Definitions:
(a) K is called a cell in Rp (or a p-cell, or a p-dimensional rectangle) if
K = I1× · · · × Ip, where Ij = [aj, bj] ⊂ R for j = 1, . . . , p.
(b) The (p-)contentc(K) of K is defined to be c(K) = (b1− a1) × · · · × (bp− ap) =
p
Y
j=1
(bj− aj).
(c) A set Z ⊂ Rp has p-content zero if ∀ > 0, ∃ a finite collection C = {Kj}mj=1 of p-cells such that
(i) Z ⊂
m
[
j=1
Kj, i.e. C = {Kj}mj=1 is a cover of Z
(ii)
m
X
j=1
c(Kj) < .
Remarks
(a) Note that the definition implies that if K is a cell (not necessarily closed) in Rp, then the boundary ∂K of K is a set of p-content zero.
(b) Note that the definition of the content for a cell is well defined since it is easy to see that the following properties are satisfied.
(i) Let K be a cell in Rp and assume that K is a finite disjoint union of cells {Ki | i = 1, . . . , l} in Rp, i.e.
K =
l
[
i=1
Ki andKi∩ Kj = ∅ for 1 ≤ i 6= j ≤ l.
Then
c(K) =
l
X
i=1
c(Ki).
(ii) Let K1, K2 be cells in Rp. Then
c(K1∪ K2) = c(K1\ K2) + c(K1∩ K2) + c(K2 \ K1).
(iii) Let x ∈ Rp, K be a cell in Rp and x + K = {x + z | z ∈ K}. Then x + K is a cell in Rp with c(x + K) = c(K), i.e. the content of a cell is invariant under translations.
(c) Definition (c) is equivalent to the following: A set Z ⊂ Rp has p-content zero if ∀ > 0, ∃ a finite collection C = {Kj}mj=1 of p-cells such that
(i) Z ⊂ Int
m
[
j=1
Kj,
(ii)
m
X
j=1
c(Kj) < .
Examples
(a) Let Z = {xj ∈ R | lim
j→∞xj = x}, a (0-dim’l) subset of R. Then (1 − dim’l) c(Z) = 0 since
∀ > 0, ∃ a 1-d cell Kxsuch that x ∈ Int(Kx), c(Kx) < /2, and xj ∈ Kx∀j ≥ L. For each j = 1, . . . , L − 1, let Kj be a 1-d cell such that xj ∈ Kj, and c(Kj) ≤ /(2L).
(b) Let Z = Q∩[0, 1], a (0-dim’l) subset of R. Then (1-dim’l) c(Z) 6= 0 since any finite collection C = {K1, . . . , Km} of 1−dimensional cells that satisfies (a) will have
m
X
j=1
c(Kj) ≥ 1.
(c) Let Z = {(x, y) | |x|+|y| = 1}, a (1-dim’l) subset of R2. Then the (2-dim’l) content c(Z) = 0.
Integrable Functions on General Measurable Sets:
In the following we shall extend the concept of content of a cell in Rnto more general measurable subsets of Rn and extend the definition of integrability of a function to general subsets of Rn.
Definition (of integrability on general Euclidaen bounded subsets.) Let A ⊂ Rn be a bounded set and let
f : A → R
be a bounded function. Let K be a closed cell containing A and define fK : K → R
by
fK(x) =
(f (x) if x ∈ A, 0 if x ∈ K \ A.
We say that f is integrable on A if fK is integrable on K, and define Z
A
f = Z
K
fK.
Remark. If A = K is a closed cell in Rn, since
fK = f on K, it is obvious
Z
A
f = Z
K
fK.
Remark. Let I be any closed cell containing A. Then K ∩ I is a closed cell containing A and
fK =
(fK∩I everywhere in K ∩ I, 0 on K \ (K ∩ I),
fI =
(fK∩I everywhere in K ∩ I, 0 on I \ (K ∩ I),
Hence,
Z
K
fK = Z
K∩I
fK∩I = Z
I
fI,
and the definition (of integrability of f ) only depends on f and A and it is independent of the choice of K.
Properties of Integrable Functions on General Sets
(a) Let f and g be integrable functions defined on a bounded set A ⊂ Rn and let α, β ∈ R.
Then the function αf + βg is integrable on A and Z
A
(αf + βg) = α Z
A
f + β Z
A
g.
Proof. For any partition P of a cell K ⊇ A, since
SP(αfK+ βgK, K) = αSP(fK, K) + βSP(gK, K)
when the same intermediate points xj are used, the function αf + βg is integrable on A.
Thus, by choosing the intermediate points from A whenever it is possible, we obtain that SP(αf + βg, A) = αSP(f, A) + βSP(g, A)
which implies that
Z
A
(αf + βg) = α Z
A
f + β Z
A
g.
(b) Let A1 and A2 be bounded sets with no pints in common, and let f be a bounded function.
If f is integrable on A1 and on A2, then f is integrable on A1∪ A2 and Z
A1∪A2
f = Z
A1
f + Z
A2
f.
Proof. Let K be a closed cell containing both A1 and A2, and let
fK(x) =
(f (x) if x ∈ A1 ∪ A2, 0 if x ∈ K \ (A1 ∪ A2) and
fKi (x) =
(f (x) if x ∈ Ai,
0 if x ∈ K \ Ai for i = 1, 2.
Since f is integrable on Ai, i = 1, 2, fKi is integrable on K and, since fK = fK1 + fK2,
and f is integrable on A1∪ A2. Also, for any partition P of K, note that SP(fK, K) = SP(fK1, K) + SP(fK2, K)
when the same intermediate points xj are used. Thus, we have Z
A1∪A2
f = Z
A1
f + Z
A2
f.
(c) If f : A → R is integrable on (bounded set) A and f (x) ≥ 0 for x ∈ A, then Z
A
f ≥ 0.
Proof. For any closed cell K ⊇ A and any partition P of K, note that SP(fK, K) ≥ 0 for any Riemann sum. Hence,
Z
A
f ≥ 0.
Remarks This implies that if f and g are integrable on A and f (x) ≤ g(x) for x ∈ A, then (i)
Z
A
f ≤ Z
A
g, (ii) |f | is integable on A, and
| Z
A
f | ≤ Z
A
|f |.
(d) Let f : A → R be a bounded function and suppose that A has content zero. Then f is integrable on A and
Z
A
f = 0.
Proof. LetK ⊇ A be a closed cell.
If > 0 is given, let P be a partition of K such that those cells in P which contain points of A have total content less than .
Now if P is a refinement of P, then those cells in P containing points of A will also have total content less than .
Hence
if |f (x)| ≤ M for x ∈ A then |SP(fK, K)−0| ≤ M for any Riemann sum corresponding to P.
Since is arbitrary, this implies thatR
Af = 0.
(e) Let f, g : A → R be bounded functions and suppose that f is integrable on (bounded set) A. Let Z ⊆ A have content zero and suppose that
f (x) = g(x) for all x ∈ A \ Z.
Then g is integrable on A and
Z
A
f = Z
A
g.
Proof. Extend f and g to functions fK, gK defined on a closed cell K ⊇ A. Thus, the function
hK = fK − gK is bounded and equals 0 except on Z.
Hence, hK is integrable on K and
Z
K
hK = 0.
Since fK is also integrable on K, we have Z
A
f = Z
K
fK = Z
K
(fK− hK) = Z
K
gK = Z
A
g.
(f) Let U be a connected, open subset of Rn and let f : U ⊂ Rn→ Rn+1
be a C1 map on U. If K is any convex, compact subset of U, then f (K) has measure (or content) zero.
Definition. Let A ⊂ Rn be a bounded set. The characteristic function of A is the function χA defined by
χA(x) =
(1 if x ∈ A, 0 otherwise.
Now, suppose A is a bounded subset of Rnand f is a bounded function on Rn. Let K be a closed cell that contains A. We say that f is integrable on A if f χA is integrable on K, and define
Z
A
f = Z
K
f χA.
Note that f χA= 0 on K \ A, so it is independent of the choice of K ⊇ A.
Question: Let f ≡ 1 on A ⊂ Rn. What does it mean when we say that f is integrable on A?
Definition. A set A ⊂ Rn is said to have content (or it is said to be (Jordan) measurable) if it is bounded and its boundary ∂A has content zero. Let
D(Rn) = {A ⊂ Rn| A has content } = {A ⊂ Rn| A is measurable } denote the set of all measurable subsets of Rn.
Remarks
(a) If A ∈D(Rn) and if K is a closed cell containing A, then the function gK defined by gK(x) =
(1 if x ∈ A 0 if x ∈ K \ A
is continuous on K except possibly at points of ∂A (which has content zero). Thus, gK is integrable on K and we define the content of A, denoted c(A), by
c(A) = Z
K
gK = Z
A
1.
(b) c(A) = Z
K
gK implies that
∀ > 0, ∃ a partition P = {Ij}mj=1 of K such that |SP(gk, K) − c(A)| < for any Riemann sum SP(gk, K).
By choosing the intermediate points in SP(gk, K) to belong to A whenever possible, we
have m
X
j=1
c(Ij) + ≥ SP(gk, K) + > c(A) > SP(gk, K) − ,
where the first inequality holds since
A ⊂
m
[
j=1
Ij.
Hence, A ⊂ Rn has content zero if and only if A has content and Z
A
1 = c(A) = 0.
Proof. ⇒ ∀ > 0, since A has content zero,
∃ closed cells I1, . . . , Imsuch that
A ⊂
m
[
j=1
Ij = K,
m
X
j=1
c(Ij) < .
Since
(i) K is bounded implies that A is bounded, (ii) K is closed implies that ∂A ⊂ K =
m
[
j=1
Ij with
m
X
j=1
c(Ij) < =⇒ c(∂A) = 0
implies that A has content and
c(A) = Z
A
1 = Z
K
gK = 0 since
0 ≤ c(A) < SP(gK, K) + ≤
m
X
j=1
c(Ij) + ≤ 2
and is arbitrary.
⇐ Suppose that A ⊂ Rn has content and that c(A) = 0 then there exists a closed cell K containing A such that the function
gK(x) =
(1 if x ∈ A 0 if x ∈ K \ A is integrable on K. ∀ > 0, let
P = {Ij}mj=1
be a partition of K such that any Riemann sum corresponding to P satisfies that 0 ≤ |SP(gK, K) − c(A)| < .
Since c(A) = 0, we have
0 ≤ SP(gK, K) < .
By taking the intermediate points in
SP(gK, K) to belong to A whenever possible, we have
A ⊂ [
1≤j≤m; Ij∩A6=∅
Ij and X
1≤j≤m; Ij∩A6=∅
c(Ij) < .
This implies that
c(A) = 0.
Theorem. Let A ∈D(Rn) and let
f : A → R be integrable on A and such that
|f (x)| ≤ M for all x ∈ A.
Then
| Z
A
f | ≤ M c(A).
More generally, if f is real valued and m ≤ f (x) ≤ M for all x ∈ A, then (∗) mc(A) ≤
Z
A
f ≤ M c(A).
Proof. Let fK be the extension of f to a closed cell K containing A. If > 0 is given, then there exists a partition
P = {Ij}hj=1
of K such that if SP(fK, K) is any corresponding Riemann sum, then SP(fK, K) − ≤
Z
K
fK ≤ SP(fK, K) + .
Note that if the intermediate points of the Riemann sum are chosen outside of A whenever possible, we have
SP(fK, K) =
0
Xf (xj)c(Ij),
where the sum is extended over those cells in P entirely contained in A. Hence, SP(fK, K) ≤ M
0
Xc(Ij) ≤ M c(A).
Therefore, we have
Z
A
f = Z
K
fK ≤ M c(A) + ,
and since > 0 is arbitrary we obtain the right side of inequality (∗). The left side is established in a similar manner.
Theorem. If A ∈D(Rn) and c(A) > 0, then there exists a closed cell J ⊆ A such that c(J ) 6= 0.
Mean Value Theorem. Let A ∈D(Rn) be a connected set and let f : A → R
be bounded and continuous on A. Then f is integrable on A and there exists a point p ∈ A such
that Z
A
f = f (p)c(A).
Proof. If c(A) = 0, the conclusion is trivial; hence we suppose that c(A) 6= 0. Let m = inf{f (x) : x ∈ A} and M = sup{f (x) : x ∈ A},
it follows from the preceding theorem that m ≤ 1
c(A) Z
A
f ≤ M.
If both inequalities are strict, the results follows from Intermediate Value Theorem (since f is continuous on A).
Suppose that
Z
A
f = M c(A).
If the supremum M is attained at p ∈ A, the conclusion also follows. Hence we assume that the supremum M is not attained on A. Since c(A) 6= 0, there exists a closed cell J ⊆ A such that c(J ) 6= 0 (prove this). Since J is compact and f is continuous on J, there exists > 0 such that
f (x) ≤ M − for all x ∈ J.
Since A = J ∪ (A \ J ) we have M c(A) =
Z
A
f = Z
J
f + Z
A\J
f ≤ (M − )c(J ) + M c(A \ J ) < M c(A), which is a contradiction. If R
Af = mc(A), then a similar argument applies.
Mean Value Theorem for Riemann-Stieltjes Integrals. Let A ∈D(Rn) be a connected set, let f : A → R
be bounded and continuous on A, and let
g : A → R
be bounded, nonnegative and continuous on A, Then there exists a point p ∈ A such that Z
A
f g = f (p) Z
A
g.
Remark (Well-definedness of (Jordan) measurability). Note that the definition for a set having content (or the definition of a set being measurable) is well defined since the the following properties hold.
Proposition. Let A, B ∈D(Rn) and let x ∈ Rn. Then
(1) A ∩ B, A ∪ B ∈D(Rn) and
c(A) + C(B) = c(A ∩ B) + c(A ∪ B).
Using induction, this concludes that if A ∈D(Rn) and A is a finite disjoint union of measurable subsets, i.e.
A =
l
[
i=1
Bi and each Bi ∈D(Rn), then
c(A) =
l
X
i=1
c(Bi).
(2) A \ B, B \ A ∈D(Rn), and
c(A ∪ B) = c(A \ B) + c(A ∩ B) + c(B \ A).
(3) If
x + A = {x + a | a ∈ A}
then x + A ∈D(Rn) and
c(x + A) = c(A), i.e. the definition of content is invariant under translations.
Proof. By using the definition of boundary points of a set, we have
∂(A ∩ B), ∂(A ∪ B), ∂(A \ B), ∂(B \ A) ⊂ ∂A ∪ ∂B.
Thus,
c(∂(A ∩ B)) = c(∂(A ∪ B)) = c(∂(A \ B)) = c(∂(B \ A)) = 0 and
A ∩ B, A ∪ B, A \ B, B \ A ∈D(Rn).
Let K ⊇ A ∪ B be a closed cell and let
fA, fB, fA∩B, fA∪B
be the functions equal to 1 on A, B, A ∩ B, A ∪ B, respectively, and equal 0 elsewhere on K.
Then they are integrable on K and , since
fA+ fB= fA∩B+ fA∪B,
we have
c(A) + c(B) = Z
K
fA+ Z
K
fB
= Z
K
(fA+ fB)
= Z
K
(fA∩B+ fA∪B)
= Z
K
fA∩B+ Z
K
fA∪B
= c(A ∩ B) + c(A ∪ B).
To prove (3), note that if > 0 is given and if J1, . . . , Jm are cells with
∂A ⊂
m
[
i=1
Ji and
m
X
i=1
c(Ji) < , then
x + J1, . . . , x + Jm are cells such that
∂(x + A) ⊂
m
[
i=1
(x + Ji) and
m
X
i=1
c(x + Ji) < .
Since > 0 is arbitrary, the set x + A belongs to D(Rn).
Let I be a closed cell containing A; hence x + I is a closed cell containing x + A. Let f1 : I → R be such that
f1(y) =
(1 for y ∈ A 0 for y ∈ I \ A, and let f2 : x + I → R be such that
f2(y) =
(1 for y ∈ x + A
0 for y ∈ x + I \ (x + A).
Then
f1(y) = f2(x + y) for each y ∈ A and
c(A) = Z
I
f1 = Z
x+I
f2 = c(x + A).
Lemma Let Ω ⊆ Rp be open, let
φ : Ω → Rp belong to class C1(Ω)
and let A be a bounded set satisfying that Cl(A) = ¯A ⊂ Ω. Then there exists a bounded open set Ω1, a collection of closed cells {Ii | 1 ≤ i ≤ m} in Ω1 and a constant M > 0 such that
A ⊂ Ω¯ 1 ⊂ ¯Ω1 ⊆ Ω and
A ⊆
m
[
i=1
Ii and
m
X
i=1
c(Ii) ≤ α,
then there exists a collection of closed cells {Ji | 1 ≤ i ≤ m} in Rp such that
φ(A) ⊆
m
[
i=1
Ji and
m
X
i=1
c(Ji) ≤ (√
pM )pα.
Proof Let
δ =
1 if Ω = Rp,
1
2inf{ka − xk | a ∈ ¯A, x /∈ Ω} if Rp\ Ω 6= ∅,
Ω1 = {y ∈ Rp | ky − ak < δ for some a ∈ A}
and
M = sup
x∈Ω1
kdφ(x)k = sup
x∈Ω1 ; 06=v∈Rp
kdφ(x)(v)k kvk < ∞.
Note that if Rp\ Ω 6= ∅, since ¯A is compact and Rp\ Ω is closed, then δ > 0 and Ω1 is open.
Let I1, . . . , Im be a collection of closed cells in Ω1 such that A ⊆
m
[
i=1
Ii and
m
X
i=1
c(Ii) ≤ α.
For each 1 ≤ i ≤ m, since Ii is convex and by the Mean Value Theorem, we have kφ(x) − φ(y)k ≤ M kx − yk ∀x, y ∈ Ii.
This implies that if Ii is a closed cell of side length 2ri then φ(Ii) is contained in a closed cell Ji ⊂ Rp of side length 2√
pM ri.
Hence, we obtain a collection of closed cells {Ji | 1 ≤ i ≤ m} such that
φ(A) ⊆
m
[
i=1
Ji and
m
X
i=1
c(Ji) ≤ (√
pM )pα.
Theorem Let Ω ⊆ Rp be open and let
φ : Ω → Rp belong to class C1(Ω).
If A ⊂ Ω has content zero and if ¯A ⊂ Ω, then φ(A) has content zero.
Proof Apply the lemma for arbitrary α > 0.
Corollary Let r < p, let Ω ⊆ Rr be open, and let
ψ : Ω → Rp belong to class C1(Ω).
If A ⊂ Ω is a bounded set with ¯A ⊂ Ω, then ψ(A) has content zero in Rp.
Proof Let Ω0 = Ω × Rp−r so that Ω0 is open in Rp, and define φ : Ω0 → Rp by φ(x1, . . . , xr, xr+1, . . . , xp) = ψ(x1, . . . , xr).
Evidently φ ∈ C1(Ω0). Let
A0 = A × {(0, . . . , 0)} ⊂ Rp, where (0, . . . , 0) ∈ Rp−r, so that ¯A0 ⊂ Ω0 and A0 has content zero in Rp. It follows that
ψ(A) = φ(A0) has content zero in Rp.
Theorem Let Ω ⊆ Rp be open and let
φ : Ω → Rp belong to class C1(Ω).
Suppose that A has content, ¯A ⊂ Ω, and the Jacobian determinant Jφ(x) = det(dφ)(x) 6= 0 for all x ∈ Int(A).
Then φ(A) has content.
Remark. If B is a subset in Rp, then
B = B ∪ B0 = IntB ∪ ∂B, where
B0 = {x ∈ Rp | ∀ r > 0 Dr(x) ∩ B \ {x} 6= ∅}, where Dr(x) = {y ∈ Rp | ky − xk < r},
= the set of limit points of B.
Int(B) = {x ∈ Rp | ∃ r > 0 such that Dr(x) ⊂ B} = the set of interior points of B.
∂B = {x ∈ Rp | ∀ r > 0 Dr(x) ∩ B 6= ∅ and Dr(x) ∩ Bc= Dr(x) ∩ Rp\ B 6= ∅},
= the set of boundary points of B.
Proof Since ¯A is compact, φ( ¯A) ⊂ Rp is compact. Thus φ( ¯A) a closed set containing φ(A) which implies that
φ(A) ⊆ φ( ¯A).
By the remark, we have
∂φ(A) ∪ Int(φ(A)) = φ(A) ⊆ φ( ¯A) = φ(Int(A) ∪ ∂A).
Since φ ∈ C1(Ω) and Jφ(x) 6= 0 for all x ∈ Int(A), the inverse function theorem implies that φ(Int(A)) ⊆ Int(φ(A)).
Hence, we infer that ∂φ(A) ⊆ φ(∂A) and that φ(A) has content since φ(A) ⊆ φ( ¯A) is bounded and ∂φ(A) has content zero.
Remark. If φ(x) ∈ Int(φ(A)), there is an open neighborhood W of φ(x) such that φ(x) ∈ W ⊂ φ(A).
Since φ : Ω → Rp ∈ C1(Ω), the set U = φ−1(W ) ∩ A is an open neighborhood of x. This implies that x ∈ Int A and Int(φ(A)) ⊂ φ(IntA). Hence, if Ω, φ : Ω → Rp, A and Jφ(x) are as in the Theorem, then we have
φ(Int(A)) = Int(φ(A)).
Corollary Let Ω ⊆ Rp be open and let
φ : Ω → Rp ∈ C1(Ω) and be injective on Ω. If A has content, ¯A ⊂ Ω, and
Jφ(x) 6= 0 for all x ∈ Int(A).
Then
∂φ(A) = φ(∂A).
Proof It suffices to show that φ(∂A) ⊆ ∂φ(A), since the reverse inclusion was established in the proof of the theorem. Let x ∈ ∂A, so that there exists a sequence {xn} in A and a sequence {yn} in Ω \ A, both of which converge to x. Since φ is continuous, then
φ(xn) → φ(x) and φ(yn) → φ(x).
Since φ is injective on Ω, then φ(yn) /∈ φ(A) and hence φ(x) ∈ ∂φ(A). Therefore φ(∂A) ⊆ ∂φ(A).
Theorem Let γ :D(Rp) → R be a function with the following properties:
(a) γ(A) ≥ 0 for all A ∈ D(Rp);
(b) if A, B ∈D(Rp) and A ∩ B = ∅, then γ(A ∪ B) = γ(A) + γ(B);
(c) if A ∈D(Rp) and x ∈ Rp, then γ(A) = γ(x + A);
(d) γ(K0) = 1 = c(K0), where K0 = [0, 1) × [0, 1) × · · · × [0, 1) ⊂ Rp. Then we have γ(A) = c(A) for all A ∈D(Rp).
Outline of the Proof If n ∈ N, let Kn be the ”half-open” cube K0 = [0, 2−n) × [0, 2−n) × · · · × [0, 2−n) ⊂ Rp. Given > 0, since A ∈D(Rp),
there is a closed cube I with side length 2M for some M ∈ N and A ⊆ Int I,
there is an n ∈ N and a partition {Ii}2i=1(M +n)p of I into small cubes of side length 2−n such that
r
[
i=1
(Ii) ⊆ IntA and c(A) − 2 ≤
r
X
i=1
c(Ii),
A ⊆¯
s
[
i=1
(Ii) and
s
X
i=1
c(Ii) ≤ c(A) +
2 for some r ≤ s.
Now each of these sets Ii differe from a translate xi+ Knby a set of content zero. Hence we have
c(A) − ≤ c
r
[
i=1
(xi+ Kn) ≤ c(A) ≤ c
s
[
i=1
(xi+ Kn) ≤ c(A) + .
Since interiors of Ii’s are disjoint, c
r
[
i=1
(xi+ Kn) =
r
X
i=1
c(xi+ Kn) =
r
X
i=1
γ(xi+ Kn) = γ
r
[
i=1
(xi+ Kn).
By (b), γ is monotone in the sense that if A ⊂ B then γ(A) ≤ γ(B). Thus we have
c(A) − ≤ γ
r
[
i=1
(xi+ Kn) ≤ γ(A) ≤ γ
s
[
i=1
(xi+ Kn) ≤ c(A) + .
Corollary Let µ : D(Rp) → R be a function satisfying properties (a), (b) and (c). Then there exists a constant m ≥ 0 such that
µ(A) = m c(A) for all A ∈ D(Rp).
Outline of the Proof If µ(K0) = 0, then µ of any bounded set is 0, whence it follows that µ(A) = 0 for all A ∈D(Rp), so we can take m = 0.
If µ(K0) 6= 0, let
γ(A) = 1
µ(K0)µ(A) for all A ∈ D(Rp).
Theorem. Let
L (Rp) = {B : Rp → Rp | B = (bij) is an p × p matrix over R}
= be the space of linear mappings on Rp, L ∈L (Rp) and let A ∈D(Rp), i.e. A has content Then
c(L(A)) = | det L| c(A).
Outline of the Proof Let µ :D(Rp) → R be defined by
µ(A) = c(L(A)) for all A ∈D(Rp).
To prove the Theorem, it suffices to show that
µ(K0) = | det L|.
If det L = 0, then L(Rp) = Rr for some r < p. It follows that c(L(A)) = 0 for all A ∈D(Rp).
If det L 6= 0 then L :D(Rp) →D(Rp) is 1 − 1, onto and the map µ = c ◦ L :D(Rp) → R satisfies the properties:
(a) if A ∈D(Rp), then µ(A) = c(L(A)) ≥ 0.
(b) suppose A, B ∈D(Rp) and A ∩ B = ∅, then
µ(A ∪ B) = c(L(A ∪ B)) = c(L(A) ∪ L(B)) = c(L(A)) + c(L(B)) = µ(A) + µ(B).
(c) if x ∈ Rp and A ∈D(Rp), then µ(x + A) = c(L(x + A)) = c(L(A)) = µ(A).
By the Corollary, there exists a constant mL≥ 0 such that
µ(A) = mLc(A) for all A ∈D(Rp).
Note that mL= µ(K0) = c(L(K0)) ≥ 0 and if L, M ∈L (Rp) are nonsingular maps, then mL◦M c(A) = c(L ◦ M (A)) = mLc(M (A)) = mLmMc(A) ∀ A ∈D(Rp) =⇒ mL◦M = mLmM. Also observe that every nonsingular map L ∈ L (Rp) is the composition of elementary row operations, i.e. L = Li1 ◦ Li2 ◦ · · · ◦ Lik =⇒ det L = det Li1 det Li2· · · det Lik, where each Lij, 1 ≤ j ≤ k, equals to one of the following linear maps :
(a) L1(x1, . . . , xp) = (αx1, x2, . . . , xp) for some α 6= 0;
(b) L2(x1, . . . , xi, xi+1, . . . , xp) = (x1, . . . , xi+1, xi, . . . , xp);
(c) L3(x1, . . . , xp) = (x1+ x2, x2, . . . , xp).
It is easy to see that if K0 = [0, 1) × · · · × [0, 1) in Rp,
mL1 = c(L1(K0)) = |α| = | det L1|, mL2 = c(L2(K0)) = 1 = | det L2|, mL3 = c(L3(K0)) = 1 = | det L3|.
To evaluate c(L3(K0)), we may assume without loss of generality that K0 = [0, 1) × [0, 1) and note that L3(K0) = {(x1+ x2, x2) | 0 ≤ x1, x2 < 1} is a right triangle of area equals to 1.
Lemma Let K ⊂ Rp be a closed cell with center 0. Let Ω be an open set containing K and let ψ : Ω → Rp ∈ C1(Ω) and be injective.
Suppose further that Jψ(x) 6= 0 for x ∈ K and that
kψ(x) − xk ≤ αkxk for x ∈ K =⇒ ψ(0) = 0, where α satisfies 0 < α < 1/√
p. Then (1 − α√
p)p ≤ c(ψ(K))
c(K) ≤ (1 + α√ p)p.
Proof If the side length of K is 2r and if x ∈ ∂K, then we have r ≤ kxk ≤ r√
p =⇒ kψ(x) − xk ≤ αkxk ≤ αr√
p x ∈ ∂K,
i.e. ψ(x) is within distance αr√
p of x ∈ ∂K. Let Ci = − (1 − α√
p)r, (1 − α√
p)r × · · · × − (1 − α√
p)r, (1 − α√ p)r
= an inner closed cell with center 0 and side length 2(1 − α√ p)r, Co = − (1 + α√
p)r, (1 + α√
p)r × · · · × − (1 + α√
p)r, (1 + α√ p)r
= an outer closed cell with center 0 and side length 2(1 + α√ p)r, Then we have
Ci ⊆ ψ(K) ⊆ Co =⇒ (1 − α√
p)p ≤ c(ψ(K))
c(K) ≤ (1 + α√ p)p.
The Jacobian Theorem Let Ω ⊆ Rp be open. Suppose that
φ : Ω → Rp belongs to class C1(Ω) is injective on Ω, Jφ(x) 6= 0 for x ∈ Ω,
A has content and ¯A ⊂ Ω.
If > 0 is given, then there exists γ > 0 such that if K is a closed cell with center x ∈ A and side length less than 2γ, then
|Jφ(x)|(1 − )p ≤ c(φ(K))
c(K) ≤ |Jφ(x)|(1 + )p.
Proof For each x ∈ Ω, let Lx = (dφ(x))−1, since
1 = det(Lx◦ dφ(x)) = (det Lx) (det dφ(x)), it follows that
det Lx = 1 Jφ(x).
Let Ω1 be a bounded open subset of Ω such that ¯A ⊂ Ω1 ⊂ ¯Ω1 ⊆ Ω and dist(A, ∂Ω1) = 2δ > 0.
Since φ ∈ C1(Ω), the entries in the standard matrix for Lx are continuous, and there exists a constant M > 0 such that
kLxk ≤ M for all x ∈ Ω1.
Let , with 0 < < 1, be given. Since dφ is uniformly continuous on Ω1, there exists β with 0 < β < δ such that
if x1, x2 ∈ Ω1 and kx1− x2k ≤ β, then kdφ(x1) − dφ(x2)k ≤ /M√ p.
Now let x ∈ A be given; hence if kzk ≤ β, then x, x + z ∈ Ω1. Hence, kφ(x + z) − φ(x) − dφ(x)(z)k ≤ kzk sup
0≤t≤1
kdφ(x + tz) − dφ(x)k ≤ M√
pkzk.
Let x ∈ A and define ψ(z) for kzk ≤ β by
(∗) ψ(z) = Lx[φ(x + z) − φ(x)].
Hence, we have
kψ(z) − zk = kLx[φ(x + z) − φ(x) − dφ(x)(z)]k
≤ M kzk sup
0≤t≤1
kdφ(x + tz) − dφ(x)k
≤
√pkzk for kzk ≤ β.
If K1 is any closed cell with center 0 and contained in the open ball with radius β, then (∗∗) (1 − )p ≤ c(ψ(K1))
c(K1) ≤ (1 + )p. It follows that if K = x + K1 then K is a closed cell with center x,
c(K1) = c(K),
c(ψ(K1)) = | det Lx|c(φ(x + K1) − φ(x)) by (∗)
= 1
Jφ(x)c(φ(K) − φ(x))
= 1
Jφ(x)c(φ(K)) Setting γ = β
√p and using (∗∗), we have if K is a closed cell centered at x ∈ A with side length less than 2γ then
(1 − )p ≤ c(φ(K))
Jφ(x) c(K) ≤ (1 + )p =⇒ |Jφ(x)|(1 − )p ≤ c(φ(K))
c(K) ≤ |Jφ(x)|(1 + )p. Change of Variables Theorem Let Ω ⊆ Rp be open. Suppose that
φ : Ω → Rp belongs to class C1(Ω) is injective on Ω, Jφ(x) 6= 0 for x ∈ Ω,
A has content and ¯A ⊂ Ω and
f : φ(A) → R is bounded and continuous.
Then Z
φ(A)
f = Z
A
(f ◦ φ)|Jφ|.
Example Find RR
S x2+ y2dA if S be the region in the first quadrant bounded by the curves xy = 1, xy = 3, x2− y2 = 1, and x2− y2 = 4. Setting u = x2− y2, v = xy, we have
Z Z
S
x2+ y2dA = Z v=3
v=1
Z u=4 u=1
1
2 dudv = 3.
Example
Z ∞ 0
e−x2dx
2
= Z ∞
0
Z ∞ 0
e−x2−y2dxdy = Z π/2
0
Z ∞ 0
re−r2drdθ = π 4.
Summary
(a) Let A ⊂ Rn. Define what it means to say that A has content (or A is Jordan measurable).
(b) Let A ⊂ Rn. Define the content c(A) of A when A has content (or A is Jordan measurable).
(c) Let A ⊂ Rn. Define what it means to say that A has content (or measure) c(A) zero.
(d) Let A be a bounded subset of Rn, and let f be a bounded function defined on A to R.
Define what it means to say that f is integrable on A. Give a class of A and a class of f from which R
Af exists.
(e) Let A be a bounded subset of Rn and let f be an integrable function defined on A to R.
Discuss the continuity of f on A.
(f) Let A be a bounded subset of Rn and let f be a continuous function defined on A to R.
Discuss the integrability of f on A?
(g) Let A be a bounded subset of Rn, let f, g be integrable functions defined on A to R, and let a, b ∈ R. Discuss the integrability of af + bg and f g on A.
(h) Let A be a bounded subset of Rn and let {fn} be a sequence of integrable function defined on A to R. Assume that f (x) = lim fn(x) exists for each x ∈ A. Discuss the integrability of f on A, and conditions on which the equality limR
Afn =R
Alim fn holds.
(i) Let Ω ⊆ Rp be open and let φ : Ω → Rp belong to class C1(Ω). Suppose that A has content (or A is measurable), ¯A ⊂ Ω. Discuss the measurability of φ(A).
(j) Let Ω ⊆ Rp be open and let φ : Ω → Rp belong to class C1(Ω). Suppose that A has content zero and if ¯A ⊂ Ω, discuss the measurability of φ(A).
(k) Let L ∈ L (Rp) = {B : Rp → Rp | B = (bij) is an p × p matrix over R} = the space of linear mappings on Rp, and let A ∈D(Rp). Find c(L(A)).
(l) Let Ω ⊆ Rp be open and let φ : Ω → Rp belong to class C1(Ω). Then dφ(x) : Rp → Rp is a linear mapping in L (Rp). If Jφ(x) 6= 0 for all x ∈ Ω and suppose that A has content (or A is measurable), ¯A ⊂ Ω. Discuss the geometric meaning of dφ(x)(v) for each v ∈ Rp, and the influence of dφ(x) on the volume element of A at x.
