The Integral as an Iterated Integral

Let K = [a1, b1] × · · · × [an, bn] in R^{n} be a closed n-cell and let f : K → R be integrable on K.

Then one can calculate the integral Z

K

f in terms of a n-fold “iterated integral”

Z bn

an

· · ·

Z b2

a2

Z b1

a1

f (x_{1}, x_{2}, . . . , x_{n}) dx_{1}

dx_{2}

· · ·

dx_{n}.
For simplicity we shall justify this for n = 2.

Theorem Let K = [a, b] × [c, d]. If f : K → R is continuous on K, then Z

K

f = Z d

c

Z b a

f (x, y)dx

dy =

Z b a

Z d c

f (x, y)dy

dx.

Proof Let F be defined for y ∈ [c, d] by F (y) =

Z b a

f (x, y)dx.

Let

c = y_{0} < y_{1} < · · · < y_{r}= d be a partition of [c, d]

a = x_{0} < x_{1} < · · · < x_{s}= b be a partition of [a, b]

and let

P = {[x_{i−1}, x_{i}] × [y_{j−1}, y_{j}] | 1 ≤ i ≤ s, 1 ≤ j ≤ r} be a partition of K.

Let y_{j}^{∗} be any point in [y_{j−1}, y_{j}] and note that
F (y_{j}^{∗}) =

Z b a

f (x, y_{j}^{∗}) dx =

s

X

i=1

Z xi

xi−1

f (x, y^{∗}_{j}) dx.

For each j and i, by the Fundamental Theorem of Calculus and the Mean Value Theorem, there
exists a point x^{∗}_{ji} ∈ [x_{i−1}, x_{i}] such that

F (y_{j}^{∗}) =

s

X

i=1

f (x^{∗}_{ji}, y_{j}^{∗})(x_{i}− x_{i−1}).

We multiply by (y_{j} − y_{j−1}) and add to obtain

r

X

j=1

F (y_{j}^{∗})(y_{j} − y_{j−1}) =

r

X

j=1 s

X

i=1

f (x^{∗}_{ji}, y^{∗}_{j})(x_{i}− x_{i−1})(y_{j}− y_{j−1}).

Now the expression on the left side of this formula is an arbitrary Riemann sum for the integral Z d

c

F (y) dy = Z d

c

Z b a

f (x, y) dx

dy.

We have shown that an arbitrary Riemann sum for F on [c, d] is equal to a particular Riemann sum of f on K corresponding to the partition P. Since f is integrable on K, the existence of the iterated integral and its equality with the integral on K is established.

A minor modification of the proof given for the preceding theorem yields the following, slightly stronger, assertion.

Theorem Let K = [a, b] × [c, d]. Suppose that f : K → R is integrable on K and for each y ∈ [c, d], the function x 7→ f (x, y) of [a, b] into R is continuous except possibly for a finite number of points, at which it has one-sided limits. Then

Z

K

f = Z d

c

Z b a

f (x, y)dx

dy.

Definition Let K = I_{1}× · · · × I_{n}, where I_{j} = [a_{j}, b_{j}] for each j = 1, . . . , n.

(a) The (n-)content or volume c(K) of K is defined to be
c(K) = (b_{1}− a_{1}) × · · · × (b_{n}− a_{n}) =

n

Y

j=1

(b_{j}− a_{j}).

(b) A set Z ⊂ R^{n} has n-content or volume zero if ∀ > 0, ∃ afinite collection C = {Kj}^{m}_{j=1} of
n-cells such that

(1) Z ⊂

m

[

j=1

K_{j}, and (2)

m

X

j=1

c(K_{j}) < ε.

Remarks

(a) If K is a cell (not necessarily closed) in R^{n}, then the boundary ∂K of K is a set of n-content
zero.

(b) (i) Let K be a cell in R^{n} and suppose that
K =

m

[

j=1

K_{j} and Int K_{i}∩ Int K_{j} = ∅ for 1 ≤ i 6= j ≤ m.

Then

c(K) =

m

X

j=1

c(K_{j}).

(ii) Let K_{1}, K_{2} be cells in R^{n}. Then

c(K_{1}∪ K_{2}) = c(K_{1}\ K_{2}) + c(K_{1}∩ K_{2}) + c(K_{2} \ K_{1}).

(iii) Let x ∈ R^{n}, let K be a cell in R^{n} and let x + K = {x + z | z ∈ K}. Then
x + K is a cell in R^{n} with c(x + K) = c(K),

i.e. the content of a cell is invariant under translations.

Examples

(a) 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, . . . , K_{m}} of 1-dimensional cells covering Z has

m

X

j=1

c(K_{j}) ≥ 1.

(b) Let Z = {(x, y) | |x| + |y| = 1}, a (1-dim’l) boundary of {(x, y) | |x| + |y| ≤ 1} ⊂ R^{2}. Then
the (2-dim’l) content c(Z) = 0.

Definition Let A ⊂ R^{n} be a bounded set and let f : A → R be a bounded function. Suppose
that K is a closed cell containing A and let f_{K} : K → R be an extension of f defined by

f_{K}(x) =

(f (x) if x ∈ A, 0 if x ∈ K \ A.

We say that f is integrable on A if f_{K} is integrable on K, and define
Z

A

f = Z

K

f_{K}.

Theorem Let A = {(x, y) : α(y) ≤ x ≤ β(y), c ≤ y ≤ d}, where α, β : [c, d] → [a, b] are continuous functions from [c, d] into [a, b]. If f : A → R is continuous on A, then f is integrable on A and

Z

A

f = Z d

c

(Z β(y) α(y)

f (x, y)dx )

dy.

Proof Let K be a closed cell containing A, and let f_{K} be the extension of f to K defined by

f_{K}(x) =

(f (x) if x ∈ A 0 if x ∈ K \ A

Since ∂A has content zero, f_{K}is integrable on K. Now for each y ∈ [c, d] the function x 7→ f_{K}(x, y)
is continuous except possibly at the two points α(y) and β(y), at which it has one-sided limits.

It follows from the preceding theorem that Z

A

f = Z

K

fK = Z d

c

Z b a

fK(x, y)dx

dy =

Z d c

(Z β(y) α(y)

f (x, y)dx )

dy.

Examples

(a) Given a prime number p, let A =

(i

p,j

p) | 1 ≤ i, j ≤ p − 1, p is a prime number

⊆ K = [0, 1] × [0, 1].

(i) Show that each horizontal and each vertical line in R^{2} intersects A in a finite number
(often zero) of points and that A does not have content zero. Let f : K → R be defined
by

f (x, y) =

(1 if (x, y) ∈ A 0 if (x, y) ∈ K \ A.

(ii) Show that f is not integrable on K. However, the iterated integrals exist and satisfy Z 1

0

Z 1 0

f (x, y)dx

dy =

Z 1 0

Z 1 0

f (x, y)dy

dx.

Proof Let {K_{j}}^{m}_{j=1} be a collection of 2-cells covering A. Since every r ∈ Q ∩ [0, 1] is a limit
point of A,

K = [0, 1] × [0, 1] = ¯A ⊆ ∪^{m}_{j=1}K¯_{j} =⇒

m

X

j=1

c(K_{i}) ≥ 1.

This implies that A does not have content zero and f is not integrable on K.

However, since each horizontal and each vertical line in R^{2} intersects A in a finite number
(often zero) of points, f is continuous on each horizontal and each vertical line in R^{2} except
possibly at a finite number of points there and hence the iterated integrals exist and satisfy

Z 1 0

Z 1 0

f (x, y)dx

dy = 0 = Z 1

0

Z 1 0

f (x, y)dy

dx.

(b) Let K = [0, 1] × [0, 1] and let f : K → R be defined by

f (x, y) =

0 if either x or y is irrational, 1

n if y is rational and x = m

n where m and n > 0 are relatively prime integers.

Show that

Z

K

f = Z 1

0

Z 1 0

f (x, y)dx

dy = 0, but that R1

0 f (x, y)dy does not exist for rational x.

Proof Divide [0, 1] into n equal length subintervals. Note that if x = p

q is a rational in a subinterval, since j − 1

n ≤ p q ≤ j

n for some 1 ≤ j ≤ n, we have q ≥ n. This implies that the Riemann sum of f with respect to partition P = i − 1

n , i n

× j − 1 n , j

n

| 1 ≤ i, j ≤ n

satisfies that

|SP(f, K)| = |

n

X

i=1 n

X

j=1

f (x^{∗}_{i}, y^{∗}_{j}) · 1
n^{2}| ≤

n

X

i=1 n

X

j=1

1 n · 1

n^{2} = 1

n → 0 as n → ∞.

Hence f is integrable on K. Similarly, f (x, y) is integrable on [0, 1] for each y ∈ [0, 1] and hence

Z

K

f = Z 1

0

Z 1 0

f (x, y)dx

dy = 0.

For each x = m

n, since the upper and lower sums of f (x, y) with respect to any partition of [0, 1] are 1

n and 0, repectively, f (x, y) is not integrable with respect to y on [0, 1].

(c) Let R denote the triangular region in the first quadrant bounded by the lines y = x, y = 0, and x = 1. Show that

Z 1 0

Z 1 y

sin x

x dxdy = Z 1

0

Z x 0

sin x x dydx.

Proof For each (x, y) ∈ R, consider the function

f (x, y) =

sin x

x if x 6= 0 1 if x = 0.

Since lim

x→0

sin x

x = 1, f is a continuous function on R and hence f is integable on R and Z

R

f = Z 1

0

Z 1 y

sin x

x dxdy = Z 1

0

Z x 0

sin x

x dydx = Z 1

0

sin xdx = − cos x|^{1}_{0}.

(d) Show that Z 2

0

Z 1 y/2

ye^{−x}^{3} dx dy = 2

3(1 − e^{−1}).

Proof Let R denote the triangular region in the first quadrant bounded by the lines y = 2x,
y = 0, and x = 1. Since f (x, y) = ye^{−x}^{3} is continuous on R, f is integrable on R and

Z

R

f = Z 2

0

Z 1 y/2

ye^{−x}^{3} dx dy =
Z 1

0

Z 2x 0

ye^{−x}^{3} dy dx.

Now

Z 1 0

Z 2x 0

ye^{−x}^{3} dy dx =
Z 1

0

y^{2}

2e^{−x}^{3}|^{2x}_{0} dx

= Z 1

0

2x^{2}e^{−x}^{3}dx

= −2e^{−x}^{3}
3 |^{1}_{0} = 2

3(1 − e^{−1}).

(e) For β > α > 0, let R = [0, ∞) × [α, β] and f (x, t) = e^{−tx}. Show that
Z ∞

0

e^{−αx}− e^{−βx}

x dx = logβ α. Proof Since f (x, t) is continuous on R,

0 < f (x, t) ≤ e^{−αx} ∀ (x, t) ∈ R and 0 <

Z

R

e^{−αx} =
Z β

α

Z ∞ 0

e^{−αx}dx dt = β − α
α < ∞,
f (x, t) is integrable on R by the M -Test and

Z

R

f = Z β

α

Z ∞ 0

e^{−tx}dx dt =
Z ∞

0

Z β α

e^{−tx}dt dx.

Hence logβ

α = Z β

α

1 t dt =

Z β α

Z ∞ 0

e^{−tx}dx dt =
Z ∞

0

Z β α

e^{−tx}dt dx =
Z ∞

0

e^{−αx}− e^{−βx}

x dx.

The Jacobian Theorem Let Ω ⊆ R^{p} be open. Suppose that

φ : Ω → R^{p} belongs to class C^{1}(Ω) is injective on Ω,
J_{φ}(x) 6= 0 for x ∈ Ω,

A has content and ¯A ⊂ Ω.

If ε > 0 is given, then there exists r > 0 such that if K is a closed cell with center x ∈ A and side length less than 2r, then

|Jφ(x)|(1 − ε)^{p} ≤ c(φ(K))

c(K) ≤ |Jφ(x)|(1 + ε)^{p}.

Proof For each x ∈ Ω, let L_{x} = (dφ(x))^{−1}, since

1 = det(L_{x}◦ dφ(x)) = (det L_{x}) (det dφ(x)),
it follows that

det L_{x} = 1
J_{φ}(x).

Let Ω_{1} be a bounded open subset of Ω such that A ⊂ Ω¯ _{1} ⊂ ¯Ω_{1} ⊆ Ω and dist(A, ∂Ω_{1}) = 2δ > 0.

Since φ ∈ C^{1}(Ω), L_{x} is continuous on the compact subset ¯Ω and there exists a constant M > 0
such that

kL_{x}k ≤ M for all x ∈ Ω_{1}.

Given 0 < ε < 1, since dφ is uniformly continuous on Ω_{1}, there exists β with 0 < β < δ such
that

if x_{1}, x_{2} ∈ Ω_{1} and kx_{1}− x_{2}k ≤ β, then kdφ(x_{1}) − dφ(x_{2})k ≤ ε
M√

p.
This implies that if x ∈ A and if kzk ≤ β, then x, x + z ∈ Ω_{1} and

kφ(x + z) − φ(x) − dφ(x)(z)k ≤ kzk sup

0≤t≤1

kdφ(x + tz) − dφ(x)k ≤ ε M√

pkzk.

For each x ∈ A, let ψ(z) be defined by

(∗) ψ(z) = L_{x}[φ(x + z) − φ(x)] for kzk ≤ β.

Since

kψ(z) − zk = kL_{x}[φ(x + z) − φ(x) − dφ(x)(z)]k

≤ M kzk sup

0≤t≤1

kdφ(x + tz) − dφ(x)k

≤ ε

√pkzk for kzk ≤ β, this implies that

ψ(z) ∈ ¯B_{εr}(z) for kzk ≤ β = √
pr.

If K_{1} is a closed cell with center 0 and contained in the open ball B_{β}(0) for β =√

pr, since ψ is
an injective, open mapping for kzk ≤ β, there exist closed cells K_{i}, K_{o} both centered 0 and with
side lengths 2(1 − ε)r and 2(1 + ε)r, respectively, such that

K_{i} ⊂ ψ(K_{1}) ⊂ K_{o},
and

(∗∗) (1 − ε)^{p} ≤ c(ψ(K_{1}))

c(K_{1}) ≤ (1 + ε)^{p}.
In general, if K = x + K_{1} is a closed cell with center x ∈ A, since

c(K_{1}) = c(K),

c(ψ(K_{1})) = | det L_{x}|c(φ(x + K_{1}) − φ(x)) by (∗)

= 1

J_{φ}(x)c(φ(K) − φ(x))

= 1

J_{φ}(x)c(φ(K))

and by substituting these into (∗∗), we have
(1 − ε)^{p} ≤ c(φ(K))

J_{φ}(x) c(K) ≤ (1 + ε)^{p} =⇒ |J_{φ}(x)|(1 − ε)^{p} ≤ c(φ(K))

c(K) ≤ |J_{φ}(x)|(1 + ε)^{p}.

Change of Variables Theorem Let Ω ⊆ R^{p} be open. Suppose that
φ : Ω → R^{p} belongs to class C^{1}(Ω) is injective on Ω,

J_{φ}(x) = det dφ(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|.

Examples

(a) Let C ⊂ R^{3} be the circular helix parametrized by

C : φ(t) = (x(t), y(t), z(t)) = (cos t, sin t, t) for t ∈ [0, 2π].

Evaluate Z

C

y sin z ds, where s = s(t) = Z t

0

|φ^{0}(u)| du.

Solution Since φ : [0, 2π] → C is C^{1} injective with |Dφ(t)| = |φ^{0}(t)| > 0 for all t ∈ [0, 2π],
we have

Z

C

y sin z ds = Z

φ([0,2π])

y sin z ds

= Z 2π

0

y sin z|(x,y,z)=φ(t) |φ^{0}(t)| dt

= Z 2π

0

sin^{2}tp

sin^{2}t + cos^{2}t + 1 dt

= √

2π.

(b) Let S^{1}(r) = {(x, y) ∈ R^{2} | x^{2} + y^{2} = r^{2}} be the counteclockwise oriented circle of radius
r with centere at (0, 0). For each k ∈ Z, evaluate

Z

S^{1}(r)

z^{k}dz, where z = x + iy and dz =
dx + idy.

Solution By identifying (x, y) ∈ R^{2} with z = x + iy ∈ C, S^{1}(r) is parametrized by
S^{1}(r) : φ(t) = re^{it} = r(cos t + i sin t) for t ∈ [0, 2π].

Since φ : [0, 2π] → S^{1}(r) is C^{1} injective with Dφ(t) = φ^{0}(t) = i r e^{it} 6= 0 for all t ∈ [0, 2π],
we have

Z

S^{1}(r)

z^{k}dz =
Z

φ([0,2π])

z^{k}dz

= Z 2π

0

z^{k}|_{z=φ(t)} φ^{0}(t) dt

= Z 2π

0

r^{k+1}e^{i(k+1)t}i dt

=

(0 if k 6= −1, 2πi if k = −1.

(c) Let S = {(x, y) | x > 0, y > 0, 1 < xy < 3, 1 < x^{2}− y^{2} < 4}. Evaluate
Z Z

S

x^{2}+ y^{2} dA
Solution Let K = {(u, v) | 1 < u < 4, 1 < v < 3} and let φ : S → K be defined by

φ(x, y) = (x^{2}− y^{2}, xy) for (x, y) ∈ S.

Then φ : S → K is C^{1} bijective with Jφ(x, y) = 2(x^{2} + y^{2}) > 0 for all (x, y) ∈ S. Hence we
have

Z Z

S

x^{2}+ y^{2} dA =
Z Z

φ^{−1}(K)

1

2J_{φ}(x, y) dA

= Z Z

K

1

2J_{φ}(x, y)|_{(x,y)=φ}^{−1}_{(u,v)} |J_{φ}^{−1}(u, v)| du dv

= Z 3

1

Z 4 1

1

2 du dv since J_{φ}(x, y)J_{φ}^{−1}(u, v) = 1 ∀ (u, v) ∈ K

= 3.

(d) Show that Z ∞

0

e^{−x}^{2}dx =

√π 2 .

Solution Let Q_{1} = {(x, y) | x > 0, y > 0}, K = {(r, θ) | r > 0, 0 < θ < π/2} and let
φ : K → Q_{1} be defined by

φ(r, θ) = (x, y) = (r cos θ, r sin θ) for (r, θ) ∈ K.

Then φ : K → Q_{1} is C^{1} bijective with J_{φ}(r, θ) = r ≥ 0 for all (r, θ) ∈ K. Hence we have

Z ∞ 0

e^{−x}^{2}dx

2

= Z ∞

0

Z ∞ 0

e^{−x}^{2}^{−y}^{2}dx dy

= Z Z

Q1

e^{−x}^{2}^{−y}^{2}dx dy

= Z Z

K

e^{−x}^{2}^{−y}^{2}|(x,y)=φ(r,θ) |J_{φ}(r, θ)| dr dθ

=

Z π/2 0

Z ∞ 0

re^{−r}^{2}dr dθ

= π

4.

Partition of Unity

Let f : R → R be defined by

f (x) =

(e^{−1/x} if x > 0,

0 otherwise.

x y

Then f is smooth, i.e. infinitely differentiable, on R. For a < b, let g : R → R be defined by g(x) = f (x − a)f (b − x). Then g is smooth and

g(x) =

(e^{−1/(x−a)}e^{−1/(b−x)} if a < x < b,

0 otherwise.

x y

Let h : R → R be defined by

h(x) = R∞

x g(t) dt R∞

−∞g(t) dt for x ∈ R.

x y

Then

h(x) = 1 if x ∈ (−∞, a], 0 < h(x) < 1 if x ∈ (a, b), h(x) = 0 if x ∈ [b, ∞).

For each p ∈ R^{n} and for any 0 < r < t, let Br(p) and Bt(p) be concentric balls of radius r and
t, respectively. Let ζ : R → R be the linear function such that ζ(r^{2}) = a and ζ(t^{2}) = b, and let

ψ(x) = h(ζ(kx − pk^{2})) for x ∈ R^{n}.

Then ψ : R^{n}→ R is smooth, 0 ≤ ψ(x) ≤ 1 for all x ∈ R^{n} and

ψ(x) =

(1 if x ∈ B_{r}(p),
0 if x /∈ B_{t}(p).

Theorem Suppose K is a compact subset of R^{n}, and {V_{α}} is an open cover of K. Then there
exist functions ψ_{1}, . . . , ψ_{s}∈ C^{∞}(R^{n}), the space of smooth functions on R^{n}, such that

(a) 0 ≤ ψ_{i} ≤ 1 for 1 ≤ i ≤ s;

(b) each ψ_{i} has its support in some V_{α}, i.e. {x ∈ R^{n}| ψ_{i}(x) 6= 0}⊂ V_{α}, and
(c)

s

X

i=1

ψ_{i}(x) = 1 for every x ∈ K.

Because of (c), {ψ} is called a partition of unity, and (b) is sometimes expressed by saying that
{ψ_{i}} is subordinate to cover {V_{α}}.

Corollary If f ∈C (R^{n}) is a continuous function in R^{n} and the support of f lies in K, then

f =

s

X

i=1

ψif.

Each ψ_{i}f has its support in some V_{α}.

Proof For each x ∈ K, since {V_{α}} is an open cover of K, there exist V_{α(x)} ∈ {V_{α}}, open balls
B(x) and W (x), centered at x, such that

(∗) B(x) ⊂ W (x) ⊂ W (x) ⊂ V_{α(x)}.

Since K is compact and {B(x) | x ∈ K} is an open cover of K, there are points x_{1}, . . . , x_{s} ∈ K
such that

K ⊂ B(x_{1}) ∪ · · · ∪ B(x_{s}).

By (∗), there are functions φ_{1}, . . . , φ_{s} ∈C (R^{n}), such that

φ_{i}(x) =

(1 if x ∈ B(x_{i})
0 if x ∈ R^{n}\ W (x_{i})
and 0 ≤ φ_{i}(x) ≤ 1 for all x ∈ R^{n} for each 1 ≤ i ≤ s.

V_{α}

W (x_{i}) W (xj)
B(xi) B(xj)

x_{i} xj

Define

ψ_{1} = φ_{1}

(†) ψ_{i+1} = (1 − φ_{1}) · · · (1 − φ_{i})φ_{i+1} for i = 1, . . . , s − 1.

Properties (a) and (b) are clear. The relation

(††) ψ_{1}+ · · · + ψ_{i} = 1 − (1 − φ_{1}) · · · (1 − φ_{i})

is trivial for i = 1. If (†) holds for some i < s, addition of (†) and (††) yields (††) with i + 1 in place of i. It follows that

s

X

i=1

ψ_{i}(x) = 1 −

s

Y

i=1

1 − φ_{i}(x)

for x ∈ R^{n}.

If x ∈ K, then x ∈ B(x_{j}) for some 1 ≤ j ≤ s, hence φ_{j}(x) = 1,

s

Y

i=1

1 − φ_{i}(x)

= 0 and

s

X

i=1

ψi(x) = 1. This proves (c).

Theorem Let A ⊂ R^{n} and let O be a collection of open subsets of R^{n} covering A. Then there is
a collection Φ of continuous functions ϕ defined in an open set containing A, with the following
properties:

(1) For each x ∈ A we have 0 ≤ ϕ(x) ≤ 1.

(2) For each x ∈ A there is an open set V containing x such that all but finitely many of ϕ ∈ Φ are 0 on V.

(3) For each x ∈ A we have X

ϕ∈Φ

ϕ(x) = 1 (by (2) for each x this sum is finite in some open set containing x).

(4) For each ϕ ∈ Φ there is an open set U ∈ O such that ϕ = 0 outside of some closed set contained in U.

A collection Φ satisfying (1) to (3) is called a continuouspartition of unity for A.If Φ also satisfies (4), it is said to besubordinate to the cover O.In this chapter we will only use continuity of the functions ϕ.

Proof

Case 1. A is compact.

We use an alternative method to construct a continuous partition of unity as follows.

Since A ⊂ R^{n} is compact, there exists an open ball BR(0) of radius R centered at 0 such that
A ⊂ B_{R}(0). By taking U ∩B_{R}(0) for each U ∈O, we may assume that O is a collection of bounded
open subsets covering A. Again since A is compact, there exist open sets U_{1}, . . . , U_{m} ∈ O such
that

(∗) A ⊂

m

[

j=1

U_{j} and A \ U_{1} ∪ · · · ∪ bU_{i}∪ · · · ∪ U_{m} 6= ∅ ∀ 1 ≤ i ≤ m,

where bU_{i} means the term U_{i} is omitted.

Since A is compact and by (∗), the set C_{1} = A \

m

[

j=2

U_{j} is a compact subset of U_{1} with

r1 = d(∂U1, C1) = inf

x∈∂U1, y∈C1

d(x, y) > 0.

Let

D_{1} = {x ∈ U_{1} | d(x, ∂U_{1}) = inf

y∈∂U1

d(x, y) ≥ r_{1}/2},
W_{1} = {x ∈ U_{1} | d(x, ∂U_{1}) = inf

y∈∂U1

d(x, y) ≥ r_{1}/4},

ψ_{1}(x) =

1 if x ∈ D_{1},

0 if x /∈ W_{1},

0 ≤ ψ_{1}(x) ≤ 1 ∀ x ∈ R^{n}.

Note that D_{1} is a compact subset of U_{1}, C_{1} ⊂ Int D_{1} and A ⊂ Int D_{1} ∪

m

[

j=2

U_{j}.

Suppose that D_{1}, . . . , D_{k} have been chosen so that A ⊂

k

[

j=1

Int D_{j} ∪

m

[

j=k+1

U_{j}. Let

C_{k+1} = A \ (Int D_{1}∪ · · · ∪ Int D_{k}∪ U_{k+2}∪ · · · ∪ U_{m}).

Then C_{k+1} ⊂ U_{k+1} is a compact with

r_{k+1} = d(∂U_{k+1}, C_{k+1}) = inf

x∈∂Uk+1, y∈Ck+1

d(x, y) > 0.

Let

D_{k+1} = {x ∈ U_{k+1} | d(x, ∂U_{k+1}) = inf

y∈∂Uk+1

d(x, y) ≥ r_{k+1}/2},
W_{k+1} = {x ∈ U_{1} | d(x, ∂U_{1}) = inf

y∈∂U1

d(x, y) ≥ r_{k+1}/4},

ψ_{k+1}(x) =

1 if x ∈ Dk+1,

0 if x /∈ Wk+1,

0 ≤ ψk+1 ≤ 1 ∀ x ∈ R^{n}.

Note that D_{k+1} is a compact subset of U_{k+1}, C_{k+1} ⊂ Int D_{k+1} and A ⊂

k+1

[

j=1

Int D_{j} ∪

m

[

j=k+2

U_{j}.
We obtain a collection of compact subsets {Di}^{m}_{i=1}, {Wi}^{m}_{i=1} and a collection of nonnegative
continuous functions {ψ_{i}}^{m}_{i=1} such that

A ⊂

m

[

i=1

Int D_{i} ⊂

m

[

i=1

D_{i} ⊂

m

[

i=1

Int W_{i} ⊂

m

[

i=1

W_{i} ⊂

m

[

i=1

U_{i},
(ψ_{i}(x) > 0 if x ∈ D_{i},

ψ_{i}(x) = 0 if x /∈ W_{i}. =⇒

m

X

j=1

ψ_{j}(x) > 0 for all x ∈

m

[

i=1

D_{i}.

∂U_{i}

Wi

Ci

D_{i}

For each 1 ≤ i ≤ m, and for x ∈

m

[

i=1

D_{i}, if we let

ϕ_{i}(x) = ψ_{i}(x)

ψ1(x) + · · · + ψm(x),

then Φ = {ϕ_{1}, . . . , ϕ_{m}} is the desired partition of unity since {D_{1}, . . . , D_{m}} covers A.

Case 2. A = A_{1}∪ A_{2}∪ A_{3}∪ · · · , where each A_{i} is compact and A_{i} ⊂ Int A_{i+1}.
For each i ∈ N, let

Bi =

(A_{1} if i = 1,

A_{i}\ Int A_{i−1} if i ≥ 2,
and

Oi =

({U ∩ Int A_{3} | U ∈O} if 1 ≤ i ≤ 2,
{U ∩ (Int A_{i+1}\ A_{i−2}) | U ∈O} if i ≥ 3.

· · ·
A_{1}

Ai−2

Ai−1

Ai

Ai+1

· · ·
B_{i}

Then

(i) each Bi is compact and A =

∞

[

i=1

Bi,

(ii) each Oi is a collection of bounded open sets covering B_{i}, i.e. B_{i} ⊂ [

U ∈Oi

U.

(iii) U ∩ V = ∅ for all U ∈Oi, V ∈Oj with j > i + 2 ⇐⇒ j − 1 > i + 1.

Thus, by case 1, there is a partition of unity Φ_{i} for B_{i}, subordinate to Oi.

Note that for each x ∈ A, since x ∈ B_{i} for some B_{i} and since Φ_{i}is a partition of unity subordinate
toOi, so there exists ϕ ∈ Φ_{i} such that ϕ(x) > 0, and by (iii),

ϕ(x) = 0 ∀ ϕ ∈ Φ_{j} with j > i + 2 ⇐⇒ j − 1 > i + 1,
and the sum

σ(x) = X

ϕ∈Φi, all i

ϕ(x)

is a finite sum in some open set containing x, and σ(x) > 0 for all x ∈ A.

Let

Φ =

∞

[

i=1

ϕ(x)

σ(x) | ϕ ∈ Φi

. Then Φ is a partition of unity subordinate to the open coverO.

Case 3. A is open.

Let

A_{i} = {x ∈ A | kxk ≤ i and d(x, ∂A) ≥ 1
i},

where d(x, ∂A) = the distance from x to the boundary ∂A. Note that A_{i}is compact, A_{i} ⊂ Int A_{i+1}
for all i ≥ 1, and

∞

[

i=1

A_{i} = lim

i→∞A_{i} = A.

By applying the case 2, we obtain a partition of unity subordinate to the open cover O.

Case 4. A is arbitrary.

Let B be the union of all U in O. By case 3 there is a partition of unity for B; this is also a partition of unity for A.

Remarks

(a) An important consequence of condition (2) of the theorem should be noted. Let C ⊂ A be
compact. For each x ∈ C there is an open set V_{x} containing x such that only finitely many
ϕ ∈ Φ are not 0 on V_{x}. Since C is compact, finitely many such V_{x} cover C. Thus only finitely
many ϕ ∈ Φ are not 0 on C.

(b) One important application of partitions of unity will illustrate their main role-piecing to- gether results obtained locally.

Definition Let A ⊆ R^{n} be an open subset and let O be an open cover of A. Then O is an
admissible open cover if every U ∈O is contained in A.

LetO be an admissible open cover of the open set A ⊆ R^{n}, Φ be a partition of unity for A
subordinate toO, and let f : A → R be bounded in some open set around each point of A.

Then we say f is integrable (in the extended sense) on A if X

ϕ∈Φ

Z

A

ϕ · |f | converges. This

implies convergence of X

ϕ∈Φ

| Z

A

ϕ · f |, and hence absolute convergence ofX

ϕ∈Φ

Z

A

ϕ · f, which we define to be

Z

A

f, i.e. if X

ϕ∈Φ

Z

A

ϕ · |f | converges, then Z

A

f :=X

ϕ∈Φ

Z

A

ϕ · f.

Furthermore,

• if {x | f is discontinuous at x} has measure 0, since {x | ϕ · |f | is discontinuous at x}

is a compact subset of supp ϕ which has content 0, each Z

A

ϕ · |f | exists,

• if {x | f is discontinuous at x} has measure 0, then f is integrable if and only if |f | is integrable on any closed cell K ⊂ A. So, f is integrable (in the extended sense) if X

ϕ∈Φ

Z

A

ϕ · |f | converges.

• these definitions do not depend on O or Φ.

Remarks

(a) Recall that a subset A of R^{n} has (n-dimensional)content 0if for every ε > 0 there is afinite
cover {K_{1}, . . . , K_{m}} of A by closed (n-dimensional) cells such that

m

X

j=1

c(K_{i}) < ε. A subset
A of R^{n} has (n-dimensional)measure 0 if for every ε > 0 there is a cover {K_{1}, K_{2}, K_{3}, . . .}

of A by closed (n-dimensional) cells such that

∞

X

j=1

c(K_{i}) < ε.

A bounded set C whose boundary has measure 0 is calledJordan-measurable.

(b) Theorem If A = A_{1}∪ A_{2}∪ A_{3}∪ · · · and each A_{i} has measure 0, then A has measure 0.

Proof Let ε > 0. Since A_{i} has measure 0, there is a cover {K_{i,1}, K_{i,2}, K_{i,3}, . . .} of A_{i} by
closed (n-dimensional) cells such that

∞

X

j=1

c(K_{i,j}) < ε
2^{i}.

Then the collection of all K_{i,j} is a cover of A. By considering the array
K_{1,1}, K_{2,1}, K_{1,2}, K_{3,1}K_{2,2}, K_{1,3}, K_{4,1}, . . .
we see that this collection can be arranged in a sequence

U_{1}, U_{2}, U_{3}, . . . .
Clearly

∞

X

i=1

c(U_{i}) ≤

∞

X

i=1

ε
2^{i} = ε.

(c) Theorem If A is compact and has measure 0, then A has content 0.

Proof Let ε > 0. Since A has measure 0, there is a cover {K_{1}, K_{2}, . . .} of A by open
rectangles such that

∞

X

i=1

v(K_{i}) < ε. Since A is compact, a finite number K_{1}, . . . , K_{n} of the

Ki also cover A and surely

n

X

i=1

v(Ki) < ε.

Remark Consider the example A = [0, 1] ∩ Q. Note that A is not compact, A has measure 0, and A does not have content 0.

Theorem

(1) If Ψ is another partition of unity, subordinate to an admissible coverF of A, thenX

ψ∈Ψ

Z

A

ψ ·

|f | also converges, and

X

ϕ∈Φ

Z

A

ϕ · f =X

ψ∈Ψ

Z

A

ψ · f.

(2) If A and f are bounded, then f is integrable in the extended sense.

(3) If A is Jordan-measurable and f is bounded, then this definition of Z

A

f agrees with the old one.

Proof

(1) Since ϕ · f = 0 except on some compact set C, then there are only finitely many ψ ∈ Ψ, which are non-zero on C, and we can write

X

ϕ∈Φ

Z

A

ϕ · f =X

ϕ∈Φ

Z

A

X

ψ∈Ψ

ψ · ϕ · f =X

ϕ∈Φ

X

ψ∈Ψ

Z

A

ψ · ϕ · f.

This result, applied to |f |, shows the convergence of X

ϕ∈Φ

X

ψ∈Ψ

Z

A

ψ · ϕ · |f |,

and hence of

X

ϕ∈Φ

X

ψ∈Ψ

| Z

A

ψ · ϕ · f |.

This absolute convergence justifies interchanging the order of summation in the above equation;

the resulting double sum clearly equals X

ψ∈Ψ

Z

A

ψ · f so that

X

ϕ∈Φ

Z

A

ϕ · f =X

ϕ∈Φ

X

ψ∈Ψ

Z

A

ψ · ϕ · f = X

ψ∈Ψ

X

ϕ∈Φ

Z

A

ϕ · ψ · f =X

ψ∈Ψ

Z

A

ψ · f.

Finally, this result applied to |f | proves convergence of X

ψ∈Ψ

Z

A

ψ · |f |.

(2) If A is contained in the closed cell B and |f (x)| ≤ M for x ∈ A, and F ⊂ Φ is finite, then X

ϕ∈F

Z

A

ϕ · |f | ≤X

ϕ∈F

M Z

A

ϕ ≤ M Z

A

X

ϕ∈F

ϕ ≤ M v(B),

since X

ϕ∈F

ϕ ≤ 1 on A.

(3) If ε > 0, since A is Jordan-measurable, there is a compact Jordan-measurable C ⊂ A such

that Z

A\C

1 < ε,

and there are only finitely many ϕ ∈ Φ which are non-zero on C. If F ⊂ Φ is any finite collection which includes these, and

Z

A

f has its old meaning, then

Z

A

f −X

ϕ∈F

Z

A

ϕ · f

≤ Z

A

f −X

ϕ∈F

ϕ · f

≤ M Z

A

1 −X

ϕ∈F

ϕ

!

= M Z

A

X

ϕ∈Φ\F

ϕ

≤ M Z

A\C

1

≤ M ε.

Sard’s Theorem Let U ⊂ R^{n} be open, g : U → R^{n}∈ C^{1}(U ), and let
B = {x ∈ U | det Dg(x) = 0}.

Then g(B) has measure 0.

Proof Let K ⊂ U be a closed (n-dimensional) cell such that all sides of K have length r, say.

Let ε > 0. If N ∈ N is sufficiently large and K is divided into N^{n} cells, with sides of length r
N,
then for each of these cells S, if x ∈ S, we have

(∗) kDg(x)(y − x) − g(y) − g(x)k < εky − xk ∀ y ∈ S.

If S ∩ B 6= ∅, we can choose x ∈ S ∩ B, since det Dg(x) = 0, the kernel space of Dg(x) is not
trivial and the range set lies in an (n − 1)-dimensional subpace V of R^{n}, i.e.

{Dg(x)(y − x) | y ∈ S} ⊂ V ⊆ R^{n−1},
and, by (∗), the set

{g(y) − g(x) | y ∈ S} lies within ε√ nr N of V, so that

{g(y) | y ∈ S} lies within ε√ nr

N of V + g(x).

On the other hand, since g ∈ C^{1}(U ) and by the Mean Value Theorem, there is an M > 0 such
that

kg(x) − g(y)k < M kx − yk ≤ M√ nr

N ∀ x, y ∈ S ⊂ K.

Thus if S ∩ B 6= ∅, the set {g(y) | y ∈ S} is contained in a cylinder

• whose height is < 2ε√ nr N , and

• whose base is an (n − 1)-dimesional sphere of radius < M√ nr

N .

Since this cylinder has volume < C

r N

n

ε for some constant C, and there are at most N^{n}such
cells S, so

g(K ∩ B) lies in a set of volume < Cr N

n

εN^{n}= Cr^{n}ε.

Since this is true for all ε > 0, the set g(K ∩ B) has measure 0. Since we cover U with a sequence of such cells K, g(B) = g(U ∩ B) has measure 0.

Tensors on Vector Space

Definition Let V be a vector space over R, we will denote a k-fold product V × · · · × V by V^{k}.
A function T : V^{k} → R is calledmultilinear if for each i with 1 ≤ i ≤ k, for all v_{1}, . . . , v_{k}, w_{i} ∈ V
and for all a ∈ R, we have

• T (v_{1}, . . . , v_{i}+ w_{i}, . . . , v_{k}) = T (v_{1}, . . . , v_{i}, . . . , v_{k}) + T (v_{1}, . . . , w_{i}, . . . , v_{k}),

• T (v_{1}, . . . , av_{i}, . . . , v_{k}) = aT (v_{1}, . . . , v_{i}, . . . , v_{k}).

A multilinear function T : V^{k}→ R is called ak-tensor on V and the set of all k-tensors, denoted
byT ^{k}(V ), becomes a vector space (over R) if for S, T ∈ T^{k}(V ), and a ∈ R, we define

• (S + T )(v_{1}, . . . , v_{k}) = S(v_{1}, . . . , v_{k}) + T (v_{1}, . . . , v_{k}),

• (aS)(v_{1}, . . . , v_{k}) = a · S(v_{1}, . . . , v_{k}).

There is also an operation connecting the various spacesT ^{k}(V ). If S ∈T ^{k}(V ) and T ∈ T ^{`}(V ),
we define the tensor product S ⊗ T ∈T^{k+`}(V ) by

S ⊗ T (v_{1}, . . . , v_{k}, v_{k+1}, . . . , v_{k+`}) = S(v_{1}, . . . , v_{k}) · T (v_{k+1}, . . . , v_{k+`})

for all v_{1}, . . . , v_{k+`} ∈ V. Note that the order of the factors S and T is crucial here since S ⊗ T
and T ⊗ S are far from equal. It is easy to check that ⊗ satisfies the following properties.

• (S1+ S2) ⊗ T = S1⊗ T + S2⊗ T,

• S ⊗ (T_{1}+ T_{2}) = S ⊗ T_{1}+ S ⊗ T_{2},

• (aS) ⊗ T = S ⊗ (aT ) = a(S ⊗ T ),

• (S ⊗ T ) ⊗ U = S ⊗ (T ⊗ U ).

Due to the above associative law of ⊗, both (S ⊗ T ) ⊗ U and S ⊗ (T ⊗ U ) are usually denoted
S ⊗ T ⊗ U ; higher-order products T_{1} ⊗ · · · ⊗ T_{r} are defined similarly.

Theorem Let V be a finite dimensional vector space over R and let V^{∗} denote the dual space
of V. Then

• T^{1}(V ) = V^{∗}.

• Let v_{1}, . . . , v_{n} be a basis for V, and let ϕ_{1}, . . . , ϕ_{n} be the dual basis such that
ϕi(vj) = δij =

(1 if i = j, 0 if i 6= j.

Then the set of all k-fold tensor products

{ϕ_{i}_{1} ⊗ · · · ⊗ ϕ_{i}_{k} | 1 ≤ i_{1}, . . . , i_{k} ≤ n}

is a basis ofT ^{k}(V ), which therefore has dimension n^{k}.

Proof If w_{1}, . . . , w_{k} are k vectors with w_{i} =

n

X

j=1

a_{ij}v_{j} and T ∈T^{k}(V ), then

T (w_{1}, . . . , w_{k}) =

n

X

j1,...,j_{k}=1

a_{1,j}_{1}· · · a_{k,j}_{k}T (v_{j}_{1}, . . . , v_{j}_{k})

=

n

X

i1,...,ik=1

T (v_{i}_{1}, . . . , v_{i}_{k}) · ϕ_{i}_{1} ⊗ · · · ⊗ ϕ_{i}_{k}(w_{1}, . . . , w_{k}).

Thus for each T ∈T ^{k}(V ), T =

n

X

i1,...,ik=1

T (v_{i}_{1}, . . . , v_{i}_{k}) · ϕ_{i}_{1} ⊗ · · · ⊗ ϕ_{i}_{k}.
Consequently, {ϕ_{i}_{1} ⊗ · · · ⊗ ϕ_{i}_{k} | 1 ≤ i_{1}, . . . , i_{k} ≤ n} span T ^{k}(V ).

Suppose now that there are numbers ai1,...,i_{k} such that

n

X

i1,...,i_{k}=1

ai1,...,i_{k}· ϕi1 ⊗ · · · ⊗ ϕi_{k} = 0.

Applying both sides of this equation to (v_{j}_{1}, . . . , v_{j}_{k}) yields a_{j}_{1}_{,...,j}_{k} = 0.

Thus {ϕ_{i}_{1} ⊗ · · · ⊗ ϕ_{i}_{k} | 1 ≤ i_{1}, . . . , i_{k} ≤ n} are linearly independent.

Let V and W be vector spaces over R and let f^{∗} : V → W be a linear transformation. Then f∗

induces a linear transformation

• f^{∗} : W^{∗} → V^{∗} is defined by

f^{∗}T (v) = T (f∗(v)) for T ∈ W^{∗} =T^{1}(W ) and v ∈ V.

• f^{∗} :T ^{k}(W ) →T ^{k}(V ) is defined by

f^{∗}T (v_{1}, . . . , v_{k}) = T (f∗(v_{1}), . . . , f∗(v_{k}))) for T ∈T^{k}(W ) and v_{1}, . . . , v_{k} ∈ V.

It is easy to verify that f^{∗}(S ⊗ T ) = f^{∗}S ⊗ f^{∗}T.

Note that theinner product h , i on R^{n}is a symmetric, positive definite 2-tensor in T^{2}(R^{n}). We
define an inner product on a vector space V over R, to be a 2-tensor T ∈ T ^{2}(V ) such that T is
symmetric, that is T (v, w) = T (w, v) for v, w ∈ V and such that T is positive definite, that is,
T (v, v) > 0 if v 6= 0.

Theorem If T is an inner product on V, there is a basis v_{1}, . . . , v_{n} for V such that
T (v_{i}, v_{j}) = δ_{ij}. (Such a basis is called orthonormal with respect to T.)
Consequently there is an isomorphism

f∗ : R^{n}→ V
such that

T (f∗(x), f∗(y)) = hx , yi for x, y ∈ R^{n}.
In other words f^{∗}T = h , i.

Proof Let w_{1}, . . . , w_{n} be any basis for V. Define
w_{1}^{0} = w_{1}

w_{2}^{0} = w_{2}− T (w_{2}, w^{0}_{1})
T (w_{1}^{0}, w^{0}_{1}) · w_{1}^{0}
w_{3}^{0} = w_{3}− T (w_{3}, w^{0}_{1})

T (w_{1}^{0}, w^{0}_{1}) · w_{1}^{0} − T (w_{3}, w^{0}_{2})
T (w^{0}_{2}, w^{0}_{2})· w^{0}_{2}

· · · ·
w^{0}_{n} = w_{n}−

n−1

X

i=1

T (w_{n}, w_{i}^{0})
T (w_{i}^{0}, w^{0}_{i}) · w^{0}_{i}
It is easy to check that

(T (w_{i}^{0}, w^{0}_{j}) = 0 if i 6= j,
T (w_{i}^{0}, w^{0}_{j}) > 0 if i = j.

Now define v_{i} = w_{i}^{0}

pT (w^{0}_{i}, w_{i}^{0}). The isomormorphism f : R^{n}→ V may be defined by f (e_{i}) = v_{i}.
Despite its importance, the inner product plays a far lesser role than another familiar, seemingly
ubiquitous function , the tensor det ∈ T^{n}(R^{n}). In attempting to generalize this function, we
recall that interchanging two rows of a matrix changes the sign of its determinant. This suggests
the following definition. A k-tensor w ∈T^{k}(V ) is called alternating k-tensor on V if

w(v_{1}, . . . ,v_{i}, . . . ,v_{j}, . . . , v_{k}) = −w(v_{1}, . . . ,v_{j}, . . . ,v_{i}, . . . , v_{k}) ∀ v_{1}, . . . , , v_{k} ∈ V.

(In this equation v_{i} and v_{j} are interchanged and all other v’s are left fixed.) The set of all
alternating k-tensors is clearly a subspace Λ^{k}(V ) of T ^{k}(V ). Since it requires considerable work
to produce the determinant, it is not surprising that alternating k-tensors are difficult to write
down . There is, however, a uniform way of expressing all of them. Recall that the sign of a
permutation σ, denoted sgn σ, is defined by

sgn σ =

(+1 if σ is even,

−1 if σ is odd.

If T ∈T ^{k}(V ), we define Alt (T ) by

Alt (T )(v_{1}, . . . , v_{k}) = 1
k!

X

σ∈S_{k}

sgn σ · T (v_{σ(1)}, . . . , v_{σ(k)}),

where S_{k} is the set of all permutations of numbers 1 to k.

Theorem

(1) If T ∈ T^{k}(V ), then Alt (T ) ∈ Λ^{k}(V ).

(2) If w ∈ Λ^{k}(V ), then Alt (w) ∈ Λ^{k}(V ).

(3) If T ∈ T^{k}(V ), then Alt (Alt (T )) = Alt (T ).

Proof

(1) Let (i, j) be the permutation that interchanges i and j and leaves all other numbers fixed. If
σ ∈ S_{k}, let σ^{0} = σ · (i, j). Since S_{k} = {σ^{0} = σ · (i, j) | σ ∈ S_{k}}, we have

Alt (T )(v_{1}, . . . , v_{j}, . . . , v_{i}, . . . , v_{k})

= 1

k!

X

σ∈Sk

sgn σ · T (vσ(1), . . . , vσ(j), . . . , vσ(i), . . . , vσ(k))

= 1

k!

X

σ∈Sk

sgn σ · T (v_{σ}^{0}_{(1)}, . . . , v_{σ}^{0}_{(i)}, . . . , v_{σ}^{0}_{(j)}, . . . , v_{σ}^{0}_{(k)})

= 1

k!

X

σ^{0}∈S_{k}

−sgn σ^{0}· T (v_{σ}^{0}_{(1)}, . . . , v_{σ}^{0}_{(i)}, . . . , v_{σ}^{0}_{(j)}, . . . , v_{σ}^{0}_{(k)})

= −Alt (T )(v_{1}, . . . , v_{k})
(2) Note that if w ∈ Λ^{k}(V ), and σ = (i, j), then

w(v_{σ(1)}, . . . , v_{σ(k)}) = sgn σ · w(v_{1}, . . . , v_{k}) ∀ v_{1}, . . . , , v_{k} ∈ V.

Since every σ ∈ Sk is a product of permutations of the form (i, j), this equation holds of all
σ ∈ S_{k}. Therefore

Alt (w)(v_{1}, . . . , v_{k}) = 1
k!

X

σ∈Sk

sgn σ · w(v_{σ(1)}, . . . , v_{σ(k)})

= 1

k!

X

σ∈Sk

sgn σ · sgn σ · w(v_{1}, . . . , v_{k})

= w(v_{1}, . . . , v_{k})
(3) follows immediately from (1) and (2).

Definition If w ∈ Λ^{k}(V ) and η ∈ Λ^{`}(V ), we define the wedge product w ∧ η ∈ Λ^{k+`}(V ) by
w ∧ η = (k + `)!

k! `! Alt (w ⊗ η).

It is easy to check that ∧ satisfies the following properties.

• (w_{1}+ w_{2}) ∧ η = w_{1}∧ η + w_{2}∧ η,

• w ∧ (η1+ η2) = w ∧ η1+ w ∧ η2,

• aw ∧ η = w ∧ aη = a(w ∧ η),

• w ∧ η = (−1)^{k`}η ∧ w,

Proof Consider τ ∈ S_{k+`} defined by

τ (j) =

1 · · · k k + 1 · · · k + `

` + 1 · · · ` + k 1 · · · k

=

(` + j if 1 ≤ j ≤ k,

j − k if k + 1 ≤ j ≤ k + `,

and note that sgn τ = (−1)^{k`} and S_{k+`}= {σ · τ | σ ∈ S_{k+`}}. Then
Alt (η ⊗ w)(v1, . . . , vk+`) = 1

(k + `)!

X

σ·τ ∈Sk+`

sgn (σ · τ ) · η ⊗ w(v_{σ·τ (1)}, . . . , v_{σ·τ (k+`)})

= 1

(k + `)!

X

σ∈Sk+`

sgn τ sgn σ · w ⊗ η(v_{σ(1)}, . . . , v_{σ(k+`)})

= (−1)^{k`}Alt (w ⊗ η)(v_{1}, . . . , v_{k+`})

• f^{∗}(w ∧ η) = f^{∗}(w) ∧ f^{∗}(η).

Proof For any v1, . . . , vk+` ∈ V,

f^{∗}(w ∧ η)(v_{1}, . . . , v_{k+`}) = w ∧ η (f∗(v_{1}), . . . , f∗(v_{k+`}))

= 1

(k + `)!

X

σ∈Sk+`

sgn σ · w ⊗ η(f∗(v_{σ(1)}), . . . , f∗(v_{σ(k+`)}))

= 1

(k + `)!

X

σ∈Sk+`

sgn σ · f^{∗}w ⊗ f^{∗}η(v_{σ(1)}, . . . , v_{σ(k+`)})

= f^{∗}(w) ∧ f^{∗}(η)(v_{1}, . . . , v_{k+`})

Theorem

(1) If S ∈T ^{k}(V ) and T ∈ T^{`}(V ) and Alt (S) = 0, then

Alt (S ⊗ T ) = Alt (T ⊗ S) = 0.

(2) Alt (Alt (w ⊗ η) ⊗ θ) = Alt (w ⊗ η ⊗ θ) = Alt (w ⊗ Alt (η ⊗ θ)).

(3) If w ∈ Λ^{k}(V ), η ∈ Λ^{`}(V ), and θ ∈ Λ^{m}(V ), then

(w ∧ η) ∧ θ = w ∧ (η ∧ θ) = (k + ` + m)!

k! `! m! Alt (w ⊗ η ⊗ θ).

Theorem If {v_{i}}^{n}_{i=1} is a basis for V and {ϕ_{i}}^{n}_{i=1} is the dual basis such that
ϕ_{i}(v_{j}) = δ_{ij} =

(1 if i = j, 0 if i 6= j.

Then the set of all

{ϕ_{i}_{1} ∧ · · · ∧ ϕ_{i}_{k} | 1 ≤ i_{1} < i_{2} < . . . < i_{k} ≤ n}

is a basis of Λ^{k}(V ), which therefore has dimension

n k

= n!

k! (n − k)!.

Proof Since {ϕ_{i}_{1} ⊗ · · · ⊗ ϕ_{i}_{k} | 1 ≤ i_{1}, . . . , i_{k}≤ n} is a basis of T ^{k}(V ), and since
ϕ_{i}_{1} ∧ · · · ∧ ϕ_{i}_{k} = k! Alt (ϕ_{i}_{1} ⊗ · · · ⊗ ϕ_{i}_{k}),

and ϕ_{i}∧ ϕ_{i} = 0 for all 1 ≤ i ≤ n, so the set of all

{ϕ_{i}_{1} ∧ · · · ∧ ϕ_{i}_{k} | 1 ≤ i_{1} < i_{2} < . . . < i_{k} ≤ n}

is a basis of Λ^{k}(V ).