Multivariable Integration (Rudin, Bartle)
Defintion: Let ai < bi for 1 ≤ i ≤ n. A n-cell in Rn is a set of the
form In = [a1 , b1 ] × · · · × [an , bn ]. Given σ ∈ Sn and a continuous function
f : In → R, we define
Z bσ(n)
Z bσ(1)
Lσ (f ) =
···
f (x1 , . . . , xn ) dxσ(1) · · · dxσ(n) .
aσ(n)
aσ(1)
When f (x1 , . . . , xn ) ∈ R[x1 , . . . , xn ] then Lσ (f ) is independent of σ. Now
choose a sequence of polynomials (fk ) such that fk → f uniformly on In ,
which is possible by the multivariable generalization of the Stone-Weierstrass
Theorem. Write k = ||fk − f ||. Then −k ≤ fk (x1 , . . . , xn ) − f (x1 , . . . , xn ) ≤
k , therefore
Z bσ(1)
−k (bσ(1) −aσ(1) ) ≤
fk (x1 , . . . , xn ) − f (x1 , . . . , xn ) dxσ(1) < k (bσ(1) −aσ(1) ).
aσ(1)
Iterating this process we eventually obtain
−k V ≤ Lσ (fk − f ) ≤ k V,
where V = (a1 − b1 ) · · · (an − bn ). Therefore
|L(σ, fk ) − L(σ, f )| ≤ ||fk − f ||V → 0.
This implies
L(σ, f ) = lim L(σ, fk ) = lim L(e, fk ) = L(e, f )
k→∞
k→∞
where e = (1)(2) · · · (n).
Definition: Let f : Rn → R be an integrable function with compact support, i.e. {x ∈ Rn : f (x) 6= 0} ⊆ [a1 , b1 ] × · · · [an , bn ]. Then
Z
Z bn
Z b1
f=
···
f (x1 , . . . , xn ) dx1 · · · dxn .
Rn
an
a1
In algebraic number theory we wish to compute the volume of the set T =
{x1 v1 + · · · + xn vn : 0 ≤ xi < 1} where {v1 , . . . , vn } are linearly independent
in Rn . T is bounded in Rn :
|x1 v1 + · · · + xn vn | ≤ |x1 ||v1 | + · · · + |xn ||vn | ≤ |v1 | + · · · + |vn |.
1
Let χT : Rn → R be defined by
(
1 x∈T
χT (x) =
0 x∈
6 T.
Then χT has compact support. We will define
Z
χT .
vol(T ) =
Rn
In order to compute this integral we will review the basics of multivariable
integation as they appear in The Elements of Real Analysis, Bartle.
A 1-cell is defined to be any interval I of the form (a, b) or (a, b] or [a, b) or
[a, b] where a ≤ b are real numbers. By definition, c(I) = b − a. If I1 , . . . , In
are
Qn 1-cells, then I1 × · · · × In is defined to pbe an n-cell and c(I1 × · · · × In ) =
i=1 C(Ii ). Given an arbitrary set Z ⊆ R , we say that c(Z) = 0 if and only
if for each > 0 there exist a finite number of p-cells J1 , . . . , Jn such that
Z ⊆ J1 ∪ · · · Jn and c(J1 ) + · · · + c(Jn ) ≤ .
Example: Let f : [0, 1] → R be continuous. Then
Z = {(x, f (x)) : 0 ≤ x ≤ 1}
satisfies c(Z) = 0. Proof: Let > 0 be given. Choose n sufficiently large
so that |x − y| ≤ n1 =⇒ |f (x) − f (y)| ≤ . For i = 0, 1, . . . , n − 1 let
Ji = [ ni , i+1
] and fi = inf x∈Jj f (x) and Ki = [fi , fi + ]. Then
n
Z⊆
n−1
[
Ji × Ki
i=0
and
n−1
X
c(Ji × Ki ) = .
i=0
Partitions and Refinements: Let I ⊆ Rp be a closed p-cell. A partition
of I is finite set partition P of I where each J ∈ P is a p-cell. A refinement of
P is any finite set partition Q of I such that each K ∈ Q there exists J ∈ P
such that K ⊆ J. When Q is a refinement of P we will write Q ≤ P .
2
Example: Let I = [0, 1] × [0, 1]. A partition is
1
1
1
1
1 1
1
1
[0, ] × [0, ], [0, ] × ( , 1], [ , 1] × [0, ], [ , 1] × ( , 1].
2
2
2
2
2
2 2
2
A refinement is
1
1 1 1
1
1
1
1
1 1
1
[0, ] × [0, ], ( , ] × [0, ], [0, ] × ( , 1], [ , 1] × [0, ], [ , 1] × ( , 1].
4
2 4 2
2
2
2
2
2 2
2
Lemma: Let P and Q be p-cells. If P ∩ Q 6= ∅ then P ∩ Q is a p-cell.
Proof: By induction on p. If [a1 , b1 ] ∩ [a2 , b2 ] 6= ∅ then
[a1 , b1 ] ∩ [a2 , b2 ] = J(a1 , b1 , a2 , b2 ) = [max(a1 , a2 ), min(b1 , b2 )].
For any other intersection of 1-cells defined by the same four numbers a1 , b1 , a2 , b2 ,
the intersection is the result of deleting a subset of the endpoints of J(a1 , b1 , a2 , b2 ).
Now assume every non-empty intersection of two p-cells is a p-cell. Let I1
and I2 be 1-cells, let J1 and J2 be p-cells, and assume (I1 ×J1 )∩(I2 ×J2 ) 6= ∅.
The result follows from the identity
(I1 × J1 ) ∩ (I2 × J2 ) = (I1 ∩ I2 ) × (J1 ∩ J2 ).
Lemma: Let P and Q be p-cells. If P \Q 6= ∅ then P \Q is a finite disjoint
union of non-empty p-cells.
Proof: By induction on p. If [a1 , b1 ]\[a2 , b2 ] 6= ∅, choose M > 0 so that
|a1 |, |b1 |, |a2 |, |b2 | < M . Then
[a1 , b1 ]\[a2 , b2 ] = ([a1 , b2 ] ∩ [−M, a2 )) ∪ ([a1 , b1 ] ∩ (b2 , M ])
is a disjoint union of 1-cells, and we can discard an empty intersection if
necessary. The general result follows from
[ [ (I1 × J1 )\(I2 × J2 ) = (I1 \J1 ) × (I2 \J2 ) =
Pi ×
Qj =
[
Pi × Q j .
3
Lemma: Let P1 , . . . , Pn be p-cells. Then P1 ∪ · · · ∪ Pn can be expressed as
a finite disjoint union of non-empty p-cells.
Proof: By induction on n ≥ 2. For n = 2 we have
P ∪ Q = (P \Q) ∪ (P ∩ Q) ∪ (Q\P ),
and by the lemmas each non-empty expression is a finite disjoint union of
p-cells. For n ≥ 3, write P2 ∪ · · · ∪ Pn = J1 ∪ · · · ∪ Jr where the latter union
is disjoint. Then
P1 ∪ · · · ∪ Pn = (P1 ∪ J1 ) ∪ · · · ∪ (P1 ∪ Jr ).
The result follows from the n = 2 case.
Definition: Given partitions P and Q of a closed p-cell I, we set
P ∧ Q = {J ∩ K : J ∈ P and K ∈ Q}\{∅}.
This is a refinement of both P and Q.
Lemma:
P If I is a p-cell and if P is a finite set partition of I into p-cells then
c(I) = J∈P c(J).
Proof: For each i ≤ p define
φi ([a1 , b1 ] × · · · × [ap , bp ]) = {ai , bi }.
Given a finite set partition P of p-cell I, set
[
φi (J)
Bi =
J∈P
and write
Bi = {bi1 , . . . , biki }
where
bi1 < · · · < biki .
We will write ∆bij = bij − bi(j−1) . Set Ii = {[bi1,i2 ], . . . , [bi,ki −1 , biki ]}. . Then
c(I) =
p
Y
i=1
(biki − bi1 ) =
p
ki
Y
X
∆bij =
X
j1 ≤k1 ,...,jp ≤kp
i=1 j=2
4
∆b1j1 · · · ∆bpjp =
X
c(R1 × R2 × · · · × Rp ).
R1 ∈I1 ,...,Rp ∈Ip
Similarly, each J ∈ P satisifes
X
c(J) =
c(R1 × R2 × · · · × Rp ).
R1 ∈I1 ,...,Rp ∈Ii
R1 ×···×Rp ⊆J
Therefore
c(P ) =
X
X
J∈P
R1 ∈I1 ,...,Rp ∈Ii
R1 ×···×Rp ⊆J
c(R1 × R2 × · · · × Rp ) =
X
c(J).
J∈P
Riemann Sums: Given a closed p-cell I and a function f : I → R and a
partition P of I and a selection function x : P → I such that x(J) ∈ J for
each J ∈ P , we set
X
f (x(J))c(J)
S(P, x, f ) =
J∈P
Then we say that
R
I
f = L if and only if
∀ > 0 : ∃P ` I : Q ≤ P =⇒ |S(Q, x, f ) − L| < for all valid selection functions x : Q → I.
Cauchy Criterion
for Integrability: Let I be a p-cell and let f : I → R
R
be given. Then I f exists if and only if for all
∀ > 0 : ∃P ` I : Q1 , Q2 ≤ P =⇒ |S(Q1 , x1 , f ) − S(Q2 , x2 , f )| < for all valid selection functions xi : Qi → I.
R
Proof: If I f exists then clearly the Cauchy Criterion is met. Conversely,
suppose this criterion is met. For each n ∈ N choose a partition Pn ` I
relative to = n1 . Then for each n set Jn = P1 ∧ · · · ∧ Pn and choose
a selection function xn : Jn → I. The sequence (S(Jn , xn , f )) is Cauchy:
When m < n, Jm and Jn are refinements of Pm , hence
|S(Jm , xm , f ) − S(Jn , xn , f )| <
5
1
.
m
Set L = limn→∞ S(Jn , xn , f ). Let > 0 be given. Choose n so that n1 < 2
and |S(Jn , xn , f ) − L| < 2 . For any refinement Q of Jn and any selection
function x : Q → I we have
|S(Q, x, f ) − L| ≤ |S(Q, x, f ) − S(Jn , xn , f )| + |S(Jn , xn , f ) − L| <
1 + < .
n 2
Integrals over a bounded set: Suppose f : Rp → R has bounded support
A. Let I be any p-cell containing A. We set
Z
Z
f = f.
A
I
This definition is well-defined: Suppose that I1 and I2 are two p-cells containing A. Let R> 0 be given. Choose P1 ` I1 so that Q ≤ P1 implies |S(Q, x, f ) − I1 f | < 2 |. Choose P2 ` I2 so that Q ≤ P2 implies
R
|S(Q, x, f ) − I2 f | < |. Let X(I1 \I2 ) be the finite set of non-empty p-cells
whose union is I1 \I2 . Let X(I2 \I1 ) be the finite set of non-empty p-cells
whose union is I2 \I1 . A refinement of P1 is
Q1 = {J ∩ K : J ∈ P1 , K ∈ P2 , J ∩ K 6= ∅} ∪ {J ∩ K : J ∈ P1 , K ∈ X(I1 \I2 )}.
A refinement of P1 is
Q2 = {J ∩ K : J ∈ P1 , K ∈ P2 , J ∩ K 6= ∅} ∪ {J ∩ K : J ∈ P2 , K ∈ X(I2 \I1 )}.
Choose selection functions x1 : Q1 → I1 and x2 : Q2 → I2 that coincide on
Q1 ∩ Q2 . Any x in I1 \I2 or I2 \I1 falls outside of A and satisfies f (x) = 0.
Hence
X
S(Q1 , x1 , f ) = S(Q2 , x2 , f ) =
x(R)c(R).
R∈Q1 ∩Q2
This implies
Z
Z Z Z f−
f ≤ S(Q1 , x1 , f ) −
f + S(Q2 , x2 , f ) −
f < .
I1
I2
Since is arbitary,
I1
R
I1
f=
R
I2
f.
Lemma: Let I be a p-cell. Then
R
I
χI = c(I).
6
I2
Proof: Every S(P, x, χI ) is equal to c(I).
Properties of the Integral: Let A be a bounded set and let f and g be
integrable on A. Assume I is a p-cell containing A.
R
R
(a) A kf = k A f for all k ∈ R.
R
R
R
(b) A (f + g) = A f + A g.
R
R
(c) If f ≤ g on A then A f ≤ A g.
(d) If f and g Rare bounded
on A and E = {x ∈ A : f (x) 6= g(x)} and
R
c(E) = 0, then A f = A g.
Proof: (a) is true if k = 0. When k 6= 0 there is a one-to-one correspondence
of S(P, x, kf ) with kS(P, x, f ).
(b): Let >R 0 be given and choose PR ` A such that Q ≤ P implies
|S(Q, x, f ) − I f | < 2 and |S(Q, x, g) − I f | < 2 . Given S(Q, x, f + g) =
S(Q, x, f ) + S(Q, x, g) and the triangle inequality, this implies
Z
Z S(Q, x, f + g) −
f + g ≤ .
I
Therefore
R
I
(f + g) =
R
I
f+
R
I
I
g.
R
R
(c): It suffices to prove A h ≥ 0 when h ≥ 0 on A. If A h < 0, then there
exists
a partition
P ` I and a selection function x such that |S(P, x, f ) −
R
R
RA h| < | A h|, which implies S(P, x, f ) < 0, a contradiction. Therefore
h ≥ 0.
A
R
(d): ItRsuffices to prove A h = 0 when h(x) = 0 for all x 6∈ E, in other
words E h = 0. Let > 0 be given. Choose p-cells J1 , . . . , Jn such that
E ⊆ J1 ∪ · · · ∪ Jn and c(J1 ) + · · · + c(Jn ) < . We can assume without loss
of generality that J1 ∪ · · · ∪ Jn ⊆ I. We have
h ≤ ||h||(χJ1 + · · · + χJn ),
Z
Z
Z
Z
h ≤ ||h|| (χJ1 + · · · + χJn ) = ||h||
χJ1 + · · · + χJn =
I
I
I
||h||(c(J1 ) + · · · + c(Jn )) < ||h||.
R
Since > 0 is arbitrary, I h = 0.
7
I
Theorem: Let A be a bounded set and let f : A → R be bounded and
integrable. If g : A → R is bounded and E = {x
R ∈ AR: f (x) 6= g(x)} and
c(E) = 0, then g is also integrable on A (hence A f = A g by the previous
theorem).
Proof: Set h = f −g. The proof of part (d) in the previous theorem depends
only on the fact that h is bounded. Hence h is integrable, hence g = f − h
is integrable.
Existence of the Integral: Let I be a closed p-cell and let f :→ RR be
bounded. Set C = {x ∈ I : f is continuous at x}. If c(I\C) = 0 then I f
exists.
Proof: We will use the Cauchy Criterion. Let > 0 be given. Write
E = I\C. Choose open p-cells J1 , . . . , Jn such that E ⊆ J1 ∪ · · · ∪ Jn and
c(J1 ) + · · · + c(Jn ) < . Then E ⊆ (J1 ∩ I) ∪ · · · (Jn ∩ I) and we have
Z
Z
c(Ji ∩ I) =
χJi ∩I ≤
χJi = c(Ji ),
Rp
Rp
therefore
c(J1 ∩ I) + · · · + c(Jn ∩ I) < .
We can write (J1 ∩ I) ∪ · · · ∪ (Jn ∩ I) = K1 ∪ · · · ∪ Km where the latter is a
p-cell partition, and we have
Z
Z
χK1 + · · · + χKm ≤
χJ1 ∩I + · · · + χJn ∩I =
c(K1 ) + · · · + c(Km ) =
Rp
Rp
c(J1 ∩ I) + · · · + c(Jn ∩ I) < .
Since the non-empty set difference of two p-cells is a disjoint union of pcells, there is a p-cell partition P of I that satisfies {K1 , . . . , Kn } ⊆ P . By
subdividing each of the sets in P \{K1 , . . . , Km } suitably we can ensure that
|f (x) − f (y)| < for all x, y ∈ J and J ∈ P \{K1 , . . . , Km } using uniform
continuity of f on the closed and bounded set E\(K1 ∪ · · · ∪ Kn ).
Now let P1 and P2 be refinements of P with selection functions x1 and
x2 . Let Q = P1 ∧ P2 . Then x1 induces a selection function y1 on P3 via
y1 (J) = x1 (K) where J ⊆ K ∈ P1 and x2 induces a selection function y2 on
Q via y2 (J) = x2 (K) where J ⊆ K ∈ P2 . We have
|S(P1 , x1 , f ) − S(P2 , x2 , f )| = |S(Q, y1 , f ) − S(Q, y2 , f )| ≤
8
X
X
|f (y1 (J)) − f (y2 (J)|c(J)+
|f (y1 (J)) − f (y2 (J)|c(J) ≤
J∈Q
J⊆I\(K1 ∪···∪Km )
J∈Q
J⊆K1 ∪···∪Km
2||f |||
X
X
c(J) + J∈Q
J⊆K1 ∪···∪Km
c(J) <
J∈Q
J⊆I\(K1 ∪···∪Km )
2||f || + c(I).
Since
> 0 is arbitrary, the Cauchy Criterion for integrability is satisfied and
R
f exists.
I
A set A ⊆ I where I is a p-cell is said to have content when c(b(A)) = 0,
in which case χA : I → R isR continuous on I\E where E ⊆ b(A). Since
c(b(A) = 0, c(E) = 0. Hence I χA exists.
Lemma:
Let A ⊆ I be given where I is a p-cell. Then c(A) = 0 if and only
R
if I χA = 0.
Proof: Suppose c(A) = 0. Let > 0 be given. Then there exists a partition of I in which the sets covering A have content
sum < . This yields
R
S(P,
x, χA ) < . Since > 0 is Rarbitrary and I χA ≥ 0, we must have
R
χ = 0. Conversely, suppose I χA = 0. Let > 0 be given. Then
I A
there exists a partition P ` I such that for Q ≤ P and selection functions
x : Q → I, |S(Q, x, χA )| < . If we choose x(J) to be an element of A
whenever J ∩ A 6= ∅, then S(Q, x, χA ) is the total content of a disjoint union
of p-cells covering A. Hence c(A) = 0.
Lemma: Let I be a p-cell. Then c(b(I)) = 0.
Proof: Write I = [a1 , b1 ] × · · · × [ap , bp ]. Then
b(I) =
p [
[
{(x1 , . . . , xp ) ∈ I : xi = ai } {(x1 , . . . , xp ) ∈ I : xi = bi } .
i=1
Hence b(I) is a union of 2p p-cells, each with content zero.
R
Definition: Let A ⊆ I satisfy c(b(A)) = 0. Then we set c(A) = I χA , and
this definition is consistent with the original definition of a set with content
zero and with the content of a p-cell.
Theorem:
(a) Let A, B ⊆ I have content. Then A ∪ B and A ∩ B have content and
c(A ∪ B) = c(A) + c(B) − c(A ∩ B).
9
(b) c(x + A) = c(A).
Proof: (a): Since A and B are subsets of a union of p-cells with content
sum arbitrarily small, so are A ∪ B and A ∩ B. Hence these two sets have
content. The result follws from χA∪B = χA + χB − χA∩B .
(b): The boundary of a translation is the translation of the boundary; translations of p-cells are p-cells; the content of a p-cell is translation invariant;
etc.
Theorem: Let γ assign a real number to each bounded subset of Rp with
content, and assume that γ is non-negative, additive, translation invariant,
and satisfies γ([0, 1)p ) = 1. Then γ(A) = c(A) for all bounded A with
content.
R
Proof: We have c(A) = I χA . Now let > 0 be given. Then there exists
P ` I such that every refinement Q ≤ P and selection function x : P → I
satisfies |S(Q, x, χA ) − c(A)| < . Choosing Q = P and choosing x(J) =
aJ ∈ A if J ⊆ A and x(J) = bJ ∈ Ac if J 6⊆ A, we have
X
S(Q, x, χA ) =
c(J).
J∈Q
J⊆A
Choosing y(J) = a0J ∈ A if J ∩ A 6= ∅ and y(J) = b0J ∈ Ac if J ∩ A = ∅, we
have
X
S(Q, y, χA ) =
c(J).
J∈Q
J∩A6=∅
Therefore
c(A) − <
X
J∈Q
J⊆A
X
c(J) ≤
c(J) < c(A) + .
J∈Q
J∩A6=∅
If we can show that γ(J) = c(J) for each p-cell J, then by additivity and
positivity of γ we have




 [ 
[ 
c(A) − < γ 
J  ≤ γ(A) ≤ γ 
J  < c(A) + .
J∈Q
J⊆A
J∈Q
J∩A6=∅
Since > 0 is arbitrary, this implies c(A) = γ(A).
10
Now we show γ(I) = c(I) for a p-cell I. The first observation to make is
that [0, 1)p is a disjoint union of translates of [0, 21n )p , and since γ([0, 1)p ) =
c([0, 1)p ) = 1, we have γ([0, 21n )p ) = c([0, 21n )p ) for all n ≥ 1. By translation
invariance, we can assume without loss of generality that
I = [0, a1 ) × · · · × [0, ap ).
Let n ∈ N be given. Set
In = [0,
j1
jp
) × · · · × [0, n )
n
2
2
and
jp + 1
j1 + 1
) × · · · × [0,
)
n
2
2n
where each ji ∈ Z and In ⊆ I ⊆ Jn . Since In and Kn are disjoint unions of
translates of [0, 21n ), γ(In ) = c(In ) and γ(Jn ) = c(Jn ). So we have
Jn = [0,
c(In ) ≤ γ(I) ≤ c(Jn ).
Since c(In ) → c(I) and c(Jn ) → c(I) as n → ∞, γ(I) = c(I).
Note that if we drop the requirement that γ([0, 1)p ) = 1, then the arguments
above imply that when γ([0, 1]p ) = m > 0, γ(A) = mc(A) for all bounded A
with content.
Theorem: Let A ⊆ Rp be bounded
with content and let f : A → R be
R
bounded and continuous. Then Rp f exists.
Proof: Let I be a p-cell containing A and define F : I → R via F (x) = f (x)
when x ∈ A, F (x) = 0 when x ∈ I\A. If F is not continuous at x then x
is not an interior point of A and is not an interior point of I\A, so every
neighborhood of x contains a point in both sets and belongs to b(A). By
hypothesis c(b(A)) = 0, so the set of points of discontinuity
R
R of F have content
zero. Hence F is integrable on I, and we have Rp f = I F .
Theorem: Let A and B be bounded sets with content and with c(A ∩ b) = 0
and let f : A ∪ B → R be bounded. If fR is integrable
onR A and integrable
R
on B then it is integrable on A ∪ B and A∪B f = A f + B f .
Proof: Since f is integrable on both A and B, there is a p-cell I containing
A ∪ B such that fA : I → R and fB : I → R are integrable, where
(
)
(
)
f (x) x ∈ A
f (x) x ∈ B
fA (x) =
,
fB (x) =
.
0
x 6∈ A
0
x 6∈ B
11
The function fA + fB is bounded on I and agrees with f outside A ∩ B, a
set of content zero, hence fA + fB is integrable and
Z
Z
Z
Z
Z
Z
Z
f = f = fA + fB = fA + fB =
f+
f.
A∩B
I
I
I
I
A
B
Theorem: Let A be bounded with content, and let f : A → R be bounded
and integrable. If m ≤ f (x) ≤ M for all x ∈ A then
Z
mc(A) ≤
f ≤ M c(A).
A
Proof: Let I be a p-cell containing I and extend f to I by setting f (x) = 0
outside A. For any partition P ` I and selection function x such
R that x(J) ∈
A ⇐⇒ J ⊆ A we have S(P, x, f ) ≤ M c(A). This implies A f ≤ M c(A).
For any partition P ` I and selection function y such
R that y(J) ∈ A ⇐⇒
J ∩ A 6= ∅ we have S(P, y, f ) ≥ mc(A). This implies A f ≥ mc(A).
Theorem: Let A be bounded with Rcontent and connected and let f : A → R
be continuous and bounded. Then A f = f (a)c(A) for some a ∈ A.
Proof: We can assume without loss of generality that
R c(A) > 0. Let m =
1
f . Then m ≤ k ≤ M .
inf x∈A f (x) and let M = supx∈A f (x). Write k = c(A)
a
k
−1
If m < k < M , suppose f = ∅. Then A = f (−∞, k) ∪ f −1 (k, ∞),
which
contradicts the assumption that A is connected. Therefore f (a) = k,
R
f = f (a)c(A).
A
Now suppose k = M . If M = f (a), we’re done. If not, then f (a) < M for
all a ∈ A. We will find a compact subset B ⊆ A and a number > 0 such
that c(B) > 0. Setting M 0 = supx∈B f (x) < M , we have
Z
Z
Z
f=
f+
f ≤ M 0 c(B)+M c(A\B) < M c(B)+M c(A\B) = M c(A),
A
B
A\B
which contradicts k = M . Let A ⊆ I and let P ` A be such that
|S(P, x, χA ) − c(A)| < c(A)
for all selection functions x. Then S(P, x, χA ) > 0 for all selection functions
x. If x(J) ∈ A iff J ⊆ A, then S(P, x, χA ) is the content sum of those J ∈ P
12
that are subsets of A. One of these must have non-zero content and must
contain a closed cell B with non-zero content.
Finally, suppose k = m. If m = f (a), we’re done. If not, then f (a) > m
for all a ∈ A. Using the compact subset B ⊆ A from before and setting
m0 = inf x∈B f (x) > m, we have
Z
Z
Z
f=
f+
f ≥ m0 c(B) + mc(A\B) > mc(B) + mc(A\B) = mc(A),
A
B
A\B
which contradicts k = m.
Theorem:
Suppose f : [a, b] → R is continuous and F 0 (x) = f (x) on (a, b).
R
Then [a,b] f = F (b) − F (a).
Proof: Let > 0 be given. Let P ` [a, b] be suchR that Q ≤ P and x : Q →
[a, b] is a selection function implies |S(Q, x, f ) − [a,b] f | < . Given J ∈ P ,
write J = [aJ , bJ ]. If aJ < bJ then F (bJ ) − F (aJ ) = f (cJ )c(J) for some
cJ ∈ J by the Mean Value Theorem. This formula remains valid even for
aJ = bJ . Using the selection function x(J) = cJ we obtain
X
X
F (bJ ) − F (aJ ) = F (b) − F (a).
f (cJ )c(J) =
S(P, x, f ) =
J∈P
J∈P
Therefore
Z
|F (b) − F (a) −
f | < .
[a,b]
Since > 0 is arbitrary,
R
[a,b]
f = F (b) − F (a).
Theorem:
R Suppose f : [a, b] → R is continuous. Define F : [a, b] → R via
F (x) = [a,x] f . Then F is continous and F 0 (x) = f (x) for all x ∈ (a, b).
Proof: For x1 < x2 ∈ [a, b] we have
Z
F (x2 ) − F (x1 ) =
f = f (x12 )(x2 − x1 )
[x1 ,x2 ]
for some x12 ∈ [x1 , x2 ]. This implies continuity of F . Now let x ∈ [a, b] be
given. For sufficiently small h > 0 we have
F (x + h) = F (x) + f (xh )h
13
for some xh ∈ [x, x + h] and
F (x) = F (x − h) + f (x0h )h
for some x0h ∈ [x − h, h], i.e.
F (x − h) = F (x) + f (x0h )(−h).
As h → 0, xh → x and x0h → x, so by continuity f (xh ) → f (x) and f (x0h ) →
f (x). This implies F 0 (x) = f (x).
Theorem: Let f : [a1 , b1 ] × [a2 , b2 ] → R be continuous. Then
Z
Z
Z
Z
Z
f=
f (∗, y) =
f (x, ∗) .
[a1 ,b1 ]×[a2 ,b2 ]
[a2 ,b2 ]
[a1 ,b1 ]
[a1 ,b1 ]
[a2 ,b2 ]
Proof: We will just prove the first equality. The second equality is proved
similarly. Let > 0 be given. Choose P ` [a1 , b1 ] × [a2 , b2 ] such that Q ≤ P
implies
Z
S(Q, x, f ) −
<
f
[a1 ,b1 ]×[a2 ,b2 ]
for any selection function x : Q → [a1 , b1 ] × [a2 , b2 ]. There is a partition Pi
of [ai , bi ] for i = 1, 2 such that
Q = {J1 × J2 : (J1 , J2 ) ∈ P1 × P2 }
is a refinement of P . For y ∈ [a2 , b2 ] and J ∈ P1 and the Mean Value Theorem
for Integrals we have
Z
f (∗, y) = f (xJ , y)c(J)
J
for some xJ ∈ J and
Z
f (xJ , y) = f (xJ , yJK )c(K)
K
for some yJK ∈ K. This yields
Z
XZ
X
f (∗, y) =
f (∗, y) =
f (xJ , y)c(J),
[a1 ,b2 ]
J∈P1
J
J∈P1
14
Z
Z
f (∗, y)
[a2 ,b2 ]
[a2 ,b2 ]
Z
K
K∈P2
Z
f (xJ , y) =
c(J)
K
(J×K)∈Q
X
=
[a1 ,b1 ]
X
X
!
Z
f (xJ , y)c(J)
=
J∈P1
!!
X
f (xJ , y)c(J)
=
J∈P1
X
c(J)c(K)f (xJ , yJK ) = S(Q, σ, f )
(J,K)∈Q
where σ(J × K) = (xJ , yJK ) ∈ J × K. Hence we have
Z
Z
Z
< .
f
(∗,
y)
−
f
[a2 ,b2 ]
[a1 ,b1 ]
[a1 ,b1 ]×[a2 ,b2 ]
Since > 0 is arbitrary, the two integrals are equal.
Remark: Clearly the result can be extended to integrals of continuous functions over p-cells. Moreover the Mean Value Theorem for Integrals over closed
1-cells is sufficient to prove that the integral is independent of the order of
integration.
Lemma: Let Ω ⊆ Rp be an open set, let φ : Ω → Rp be continuously
differentiable, and let I ⊆ Ω be a p-cube. Let M > 0 satisfy ||φ0 (x)|| ≤ M
√
for all x ∈ I. Then φ(I) is contained in a p-cube with content (2M p)p c(I).
Proof: Given x, y ∈ I, set
γ(t) = (φ(y) − φ(x)) • φ((1 − t)x + ty),
0 ≤ t ≤ 1. Then
||φ(y) − φ(x)||2 = γ(1) − γ(0) = γ 0 (t∗ ) =
(φ(y) − φ(x)) • φ0 ((1 − t)x + ty)(y − x) ≤
||φ(y) − φ(x)|| · ||φ0 ((1 − t)x + ty)(y − x)|| ≤
||φ(y) − φ(x)|| · ||φ0 (1 − t)x + ty|| · ||y − x|| ≤
||φ(y) − φ(x)|| · M ||y − x||
for some 0 < t∗ < 1, hence
|φ(y) − φ(x)| ≤ M ||y − x||.
15
In particular, if I has sides of length r and we fix x ∈ I and, then for any
√
√
y ∈ I we have ||y − x|| ≤ pr, hence |φ(y) − φ(x)| ≤ M pr. Therefore φ(y)
√
belongs to the cube J with center φ(x) and sides of length of length 2M pr,
√
√
content (2M pr)p = (2M p)p c(I), and φ(I) ⊆ J.
Theorem: Let A ⊆ Rp be a bounded set with content 0. If Ω is an open
set containing A and φ : Ω → R is continuously differentiable then φ(A) has
content zero.
Proof: By the Lemma it suffices to find M > 0 such that for all > 0, A
can be covered inside Ω by a finite number of p-cubes with content sum < ,
within each of which ||φ0 (x)|| ≤ M .
We will first find δ > 0 such that Bδ (x) ⊆ Ω for all x ∈ A. If this is not the
case, then for each n ∈ N there exists xn ∈ A and yn ∈ Ωc and ||xn −yn || < n1 .
By compactness of An , a subsequence (xnk ) converges to x ∈ A, hence (ynk )
converges to x. Hence x is a limit point of Ωc , hence x ∈ Ωc since Ωc is
closed, which contradicts x ∈ A ⊆ Ω. So δ exists.
S
Next, set Ω1 = x∈A B δ (x). Then A ⊆ Ω1 and Ω1 ⊆ Ω. By compactness of
2
Ω1 we can set M = maxx∈Ω1 ||φ0 (x)||.
Let > 0 be given. We can cover A by a disjoint union of closed p-cells
with total content less than 2 . We can diadically approximate this union
by a disjoint unon of p-cubes with side length < 2√δ p and total content < .
Discarding those that don’t intersect A, any remaining cube J contains a
point aJ ∈ A, and every y ∈ J satisfies ||y−aJ || < 2δ , hence y ∈ B δ (aJ ) ⊆ Ω1 .
2
Since ||φ0 (x)|| ≤ M for all x ∈ J, we have found our desired p-cube cover of
A relative to M and .
Theorem: Let A ⊆ Rk be a bounded set where k < p. Assume A ⊆ Ω,
Ω is open in Rk , and φ : Rk → Rp is continuously differentiable. Then
c(φ(A)) = 0.
Proof: Lift A to A1 = A × {v} for an arbitrary v ∈ Rp−k . Lift Ω to
Ω1 = Ω × Rp−k . Then A1 is bounded and A1 ⊆ Ω1 . Define φ1 : Ω1 → Rp via
φ1 (x1 , . . . , xp ) = φ(x1 , . . . , xk ). Then φ1 is continuously differentiable. Since
A is bounded, it is contained inside a finite k-cell I. Hence A1 ⊆ I × {v}.
Since c(I × {v}) = 0, c(A1 ) = 0, hence φ1 (A1 ) has content zero. Since
φ1 (A1 ) = φ(A), φ(A) has content zero.
16
Theorem: Let Ω ⊆ Rp be open, let φ : Ω → Rp be continuously differentiable, and let A be a bounded set that satisfies A ⊆ Ω. Then b(φ(A)) ⊆
φ(b(A)). Moreover if φ is injective then b(φ(A)) = φ(b(A)).
Proof: Let y ∈ b(φ(A)) be given. Then φ(ai ) → y for some sequence (ai )
in A. By compactness of A, some subsequence (aki ) converges to x ∈ A.
Therefore y = limi→∞ φ(aki ) = φ(x). We wish to show that x ∈ b(A).
Every neighborhood of x contains some ai ∈ A. We wish to show that every
neighborhood of x is not a subset of A. If this is false, then x ∈ U ⊆ A where
U is open. Since φ is continously differentiable at x, there is some open set V
satisfying x ∈ V ⊆ U such that φ(V ) is open. We have y = φ(x) ∈ φ(V ) ⊆
φ(A), which contradicts the fact that y is a boundary point of φ(A). Hence
y = φ(x) ∈ φ(b(A)).
Now suppose φ is injective. Let y ∈ φ(b(A)). Then y = φ(x) for some x ∈
b(A). We have ai → x, therefore φ(ai ) → y, therefore every neighborhood
of y contains a point in φ(A). We also have xi → x where each xi ∈ Ω\A,
therefore φ(xi ) → y, therefore every neighborhood of y contains a point in
φ(Ω\A). We wish to show φ(Ω\A) ⊆ φ(A)c , i.e. φ(Ω\A) ∩ φ(A) = ∅. This is
true by injectivity of φ on Ω.
Example: Let T = {x1 v1 + · · · + xn vn : 0 ≤ xi < 1} where v1 , . . . , vn ∈ Rn .
We will prove that b(T ) has content zero. Define φ : Rn → Rn via
φ(x1 , . . . , xn ) = x1 v1 + · · · + xn vn .
Then φ is continuously differentiable. Set I = [0, 1)n . Then T = φ(I),
b(T ) = b(φ(I)) ⊆ φ(b(I)). Since b(I) has content zero, b(T ) has content zero.
Note also that if v1 , . . . , vn are linearly independent then φ is injective, hence
b(T ) = φ(b(I)).
Theorem: Let φ : Rp → Rp be linear. Let A ⊆ Rp be bounded and have
boundary of content zero. Then φ(A) is is bounded, b(φ(A)) has boundary
of content zero, and c(φ(A)) = | det φ|c(A).
Proof: A is compact and L is continuous, hence φ(A) is compact. This
implies φ(A) is bounded. Since φ is continuously differentiable, b(φ(A)) ⊆
φ(b(A)). Since b(A) has content zero, b(φ(A)) has content zero. It remains
to compute c(φ(A)).
Suppose det(φ) = 0. Then φ(Rp ) = span(v1 , . . . , vk ) for some k < p. Therefore φ(A) = {x1 v1 + · · · + xk vk : (x1 , . . . , xk ) ∈ S} for some S ⊆ Rk .
17
Since φ(A) is bounded, S is bounded. If we define ψ : Rk → Rp via
ψ(x1 , . . . , xk ) = x1 v1 + · · · + xk vk , then ψ is continuously differentiable and
ψ(S) = φ(A). Since ψ(S) has content zero, φ(A) has content zero. So we
have c(φ(A)) = | det φ|c(A).
Now assume det(φ) 6= 0. Given a bounded set A with content 0 in Rp ,
define λφ (A) = c(φ(A)). Since the content function is positive, so is λφ .
Since φ is injective and additive and the content function is additive, λφ is
additive. Since φ is linear and the content function is translation-invariant,
λφ is translation-invariant. Suppose λφ ([0, 1)p ) = 0. This implies λφ (I) = 0
where I is any half-open p-cell with integer vertices. Given any bounded
set A with boundary of content 0, B = φ−1 (A) has the same properties,
hence B ⊆ I where I is a half-open p-cell with integer vertices. Therefore
c(A) = c(φ(B)) = λφ (B) ≤ λφ (I) = 0, which implies 1 = c([0, 1)p ) = 0, a
contradiction. Therefore λφ ([0, 1)p ) 6= 0. So now we know λφ (A) = mφ c(A)
for some mφ > 0. Our goal is to show that mφ = | det φ|.
Given any linear and invertible φ, ψ : Rp → Rp we have
mφ◦ψ = λφ◦ψ ([0, 1)p ) = c(φ(ψ([0, 1)p )) =
λφ (ψ([0, 1)p )) = mφ c(ψ([0, 1)p )) = mφ λψ ([0, 1)p ) = mφ mψ .
It will suffice to show that any linear, invertible φ is a composition of the
form φ = φ1 ◦ · · · ◦ φn where each φi is linear and invertible and satisfies
mφi = | det φi |.
From elementary matrix theory, for any invertible matrix A there are elementary matrices E1 , . . . , Em and F1 , . . . , Fn such that
(
m
Y
i=1
n
Y
Ei )A( Fi ) = I.
i=1
This yields
(
m
Y
i=1
n
Y
mEi )mA ( mFi ) = mI = 1.
i=1
It will suffice to show that mE = | det E| for every elementary matrix E. We
can assume without loss of generality that p ≥ 2.
If E is used to create a row permutation then E is a permutation matrix
with determinant ±1. Moreover E(x1 , . . . , xp ) = (xσ(1) , . . . , xσ(p) ) for some
18
permutation σ, hence E fixes [0, 1)p , hence mE = 1 = | det E|. We get the
same conclusion if E is used to create a column permutation.
If E is used to multiply row 1 by a 6= 0, then det(E) = a and
(
[0, a) × [0, 1]p−1 a > 0
E([0, 1)p ) =
(a, 0] × [0, 1]p−1 a < 0
hence mE = |a|, therefore mE = | det E|. We get the same conclusion if E is
used to multiply column 1 by a =
6 0.
If E is used to add row 1 to row 2 then
E(x1 , x2 , . . . , xp ) = (x1 + x2 , x2 , . . . , xp ).
Let S1 = {a ∈ [0, 1)p : a1 ≥ a2 } and S2 = {a ∈ [0, 1)p : a1 < a2 }. Then
E([0, 1]p ) = S1 ∪ (e1 + S2 ),
c(E([0, 1]p ) = c(S1 ) + c(S2 ) = c([0, 1)p ) = 1,
mE = 1 = | det E|.
If F is used to add column 1 to column 2 then
F (x1 , x2 , . . . , xp ) = (x1 , x1 + x2 , . . . , xp ) = E(12) ◦ E ◦ E(12) (x1 , . . . , xp ),
so F is a composition of the elementary matrices already considered.
Theorem: Let Ω ⊆ Rp be open and let φ : Ω → Rp be continuously
differentiable, injective, and satisfy φ0 (x) 6= 0 for all x ∈ Ω. Let A have
content and satisfy A ⊆ Ω. Then for each ∈ (0, 1) there exists γ > 0
such that for all a ∈ A and all non-trivial closed p-cells K with center a and
side-length 2γ,
(1 − )p c(φ(K)) ≤ | det φ0 (a)|c(K) ≤ (1 + )p c(φ(K)).
Proof:
S As above, find δ > 0 such that Bδ (x) ⊆ Ω for all x ∈ A. Then set
Ω1 = x∈A B δ (x) so that A ⊆ Ω1 and Ω1 ⊆ Ω. Set M = minx∈Ω1 ||φ0 (x)−1 || >
2
0.
19
Let ∈ (0, 1) be given. Set 0 = 2√pM . By uniform continuity of φ0 on Ω1
there exists β > 0 such that x, y ∈ Ω1 and ||x − y|| < β implies ||φ0 (x) −
φ0 (y)|| < 0 . We claim that for all a ∈ A and all ||h|| < min( 2δ , β) we have
||φ(a + h) − φ(a) − φ0 (a)h|| ≤ 0 ||h||.
To prove this, observe that we have a + h ∈ Ω1 and ||φ0 (a + h) − φ0 (a)|| < 0 .
Set
f (t) = (φ(a + h) − φ(a) − φ0 (a)h) • (φ(a + th) − φ(a) − φ0 (a)th)
for 0 ≤ t ≤ 1. Then
||φ(a + h) − φ(a) − φ0 (a)h||2 = f (1) − f (0) = f 0 (t∗ ) =
(φ(a + h) − φ(a) − φ0 (a)h) • (φ0 (a + th)h − φ0 (a)h) =
||φ(a + h) − φ(a) − φ0 (a)h|| · ||φ0 (a + th) − φ0 (a)||||h|| ≤
||φ(a + h) − φ(a) − φ0 (a)h|| · 0 ||h||.
Hence the claim is true. Now we can write
φ(a + h) = φ(a) + φ0 (a)h + ||h||V (h)
where ||V (h)|| ≤ 0 . Write ψ(x) = φ−1 (a)φ(x). Then we have
ψ(a + h) = ψ(a) + h + ||h||W (h)
where ||W (h)|| ≤ M 0 .
Now choose γ > 0 so that the closed p-cube K of side length 2γ centered
at a satisfies K ⊆ Bmin( δ ,β) (a). Since ψ is continuously differentiable and
2
injective, b(ψ(K)) = ψ(b(K)). Let a + h ∈ K be given and write
ψ(a + h) = ψ(a) + h + k
where ||k|| ≤ ||h||M 0 . Writing h = (h1 , . . . , hp ) and k = (k1 , . . . , kp ), we
√
√
have |hi | ≤ γ for each i, hence ||h|| ≤ pγ, hence |ki | ≤ pγM 0 and
|hi + ki | ≤ γ(1 +
20
√
pM 0 ).
Therefore ψ(K) is contained in the interior of a p-cube of side length 2γ(1+)
centered at ψ(a). Moreover if |hi | = γ then
√
|hi + ki | ≥ γ − |ki | ≥ γ − ||k|| > γ − 2 pγM 0 = γ(1 − ),
therefore b(ψ(K)) does not intersect a p-cube of side length 2γ(1−) centered
at ψ(a).
Let K1 be the cube of side length 2γ(1 − ) and let K2 be the cube of side
length 2γ(1 + ). We know that ψ(K) ⊆ K1 .Then K1 ∩ b(ψ(K)) = ∅ implies
K1 is contained in the union of the interior of ψ(K) and the complement
of ψ(K), both of which are open sets. Since K1 is connected, it must be a
subset of one of these sets. Since ψ(a) ∈ K1 and ψ(a) is in the interior of
ψ(K), K1 ⊆ ψ(K). This implies
c(K1 ) ≤ c(ψ(K)) ≤ c(K2 ),
(1 − )p c(K) ≤ | det φ0 (a)−1 |c(φ(K)) ≤ (1 + )p c(K),
(1 − )p | det φ0 (a)|c(K) ≤ c(φ(K)) ≤ (1 + )p | det φ0 (a)|c(K).
Theorem: Let Ω ⊆ Rp be open. Let φ : Ω → Rp be continuously differentiable, injective, and satisfy φ0 (x) 6= 0 for all x ∈ Ω. Let A be bounded, have
boundary of content zero, and satisfy A ⊆ Ω. Let f : φ(A) → R be bounded
and continuous. Then
Z
Z
f = (f ◦ φ)| det φ0 |.
φ(A)
A
Proof: Let I be a half-open p-cell with integer vertices containing A. Let
Ω1 be a bounded open set containing A. Fix M > 0 so that ||φ0 (x)|| ≤ M
for all x ∈ Ω1 and and | det φ0 (a)| ≤ M and |f (φ(a))| ≤ M for all a ∈ A. Let
Pn be the partition of I into cells of the form [a1 , b1 ) × · · · × [ap , bp ) where
for each i there exists ki ∈ Z such that ai = 2kni , bi = ki2+1
.
n
Since f = 12 (|f | + f ) − 12 (|f | − f ), we can assume without loss of generality
that f ≥ 0. Let ∈ (0, 1) be given. Then
S there exists a disjoint set of p-cells
P contained in A with c(A) − < c( P ) ≤ c(A), and forSall sufficiently
S
large n() ∈ N there exists a subset B,n() of Pn() such that B,n() ⊆ P
21
S
S
S
and c( P ) − < c( B,n() ) ≤ c( P ). In short, the content of A is
approximated by the content sum of B,n() , a union of half-open p-cubes
with side length 21n contained in A, and
[
c(A) − 2 < c( B,n() ) ≤ c(A).
Similarly, the content of A is approximated by the content sum of C,n() , a
union of half-open p-cubes with side length 21n contained in Ω1 , including all
those in B,n() , and
[
c(A) ≤ c( C,n() ) < c(A) + 2.
Now choose n() large enough so that
(1 − )p c(φ(K)) ≤ | det φ0 (a)|c(K) ≤ (1 + )p c(φ(K))
is satisfied for all K ∈ B,n() , where aK is the center of K, and
|f (φ(x))| det φ0 (x)| − f (φ(y))| det φ0 (y)|| < for all for all x, y ∈ K ∈ C,n() . For each K ∈ B,n() choose yK = φ(xK ) ∈
φ(K) such that
Z
f = f (yK )c(φ(K)) = f (φ(xK ))c(φ(K))
φ(K)
where xK ∈ K. These choices yield the following series of inequalities:
1.
X
(1 − )p
f (φ(aK ))c(φ(K)) ≤
K∈B,n()
X
f (φ(aK ))| det φ0 (a)|c(K) ≤
K∈B,n()
(1 + )p
X
f (φ(aK ))c(φ(K)).
K∈B,n()
2.
√
c(φ(K)) ≤ (2M p)p c(K)
22
for all K ∈ C,n() .
3.
Z
Z
Z
0
0 (f ◦ φ)| det φ | ≤
(f ◦ φ)| det φ0 | ≤ 2M 2 .
(f ◦ φ)| det φ | − S
S
A
B,n()
A\ Bn,
4.
Z
Z
Z
√
f− S
f ≤
f ≤ 4(2M p)p .
S
S
φ(A)
φ( B,n() ) φ( C,n() \ B,n() )
5.
Z
X
0
0
≤ c(A).
(f
◦
φ)|
det
φ
|
−
f
(φ(a
))|
det
φ
(a
)|c(K)
K
K
S
B,n()
K∈B,n()
6.
Z
X
=
f
−
f
(φ(a
))c(φ(K))
K
S
φ( B,n() )
K∈B,n()
X
X
≤ (2M √p)p c(A).
f
(φ(x
))c(φ(K))
−
f
(φ(a
))c(φ(K))
K
K
K∈B,n()
K∈B,n()
Letting → 0, inequalities 6 and 4 imply
Z
X
f (φ(aK ))c(φ(K)) →
f.
φ(A)
K∈B,n()
Therefore inequality 1 implies
X
Z
0
f (φ(aK ))| det φ (a)|c(K) →
f.
φ(A)
K∈B,n()
Therefore inequalities 5 and 3 imply
Z
Z
0
(f ◦ φ)| det φ | =
A
φ(A)
23
f.
Example: Let
E = {(x, y) : x2 + y 2 ≤ 1}.
We wish to compute c(E). Let
1
1
1
1
En = {(r cos θ, r sin θ) : (r, θ) ∈ [ , 1 − ] × [−π + , π − ]}.
n
n
n
n
Then
1
1
), sin(π − )]
n
n
and c(E\En ) ≤ 2 sin( n1 ) → 0 as n → ∞. Therefore c(E) = limn→∞ c(En ).
E\En ⊆ [−1, 0] × [− sin(−π +
Define φ : (0, 1) × (−π, π) → R2 via φ(r, θ) = (r cos θ, r sin θ). Then
Z
Z
1=
1=
c(En ) =
1
1
1
1
,1− n
]×[−π+ n
,π− n
])
φ([ n
En
 ∂(r cos θ)
∂r
det 
1
1
1
1
∂(r sin θ)
[n
,1− n
]×[−π+ n
,π− n
]
Z
∂r
Z
Z
|r| =
1
1
1
1
[n
,1− n
]×[−π+ n
,π− n
]
1
π− n
Z
∂(r cos θ) 
∂θ
 =
∂(r sin θ)
1
1− n
r dr dθ =
1
−π+ n
1
n
hence c(En ) → π as n → ∞.
24
∂θ
1 1
−
2 n
2
2π −
n