ≤
|ϕj f | ≤ M v(Q),
j=1
Q
j=1
Q
where M = sup |f | and we have used
Ω
N
∑
ϕj ≤ 1.
j=1
In particular, the order in the sum is not important.
Claim. The definition of J(f ) is independent of the choice of partition of unity.
6
MATANIA BEN-ARTZI
∞
Proof. Let {0 ≤ ψk }k=1 ⊆ C0∞ (Ω) be another partition of unity. Then for every j,
∫
∞ ∫
∑
ϕj f =
ϕj ψk f,
Q
k=1
Q
and because of absolute convergence
∞ ∫
∞ ∑
∞ ∫
∑
∑
J(f ) =
ϕj f =
ϕj ψk f,
j=1
Q
k=1 j=1
Q
which is symmetric with respect to the two partitions.
The following theorem shows that our new definition of J(f ) is consistent with
the previous definition of the integral in the case of Jordan domains.
Theorem. If Ω is a Jordan domain then
∫
∫
J(f ) =
f=
fe
Ω
Q
(see the definition of fe above).
• Step 1. For any integer m > 0 let ∂Ω b Dm , where Dm is a finite
union of open boxes of total volume less than 1/m (∂Ω is of zero content).
∞
• Step 2. Take an exhaustive sequence {Vm }m=1 as in Lemma 2. Each Vm
can be taken as a union of open boxes (with closures in the next Vm+1 ). In
particular, it is a Jordan domain and we can assume further that Ω \ Dm b
Vm , so that
1
v(Ω \ Vm ) < .
m
∞
• Step 3. Let {Uj }j=1 be a locally finite cover of Ω, so that Uj b Ω for all j,
∞
and let {0 ≤ ϕj }j=1 ⊆ C0∞ (Ω) be a partition of unity subject to this cover.
• Step 4. If M = supΩ |f |, clearly
∫
∫
1
| fe − χVm f | ≤ M .
m
Proof.
Q
Q
(χA is the characteristic function of the set A).
• Step 5. Let
∑ ∫
ϕj f.
Im =
Uj ∩Vm ̸=∅ Q
Then
∫
|
∫
∑
χVm f − Im | = |
Q
ϕj (χVm − 1)f | ≤ M
Q Uj ∩Vm ̸=∅
• Step 6. Also
∫
|J(f ) − Im | = |
∑
Q Uj ∩Vm =∅
ϕj f | ≤ M
1
.
m
1
.
m
RIEMANN INTEGRATION ON GENERAL DOMAINS, PARTITION OF UNITY
7
• Step 7. From Steps 4-6 we get, for all m > 0,
∫
1
| fe − J(f )| ≤ 3M .
m
Q
We conclude that for Jordan domains, the definition of J(f ) is consistent with
the previous definition of the integral. We therefore have the following definition.
Definition. Let Ω ⊆ Rn be any bounded, open domain.
Then:
• The space of integrable functions is the space of all bounded functions
whose set of discontinuities is of measure zero (and of course can be empty).
• For any integrable function the integral is defined as
∫
f = J(f ).
Ω
Corollary. For every bounded, open domain Ω, we can define the volume by
∫
v(Ω) =
1.
Ω
From the proof of the last theorem we can extract the following proposition,
where Ω is not required to be a Jordan domain.
∞
Proposition. Let Ω be any open, bounded domain. Let {Vm }m=1 be an exhaustive sequence as in Lemma 2, where each Vm is a Jordan domain. Then, for
any f integrable in Ω,
∫
∫
lim
m→∞
Vm
f=
f.
Ω
Proof. The proof is similar to the previous proof, only that now we do not assume
that the boundary has zero measure.
The main point is that, given partition of unity as before,
∞ ∫
∑
lim
ϕj |f | = 0,
J→∞
j=J Q
since the series is absolutely convergent.
Thus
∫
∫
∫
f = χVm f =
Vm
f=
Ω
ϕj χVm f,
Q Uj ∩Vm ̸=∅
Q
∫
∫
and
∑
∑
∫
ϕj f +
Q Uj ∩Vm ̸=∅
∑
ϕj f.
Q Uj ∩Vm =∅
The difference can be estimated by
∫
∫
∞ ∫
∑ ∫
∑
ϕj |f |,
|
f−
f| ≤ 2
ϕj |f | ≤ 2
Ω
Vm
Uj *Vm Q
j=Jm Q
where Jm is the first index for which Uj * Vm .
∞
Since {Vm }m=1 is exhaustive, Jm → ∞ as m → ∞.
8
MATANIA BEN-ARTZI
Actually, we can improve this proposition and drop the requirement that the
domains in the sequence are Jordan domains:
∞
Theorem. Let Ω be any open, bounded domain. Let {Ωj }j=1 ↑ Ω be an
increasing sequence of domains (not necessarily Jordan) whose union is Ω. Then,
for any f integrable in Ω,
∫
∫
lim
j→∞
Ωj
f=
f.
Ω
Proof. Note that every compact K b Ω is contained in some Ωj .
∞
Let {Vm }m=1 be an exhaustive sequence (for Ω) as in the proposition, where
each Vm is a Jordan domain. By the proposition we can assume that
∫ ∫
f−
f < 2−m , m = 1, 2, 3...
Vm
Ω
{ j }∞
In addition, for every index j let Wm
be an exhaustive sequence of Jordan
m=1
domains for Ωj . We can assume again
∫
∫
f
−
f < 2−m , m = 1, 2, 3..., j = 1, 2, 3...
j
Wm
Ωj
By a diagonal process we now take an exhaustive sequence
{
}∞
j
Wm
for Ω so
j
j=1
that
j
j+1
mj+1 > mj , j = 1, 2, 3...
• Wm
j b Wmj+1 ,
• If jk is the first index j so that Vk b Ωj then we require also mjk is such
that
jk
V k b Wm
, k = 1, 2, 3...
j
k
Given ε > 0, take jε such that 2−jε < ε. Take any j > jε and let
l > k > j be such that
j
l
Wm
j b Vk ⊆ Vk b Wml .
Now
∫
∫
f−
Vk
j
Wm
j
∫
f ≤
∫
l \W j
Wm
mj
Ωj
|f | −
l
Wm
l
Since k > j > jε and mj > j, we have
∫
∫
∫
∫
f−
f ≤ f−
Ω
∫
|f | =
Vk
j
Wm
j
l
l
Wm
Ωj
l
According to the Proposition above , since
sequence,
∫
∫
|f | ↑
|f |,
j
Wm
j
Ω
|f |.
f + 2ε,
and combining with the previous inequality
∫
∫
∫
∫
f−
f ≤ 2ε +
|f | −
Ω
j
Wm
j
j
Wm
j
|f |.
{
}∞
j
Wm
j
j=1
is an exhaustive
RIEMANN INTEGRATION ON GENERAL DOMAINS, PARTITION OF UNITY
∫
∫
lim sup f−
hence
j→∞
Ω
9
f ≤ 2ε,
Ωj
for any ε > 0 >
∞
{Ωj }j=1
Corollary. Let Ω be any open, bounded domain. Let
↑ Ω be an
increasing sequence of domains (not necessarily Jordan) whose union is Ω. Then
v(Ωj ) ↑ v(Ω).
Note. If Ω1 ⊆ Ω2 ⊆ Ω and f ≥ 0, then
∫
∫
f≤
Ω1
f.
Ω2
For later use, we need to deal with partitions of unity with functions ϕj that are
only continuous.
∞
Lemma 8. Let {Wj }j=1 be a locally finite cover of Ω, so that Wj b Ω for all j.
∞
Let the sequence of nonnegative functions {0 ≤ ϕj }j=1 ⊆ C0 (Ω) be a partition
of unity (subject to Wj ) by continuous, compactly supported functions, namely:
(1) supp ϕj b Wj , j = 1, 2, . . .
∞
∑
(2)
ϕj (x) = 1, x ∈ Ω.
j=1
Let f be an integrable function on Ω. Then
∫
∞ ∫
∑
f=
ϕj f.
Ω
j=1
Ω
Proof. Note that the right-hand side is absolutely convergent since for every integer
N,
N ∫
N ∫
∑
∑
ϕj f ≤
ϕj |f | ≤ M v(Ω).
Ω
j=1
Ω
j=1
Let ε > 0. By the above Proposition, we have a Jordan domain V ⊆ V b Ω so that
∫ ∫
f−
f < ε.
Ω
Now
∫
f=
V
∞ ∫
∑
χV ϕj f =
Ω
j=1
V
∑ ∫
Wj ⊆V
ϕj f +
Q
∑ ∫
Wj *V
χV ϕj f.
Ω
The sums are all finite.
We conclude
∞ ∫
∫
∑ ∫
∑
ϕj |f |,
ϕ
f
≤
2
f
−
j V
Ω
j=1
Wj *V
Ω
which can be made arbitrarily small (take V as element of exhaustive sequence and
note that the tail of the series shrinks to zero).
Properties of the integral
∫
Ω
.
10
MATANIA BEN-ARTZI
• The space of integrable functions is linear and the integral is a linear functional on it.
• The integral is order-preserving:
∫
∫
f ≤g⇒
f≤
g.
Ω
Ω
***************
We conclude this section by a generalization of the idea of Riemann sums.
∞
Definition. A curved partition of the domain Ω is a countable set {Dj }j=1
of open, connected subsets of Ω, so that
• Every Dj is a Jordan domain, and Dj b Ω.
∞
∪
•
Dj = Ω.
j=1
• The subsets do not have common interior points: D̊j ∩ D̊l = ∅ for j ̸= l.
• Every compact K b Ω intersects at most a finite number of Dj .
Note that in a regular partition of a rectangular box Q, the partition boxes satisfy
these requirements (with the compact Q replacing the open Ω).
Lemma 9. Let f be an integrable function with compact support in Ω. For
every ε > 0 there exists δ > 0 so that:
∞
If {Dj }j=1 is a curved partition in Ω with
sup diam(Dj ) < δ,
j
and xj ∈ Dj , j = 1, 2, 3 . . . , then
∫ ∞
∑
f (xj )v(Dj ) −
f < ε.
Ω
j=1
Note that the sum is finite.
Proof. Since f is compactly supported we can extend it as zero to a box Q ⊇ Ω.
N
Let P = {Sk }k=1 be a partition of Q, so that
U (f, P ) − L(f, P ) < ε.
Let C =total area of faces of the boxes Sk . Clearly if diam(Dj ) < δ for all j, then
the total volume of the Dj that intersect faces of the partition P is less than Cδ.
We set
Mj = sup f, mj = inf f.
Dj
Dj
Counting separately the Dj that intersect faces of P, and all others (that are contained in the interior of some Sk ) we obtain (note that the sum is actually finite)
∞
∑
(Mj − mj )v(Dj ) ≤ U (f, P ) − L(f, P ) + M Cδ ≤ ε + M Cδ.
j=1
Clearly
∞
∑
j=1
∫
mj v(Dj ) ≤
f≤
Ω
∞
∑
Mj v(Dj ).
j=1
RIEMANN INTEGRATION ON GENERAL DOMAINS, PARTITION OF UNITY
11
*************************************************************
E) CHANGE OF VARIABLES IN INTEGRATION.
************************************************************
We start with basic facts about change of volumes under linear transformations.
Then we proceed to general maps.
The definition of the integral, and in particular the volume of a domain, was
made with respect to a specific system of coordinates in Rn . Clearly, the definition
is invariant under translation, because the volume of rectangular boxes is invariant under such translations. However, we need to verify that the integral is also
invariant under rotations.
Claim. Let Ω ⊂ Rn be an open, bounded domain. Let R be a rotation (expressed by an orthogonal matrix) and let Γ = RΩ be the image domain. Then
v(Ω) = v(Γ).
Proof. It is sufficient to see that the volume of every rectangular box is invariant
under the rotation.
More generally: {
}
Lemma 10. Let a1 , a2 , . . . , an ⊆ Rn and let Ω be the parallelepiped formed
by these vectors:
{ n
}
∑
Ω=
ti ai , ti ∈ (0, 1), i = 1, 2, . . . , n .
i=1
Then
v(Ω) = |det(A)|,
{
}
where A is the matrix whose rows are a1 , a2 , . . . , an .
Proof. By induction on n. For n = 2 (the plane) this is a direct computation (the
area of the triangle can be computed by Fubini’s theorem or adding and subtracting
areas of trapezoids).
{
}
Suppose the statement is true for n−1. By rotation we can assume that a1 , a2 , . . . , an−1
span Rn−1 , so in the new system they are given by
ãi = (ãi1 , . . . , ãin−1 , 0),
i = 1, . . . , n − 1.
The (n − 1) volume spanned by them is, by the induction hypothesis, |det(An−1 )|,
where

 1
ã1
... ã1n−1
...
...  .
An−1 =  ...
n−1
ãn−1
...
ã
1
n−1
The remaining vector is now ãn = (ãn1 , . . . , ãin−1 , ãnn ).
By Fubini’s theorem, the n volume is given by |ãnn ||det(An−1 )|, which is equal to
|det(An )|, with
 1

ã1
... ã1n−1 , 0
 ...
...
... 
.
An = 
n−1
ãn−1
... ãn−1 , 0 
1
ãn1
... ãnn−1 , ãnn
But |det(An )| = |det(A)|, since the transformation matrix was orthogonal (rotation).
12
MATANIA BEN-ARTZI
Corollary. If A ∈ Hom(Rn , Rn ) is a regular matrix, then for every domain Ω,
v(AΩ) = v(Ω)|det(A)|.
*****************************************
General maps
******************************************
Let Ω, Γ ⊂ Rn be open, connected, bounded domains. Let
Φ : Ω → Γ,
be a 1 − 1 map of Ω onto Γ so that Φ ∈ C 1 .
For every x ∈ Ω, the linear operator J(x) = Φ′ (x) ∈ Hom(Rn , Rn ) is called the
Jacobian of Φ at x.
Recall that the corresponding matrix is
 ∂Φ1

1
... ∂Φ
∂x1
∂xn
J(x) =  ... ... ...  .
∂Φn
n
... ∂Φ
∂x1
∂xn
We assume that J(x) is nonsingular and that the map x → J(x) is continuous on
Ω. Furthermore:
Assumption. Both det(J(x)) and det(J −1 (x)) are bounded on Ω.
Remark. Note in particular that it implies that J(x) is nonsingular for x ∈ Ω
and, since Φ ∈ C 1 , the maps
x → J(x),
x → J −1 (x),
x ∈ Ω,
are continuous (into the space of matrices).
The next Lemma is crucial; it gives the local effect of the nonlinear map Φ on
volumes.
Lemma 11. Let x ∈ Ω and let Qh (x) be a cubic box (all sides equal to h)
centered at x. Then
v(Φ(Qh (x)))
lim
= |det(J(x))|.
h→0 v(Qh (x))
Furthermore, if K b Ω, then the limit is attained uniformly for x ∈ K.
Proof. The (first-order) Taylor approximation at x can be written as
Φ(y) − Φ(x) = J(x) · (y − x) + R(x, y),
where
lim
y→x
|R(x, y)|
= 0.
|y − x|
y − x = J −1 (x) · (Φ(y) − Φ(x)) − J −1 (x) · R(x, y).
By assumption det(J −1 (x)) is uniformly bounded for x ∈ K, hence
|J −1 (x) · (Φ(y) − Φ(x)) − (y − x)|
= 0.
y→x
|y − x|
lim
We conclude that there exists δ(h) > 0, with limh→0 δ(h) = 0, so that, for
y ∈ ∂Qh (x),
J −1 (x) · Φ(y) ∈ Qh(1+δ(h)) (J −1 (x) · Φ(x)) \ Qh(1−δ(h)) (J −1 (x) · Φ(x)),
RIEMANN INTEGRATION ON GENERAL DOMAINS, PARTITION OF UNITY
13
hence
Qh(1−δ(h)) (J −1 (x) · Φ(x)) ⊆ J −1 (x) ◦ Φ(Qh (x)) ⊆ Qh(1+δ(h)) (J −1 (x) · Φ(x)).
By Lemma 10 and its Corollary
v(J −1 (x) ◦ Φ(Qh (x))) = |det(J(x))|−1 v(Φ(Qh (x))).
This lemma has an important consequence, namely, that sets of measure zero
are preserved under a transformation having the properties of Φ :
Lemma 12. If C ⊆ Ω is a set of measure zero, then so is Φ(C) ⊆ Γ, the image
of C under Φ.
Proof. It is sufficient to deal with the case that C is contained in a compact set
K b Ω.
We can now state the main theorem:
Theorem. Let f be integrable on Γ. Then the function f ◦ Φ is integrable on
Ω, and
∫
∫
(Cof V )
(f ◦ Φ)|det(J(x))| =
f.
Ω
Γ
• Step 1. (f ◦ Φ) is integrable in Ω, since its set of discontinuities has
measure zero by Lemma 12.
∞
• Step 2. Let {Wj }j=1 be a locally finite cover of Γ, so that Wj b Ω for all j.
∞
Let {0 ≤ ϕj }j=1 ⊆ C0∞ (Ω) be a partition of unity (subject to Wj .)
{
}∞
∞
Then Φ−1 (Wj ) j=1 is a locally finite cover of Ω, and {0 ≤ ϕj ◦ Φ}j=1 ⊆
C0 (Ω) is a partition of unity (subject to it) .
• Step 3. By Lemma 8 we know that
∫
∞ ∫
∑
(f ◦ Φ)|det(J(x))| =
(f ◦ Φ)(ϕj ◦ Φ)|det(J(x))|,
Proof.
Ω
and
j=1
∫
f=
Γ
Ω
∞
∑
ϕj f.
j=1
Thus, it is sufficient to consider the case that supp(f ) b Γ is compact, and
therefore so is supp(f ◦ Φ) b Ω.
• Step 4. In particular, we can assume that all elements of J(x), J −1 (x) are
uniformly bounded in a neighborhood of supp(f ◦ Φ) b Ω.
N
• Step 5. Let P = {Sk }k=1 be a partition of the box Q (containing Ω) to
boxes Sk . Given ε > 0, we can take the parameter of P sufficiently small,
so that
∫
N
∑
k
k
(f ◦ Φ)(x )|det(J(x ))|v(Sk ) − (f ◦ Φ)|det(J(x))| < ε,
k=1
for any choice xk ∈ Sk .
Ω
14
MATANIA BEN-ARTZI
• Step 6. We can refine the partition further, if needed, so that, by Lemma
11,
|det(J(xk ))|v(Sk ) − v(Φ(Sk )) < εv(Sk ).
• Step 7. It follows by summation and the previous inequality that
∫
N
∑
(f ◦ Φ)(xk )v(Φ(Sk )) − (f ◦ Φ)|det(J(x))| < ε + M εv(Q),
Ω
k=1
where M = supΓ |f |.
N
• Step 8. Note that {Φ(Sk )}k=1 is a curved partition of Γ. Hence, refining
further the partition P if needed, we obtain from Lemma 9,
∫ N
∑
(f ◦ Φ)(xk )v(Φ(Sk )) − f < ε.
k=1
Γ
Combining the results in Steps 5,7-8 equality (CofV) follows.
Institute of Mathematics, Hebrew University, Jerusalem 91904, Israel
E-mail address: [email protected]