Riemann Integration

Partition

A partition of

[a, b]

is a finite tuple sorted in ascending order

P = (x_{1}, x_{2}, \dots x_{n + 1})

such that

a = x_{1}

and

b = x_{n + 1}

i-th Section of a Partition

Suppose that

P = (x_{1}, \dots x_{n + 1})

is a partition, then for each

i \in [1, \dots, n]

we define

s_{i} : = [x_{i}, x_{i + 1}]

Supremum of a Bounded Function on a Section

Suppose that

f : [a, b] \to ℝ

and that

P = (x_{1}, \dots, x_{n + 1})

is a partition of

[a, b]

then for each

i \in [1, \dots, n]

we have

M_{i} (f, P) = \sup ({f (x) : x \in [x_{i}, x_{i + 1}]})

i-th Delta of a Partition

Suppose that

P = (x_{1}, \dots, x_{n + 1})

is a partition, then for each

i \in [1, \dots n]

we define:

Δ_{i} = x_{i + 1} - x_{i}

Note that if a partition has $k$ elements then there will be $k - 1$ deltas.

A Delta Sum Telescopes

Suppose that

P = (x_{1}, \dots, x_{n + 1})

is a partition of

[a, b]

then

\sum_{i = 1}^{n} Δ_{i} = b - a

Recall that

\sum_{i = 1}^{n} Δ_{i} = \sum_{i = 1}^{n} (x_{i + 1} - x_{i}) \overset{1}{=} x_{n + 1} - x_{1} = b - a

where we've used fact 1.

Mesh of a Partition

Suppose that

P

is a partition of

[a, b]

such that

| P | = n + 1

for some

n \in ℕ_{0}

then

mesh (P) = \max ({Δ_{i} : i \in [1, \dots, n]})

Upper Sum of a Bounded Function over a Partition

Suppose that

f : [a, b] \to ℝ

is bounded, and that

P

is a partition of

[a, b]

then we define

U (f, P) : = \sum_{j = 1}^{n} M_{j} (f, P) Δ_{j}

Lower Sum of a Bounded Function over a Partition

Suppose that

f : [a, b] \to ℝ

is bounded, and that

P

is a partition of

[a, b]

then we define

L (f, P) : = \sum_{j = 1}^{n} m_{j} (f, P) Δ_{j}

Riemann Integrable

Suppose that

f : [a, b] \to ℝ

is bounded, then we say that it is Riemann integrable if

\sup_{P} (L (f, P)) = \inf_{Q} (U (f, Q))

In that case we write

\int_{a}^{b} f (x) d x

as their common value

Riemann Integrable Characterizations

Let

f (x)

be bounded on

[a, b]

then the following are equivalent

$f$ is Riemann integrable
For each $ϵ \in ℝ^{+}$ there is a partition $P$ such that

U (f, P) - L (f, P) < ϵ

For every $ϵ \in ℝ^{+}$ there is some $δ \in ℝ^{+}$ such that for every partition $Q$ such that $mesh (Q) < δ$ wherein

U (f, Q) - L (f, Q) < ϵ

There exists an $I \in ℝ$ such that for every $ϵ \in ℝ^{+}$ there is a $δ \in ℝ^{+}$ such that for every partition $Q$ with $mesh (Q) < δ$ and every evaluation sequence $X$ of $Q$ we have

| I (f, Q, X) - I | < ϵ

I = \int_{a}^{b} f (x) d x

Note that bullet point 2 is a the most tractable for explicit functions, as one can construct a partition that works for a given function.

Jump Discontinuity

Suppose

f : [a, b] \to ℝ

, and

c \in (a, b)

then if

\lim_{x \to c^{+}} f (x) \neq \lim_{x \to c^{-}} f (x)

then we say that

f

has a jump discontinuity at

c

Function Difference Set is the Same as its Negative

Suppose that

f : A \to ℝ

and define

F = {f (x) - f (y) : x, y \in A}

, then

F = - F

Note that

F = {f (x) - f (y) : x, y \in A} = {- (f (y) - f (x)) : y, x \in A} = - F

as needed.

Supremum of the Function Difference Set is the Same with Absolute Values

Let

f : A \to ℝ

and define

F = {f (x) - f (y) : x, y \in A}

then

\sup (F) = \sup (| F |)

Recall that if

F = - F

then we know

\sup (F) = \sup (| F |)

Supremum of the Sum of Two Functions is Less than or Equal to the Sum of the Supremums

Suppose that

f, g : A \to ℝ

then

\sup ({f (x) + g (x) : x \in A}) \leq \sup ({f (x) : x \in A}) + \sup ({g (x) : x \in A})

Let $α = \sup ({f (x) : x \in A}), β = \sup ({g (x) : x \in A})$ and $γ = \sup ({f (x) + g (x) : x \in A})$ .

We know that for any $x \in A$ we have that $f (x) \leq α$ and $g (x) \leq β$ therefore we have that $f (x) + g (x) \leq α + β$ , this proves that $α + β$ is an upper bound to the set ${f (x) + g (x) : x \in A}$ therefore we have $\sup ({f (x) + g (x) : x \in A}) \leq \sup ({f (x) : x \in A}) + \sup ({f (x) : x \in A})$ As $γ$ is the least upper bound.

Supremum of the Sum of two Functions is the Same as the Sum of the Supremums if the Variables are Independent

Suppose that

f, g : A \to ℝ

then

\sup ({f (x) + g (y) : x, y \in A}) = \sup ({f (x) : x \in A}) + \sup ({g (x) : x \in A})

Let $α = \sup ({f (x) : x \in A}), β = \sup ({g (x) : x \in A})$ and $γ = \sup ({f (x) + g (y) : x, y \in A})$ .

For any $x \in A$ we have that $f (x) \leq α$ and for any $y \in A$ we have $g (x) \leq β$ thus we have $f (x) + g (y) \leq α + β$ so that $α + β$ is an upper bound of the set $γ$ is supping over there fore we have that $γ \leq α + β$

... [TODO] not sure how to prove the other direction ...

Bound on the Upper minus Lower Sum for The Same Partition

Suppose that

f : [a, b] \to ℝ

is bounded, and let

P = {x_{1}, \dots, x_{n + 1}}

be a partition of

[a, b]

then we have that

U (f, P) - L (f, P) \leq 2 {‖ f ‖}_{\infty} (b - a)

\begin{matrix} U (f, P) - L (f, P) & = \sum_{i = 1}^{n} (\sup_{x \in [x_{i}, x_{i + 1}]} f (x) - \inf (x \in [x_{i}, x_{i + 1}] f (x))) Δ_{i} \\ \overset{1}{=} \sum_{i = 1}^{n} (\sup_{x \in [x_{i}, x_{i + 1}]} f (x) + \sup (x \in [x_{i}, x_{i + 1}] - f (x))) Δ_{i} \\ \overset{2}{=} \sum_{i = 1}^{n} (\sup_{x, y \in [x_{i}, x_{i + 1}]} (f (x) - f (y))) Δ_{i} \\ \overset{3}{=} \sum_{i = 1}^{n} (\sup_{x, y \in [x_{i}, x_{i + 1}]} (| f (x) - f (y) |)) Δ_{i} \\ \overset{4}{\leq} \sum_{i = 1}^{n} (2 \sup_{x, y \in [x_{i}, x_{i + 1}]} (| f (x) |)) Δ_{i} \\ \overset{5}{\leq} \sum_{i = 1}^{n} (2 {‖ f ‖}_{\infty}) Δ_{i} \\ = 2 {‖ f ‖}_{\infty} \sum_{i = 1}^{n} Δ_{i} \\ \overset{6}{=} 2 {‖ f ‖}_{\infty} (b - a) \end{matrix}

Where we've used

fact 1
fact 2
fact 3
fact 4
fact 5: The supremum gets bigger on a superset
fact 6

Piecewise Continuous

A function

f : [a, b] \to ℝ

is called piecewise continuous if for every

[c, d] \subseteq [a, b]

if it only has a finite number of discontinuities all of which are jump discontinuities

Riemann Integrable on Two Parts of the Interval Means Riemann Integrable on the Whole Interval

Suppose that

f : [a, b] \to ℝ

and let

x \in [a, b]

then if

f

restricted to

[a, x]

and

f

restricted to

[x, b]

(which we denote as

g_{1}, g_{2}

) are both riemann integrable then so is

f

[a, b]

and then

\int_{a}^{b} f (x) d x = \int_{a}^{x} f (x) d x + \int_{x}^{b} f (x) d x

Let

ϵ \in ℝ^{+}

using bullet point 2 from the RMI characterization we know that there are partitions

P_{1}, P_{2}

[a, x], [x, b]

such that

(U (g_{1}, P_{1}) - L (g_{1}, P_{1})) + (U (g_{2}, P_{2}) - L (g_{2}, P_{2})) < ϵ

Then note that

U (g_{1}, P_{1}) + U (g_{2}, P_{2}) = U (f, P_{1} \cup P_{2})

which also holds for the lower sums, and therefore from the previous inequality we have

U (f, P_{1} \cup P_{2}) - L (f, P_{1} \cup P_{2}) < ϵ

therefore by (2) of the RMI characterization

f

is Riemann integrable.

Every Piecewise Continuous Function is Riemann Integrable

Every piecewise continuous function is Riemann integrable

Since $f$ is piecewise continuous, then suppose its jump discontinuity points are given by $D = {d_{1}, d_{2}, \dots d_{n}}$ . We will show this by using strong induction on the number of discontinuities within $[a, b]$ .

Note that one can manage to prove that if $f$ is RMI over $[a, b]$ with $1 \leq i \leq k$ jump discontinuities, then it is RMI over $[a, b]$ if it has $k + 1$ discontinuities. This can be done because we can break $[a, b]$ into two intervals $I_{1}, I_{2}$ such that $I_{1} ⊔ I_{2} = [a, b]$ , where $I_{1}, I_{2}$ each have a number of discontinuities within the range $[1, \dots, k]$ so that $f ↾ I_{1}$ and $f ↾ I_{2}$ are both integrable on their respective intervals, so that $f$ is RMI on $[a, b]$ .

Therefore all that remains is the base case, which is that $f$ is RMI if there is exactly $1$ jump discontinuity of $f$ on the set $[a, b]$ , let $d$ be this discontinuity, since $f$ is continuous on $L : = [a, d - δ]$ and $R : = [d + δ, b]$ for any small $δ$ , then we know that $f$ is RMI when restricted to these intervals.

Let $ϵ \in ℝ^{+}$ by (2) of the RMI characterization we obtain $P_{L}, P_{R}$ which are partitions of $L, R$ respectively (note that $P_{L} \cap P_{R} = \emptyset$ such that $U (f ↾ L, P_{L}) - L (f ↾ L, P_{L}) < \frac{ϵ}{3}$ and $U (f ↾ R, P_{R}) - L (f ↾ R, P_{R}) < \frac{ϵ}{3}$

Finally Consider $P_{L} \cup P_{R}$ which is a partition of $[a, b]$ which contains the section $[d - δ, d + δ]$ along with the left and right partitions. Focusing on this section we can see it will contribute $(\sup_{x \in [d - δ, d + δ]} f (x) - \sup_{y \in [d - δ, d + δ]} f (y)) 2 δ$ to $U (f, P_{L} \cup P_{R}) - L (f, P_{L} \cup P_{R})$ , now:

\begin{matrix} U (f, P_{L} \cup P_{R}) - L (f, P_{L} \cup P_{R}) & = (U (f ↾ L, P_{L}) + (\sup_{x \in [d - δ, d + δ]} f (x)) 2 δ + U (f ↾ R, P_{R})) - (L (f ↾ L, P_{L}) + (\inf_{x \in [d - δ, d + δ]} f (x)) 2 δ + L (f ↾ R, P_{R})) \\ = (U (f ↾ L, P_{L}) - L (f ↾ L, P_{L})) + (\sup_{x \in [d - δ, d + δ]} f (x) - \inf_{x \in [d - δ, d + δ]} f (x)) 2 δ + (U (f ↾ R, P_{R}) - L (f ↾ L, P_{L})) \\ < \frac{2 ϵ}{3} + \sup_{x, y \in [d - δ, d + δ]} (| f (x) - f (y) |) 2 δ \\ \leq \frac{2 ϵ}{3} + 2 \sup_{x \in [d - δ, d + δ]} | f (x) | \\ \leq \frac{2 ϵ}{3} + 2 {‖ f ‖}_{\infty} 2 δ \\ = \frac{2 ϵ}{3} + 4 {‖ f ‖}_{\infty} δ \end{matrix}

Thus a selection of

δ = \frac{ϵ}{4 {‖ f ‖}_{\infty} 3}

shows that

U (f, P_{L} \cup P_{R}) - L (f, P_{L} \cup P_{R}) < ϵ

as needed.

Volume of a Rectangle

Suppose that

A = [a_{1}, b_{1}] \times \dots \times [a_{n}, b_{n}] \subseteq ℝ^{n} .

The volume of the closed rectangle

A

v (A) = \prod_{i = 1}^{n} (b_{i} - a_{i}) .

We use the same formula for the corresponding open rectangle

(a_{1}, b_{1}) \times \dots \times (a_{n}, b_{n})

Partition of a Closed Rectangle

Suppose that

A = [a_{1}, b_{1}] \times \dots \times [a_{n}, b_{n}]

is a closed rectangle. A partition of

A

is a tuple

P = (P_{1}, \dots, P_{n}),

where each

P_{i}

is a partition of the interval

[a_{i}, b_{i}]

Subrectangle of a Partition

Let

P = (P_{1}, \dots, P_{n})

be a partition of a closed rectangle

A

, and write

P_{i} = (t_{i, 0}, t_{i, 1}, \dots, t_{i, N_{i}}) .

A subrectangle of

P

is a rectangle of the form

[t_{1, j_{1} - 1}, t_{1, j_{1}}] \times \dots \times [t_{n, j_{n} - 1}, t_{n, j_{n}}],

where

1 \leq j_{i} \leq N_{i}

for every

i

Multivariable Upper and Lower Sums

Suppose that

A \subseteq ℝ^{n}

is a closed rectangle,

f : A \to ℝ

is bounded, and

P

is a partition of

A

. For each subrectangle

S

P

, define

m_{S} (f) = \inf {f (x) : x \in S}

and

M_{S} (f) = \sup {f (x) : x \in S} .

The lower sum and upper sum of

f

over

P

are

L (f, P) = \sum_{S} m_{S} (f) v (S)

and

U (f, P) = \sum_{S} M_{S} (f) v (S),

where

v (S)

is the volume of the rectangle

S

, and the sums range over all subrectangles

S

P

Riemann Integrable on a Closed Rectangle

Suppose that

A \subseteq ℝ^{n}

is a closed rectangle and

f : A \to ℝ

is bounded. We say that

f

is Riemann integrable on

A

\sup_{P} L (f, P) = \inf_{P} U (f, P),

where

P

ranges over all partitions of

A

. The common value is denoted

\int_{A} f

\int_{A} f (x^{1}, \dots, x^{n}) d x^{1} \dots d x^{n} .

Multivariable Lower Sums are Below Upper Sums

P

and

Q

are partitions of a closed rectangle

A

, then

L (f, P) \leq U (f, Q) .

Multivariable Riemann Integrability Criterion

Suppose that

A \subseteq ℝ^{n}

is a closed rectangle and

f : A \to ℝ

is bounded. Then

f

is Riemann integrable on

A

if and only if for every

ϵ \in ℝ^{+}

, there is a partition

P

A

such that

U (f, P) - L (f, P) < ϵ .

Measure Zero in Euclidean Space

A set

E \subseteq ℝ^{n}

has

n

-dimensional measure zero if for every

ϵ \in ℝ^{+}

, there is a sequence

U_{1}, U_{2}, \dots

of closed rectangles such that

E \subseteq ⋃_{i = 1}^{\infty} U_{i}

and

\sum_{i = 1}^{\infty} v (U_{i}) < ϵ .

Here

v (U_{i})

is the volume of the rectangle

U_{i}

Content Zero in Euclidean Space

A set

E \subseteq ℝ^{n}

has

n

-dimensional content zero if for every

ϵ \in ℝ^{+}

, there are finitely many closed rectangles

U_{1}, \dots, U_{k}

such that

E \subseteq U_{1} \cup \dots \cup U_{k}

and

\sum_{i = 1}^{k} v (U_{i}) < ϵ .

Here

v (U_{i})

is the volume of the rectangle

U_{i}

Countable Union of Measure Zero Sets has Measure Zero

E = ⋃_{i = 1}^{\infty} E_{i}

and each

E_{i} \subseteq ℝ^{n}

has measure zero, then

E

has measure zero.

Compact Measure Zero Implies Content Zero

E \subseteq ℝ^{n}

is compact and has measure zero, then

E

has content zero.

Lebesgue Criterion for Multivariable Riemann Integrability

Suppose that

A \subseteq ℝ^{n}

is a closed rectangle and

f : A \to ℝ

is bounded. Let

D = {x \in A : f is not continuous at x} .

Then

f

is Riemann integrable on

A

if and only if

D

has measure zero.

Jordan Measurable Set

A bounded set

C \subseteq ℝ^{n}

is Jordan measurable if its characteristic function

χ_{C}

is Riemann integrable on some closed rectangle

A

with

C \subseteq A

. In that case its Jordan content is

| C | = \int_{A} χ_{C} .

Jordan Measurable iff Boundary has Measure Zero

A bounded set

C \subseteq ℝ^{n}

is Jordan measurable if and only if its boundary has measure zero.

Integral over a Jordan Measurable Set

Suppose that

C \subseteq ℝ^{n}

is bounded,

C \subseteq A

A

is a closed rectangle, and

f : A \to ℝ

is bounded. If

f χ_{C}

is Riemann integrable on

A

, define

\int_{C} f = \int_{A} f χ_{C} .

Removing a Measure-Zero Set Does Not Change a Riemann Integral

Suppose

C \subseteq ℝ^{n}

is Jordan measurable,

E \subseteq C

has measure zero, and

f : C \to ℝ

is bounded and integrable over

C

. Then

\int_{C} f = \int_{C ∖ E} f .

Choose a closed rectangle

A

containing

C

. The two integrals are

\int_{A} f χ_{C} and \int_{A} f χ_{C ∖ E} .

These two functions differ only on

E

. Since

E

has measure zero, changing a bounded function only on

E

does not change its upper or lower Riemann integrals: rectangles covering

E

can be chosen with arbitrarily small total volume, so the possible change in any Riemann sum is bounded by the bound for

f

times an arbitrarily small total volume. Therefore the two integrals are equal.

Fubini for Riemann Integrals on Rectangles

Suppose that

A \subseteq ℝ^{n}

and

B \subseteq ℝ^{m}

are closed rectangles and

f : A \times B \to ℝ

is Riemann integrable. If for each

x \in A

, the function

y \mapsto f (x, y)

is Riemann integrable on

B

, and the function

x \mapsto \int_{B} f (x, y) d y

is Riemann integrable on

A

, then

\int_{A \times B} f = \int_{A} (\int_{B} f (x, y) d y) d x .

Support of a Function

The support of a function

φ : A \to ℝ

is the closure of the set where

φ

is nonzero:

supp (φ) = {x \in A : φ (x) \neq 0} .

Partition of Unity

Let

A \subseteq ℝ^{n}

and let

𝒰

be an open cover of

A

. A partition of unity for

A

, subordinate to

𝒰

, is a collection

Φ

of nonnegative continuous functions

φ : A \to ℝ

such that:

for each $φ \in Φ$ , there is some $U \in 𝒰$ with $supp (φ)$ $\subseteq U$ ,
each point of $A$ has a neighborhood on which only finitely many $φ \in Φ$ are nonzero,
for every $x \in A$ , $\sum_{φ \in Φ} φ (x) = 1$ .

Existence of Partitions of Unity in Euclidean Space

A \subseteq ℝ^{n}

and

𝒰

is an open cover of

A

, then there is a partition of unity for

A

subordinate to

𝒰

First suppose that

A

is compact. Choose a finite subcover

U_{1}, \dots, U_{N}

A

from

𝒰

. We construct compact sets

D_{i} \subseteq U_{i}

whose interiors cover

A

Suppose $D_{1}, \dots, D_{k}$ have been chosen so that $int D_{1}, \dots, int D_{k}, U_{k + 1}, \dots, U_{N}$ still cover $A$ . Let $C_{k + 1} = A ∖ (int D_{1} \cup \dots \cup int D_{k} \cup U_{k + 2} \cup \dots \cup U_{N}) .$ Then $C_{k + 1}$ is compact and $C_{k + 1} \subseteq U_{k + 1}$ . Since $U_{k + 1}$ is open, choose a compact neighborhood $D_{k + 1}$ with $C_{k + 1} \subseteq int D_{k + 1} and D_{k + 1} \subseteq U_{k + 1} .$ Continuing this process gives compact sets $D_{1}, \dots, D_{N}$ whose interiors cover $A$ .

For each $i$ , define a nonnegative continuous function $ψ_{i}$ that is positive on $D_{i}$ and zero outside $U_{i}$ . One explicit choice uses the Euclidean distance to a set: $ψ_{i} (x) = \frac{d (x, ℝ^{n} ∖ U_{i})}{d (x, ℝ^{n} ∖ U_{i}) + d (x, D_{i})} .$ This is continuous, equals $0$ off $U_{i}$ , and is positive on $D_{i}$ . Since the interiors of the $D_{i}$ cover $A$ , the sum $σ (x) = ψ_{1} (x) + \dots + ψ_{N} (x)$ is positive on $A$ . Set $φ_{i} (x) = \frac{ψ_{i} (x)}{σ (x)} .$ Then the $φ_{i}$ are nonnegative continuous functions, their sum is $1$ on $A$ , and the support of $φ_{i}$ is contained in $U_{i}$ . Thus they form the required partition in the compact case.

Next suppose that $A = A_{1} \cup A_{2} \cup \dots$ , where each $A_{i}$ is compact and $A_{i} \subseteq int A_{i + 1}$ . Put $A_{0} = A_{- 1} = \emptyset$ and $B_{i} = A_{i} ∖ int A_{i - 1} .$ Each $B_{i}$ is compact. Cover $B_{i}$ by the sets $U \cap (int A_{i + 1} ∖ A_{i - 2}), U \in 𝒰 .$ By the compact case, choose a partition of unity $Φ_{i}$ for $B_{i}$ subordinate to this cover. At each $x \in A$ , only finitely many of the functions from all the $Φ_{i}$ are nonzero near $x$ , because the $i$ -th family is supported in $int A_{i + 1} ∖ A_{i - 2}$ . Therefore $σ (x) = \sum_{i} \sum_{φ \in Φ_{i}} φ (x)$ is locally finite and positive on $A$ . Replacing every $φ$ by $φ / σ$ gives a partition of unity on $A$ subordinate to $𝒰$ .

If $A$ is open, define $A_{i} = {x \in A : | x | \leq i and d (x, ℝ^{n} ∖ A) \geq \frac{1}{i}} .$ These sets are compact, their union is $A$ , and $A_{i} \subseteq int A_{i + 1}$ . Hence the previous case applies.

Finally let $A$ be arbitrary and let $B = ⋃_{U \in 𝒰} U$ . The set $B$ is open and $A \subseteq B$ . Apply the open case to $B$ and then restrict the resulting functions to $A$ . The restricted functions still sum to $1$ on $A$ , are locally finite on $A$ , and have supports contained in elements of $𝒰$ . Thus they form the desired partition of unity.

Change of Variables for Riemann Integrals

Let

A \subseteq ℝ^{n}

be open and let

g : A \to ℝ^{n}

be one-to-one and continuously differentiable with

\det {[D g (x)]}_{E} \neq 0

for every

x \in A

. If

f : g (A) \to ℝ

is integrable, then

\int_{g (A)} f = \int_{A} (f \circ g) | \det {[D g]}_{E} | .

First reduce the result to local pieces. Suppose $𝒱$ is an open cover of $A$ such that the formula holds on each $V \in 𝒱$ . The sets $g (V)$ form an open cover of $g (A)$ . Choose a partition of unity $Φ$ on $g (A)$ subordinate to this cover. For each $φ \in Φ$ , choose $V_{φ} \in 𝒱$ with $supp (φ) \subseteq g (V_{φ}) .$ Since $g$ is one-to-one, the function $(φ f) \circ g$ is zero outside $V_{φ}$ . Applying the local formula to $φ f$ gives $\int_{g (A)} φ f = \int_{A} ((φ f) \circ g) | \det {[D g]}_{E} | .$ Summing over the locally finite family $Φ$ , and using $\sum_{φ \in Φ} φ = 1$ , gives $\int_{g (A)} f = \int_{A} (f \circ g) | \det {[D g]}_{E} | .$

It is enough to prove the formula for constant functions on rectangles. Indeed, let $R \subseteq g (A)$ be a closed rectangle and let $P$ be a partition of $R$ . For each subrectangle $S$ , let $m_{S} (f)$ and $M_{S} (f)$ be the lower and upper values of $f$ on $S$ . Applying the constant-function case to the characteristic pieces of $S$ gives the lower-sum inequality $L (f, P) \leq \int_{g^{- 1} (R)} (f \circ g) | \det {[D g]}_{E} |$ and similarly the upper-sum inequality in the other direction. Taking the supremum of lower sums and the infimum of upper sums, using the definition of Riemann integrability, proves the formula for $f$ .

The formula is compatible with composition. If it holds for $g : A \to ℝ^{n}$ and for $h : B \to ℝ^{n}$ , with $g (A) \subseteq B$ , then $\int_{h (g (A))} f = \int_{g (A)} (f \circ h) | \det {[D h]}_{E} |$ and applying the formula again to $g$ gives $\int_{A} (f \circ h \circ g) | \det {[D h (g (x))]}_{E} | | \det {[D g (x)]}_{E} | .$ By the chain rule and multiplicativity of determinants, this is $\int_{A} (f \circ h \circ g) | \det {[D (h \circ g)]}_{E} | .$

The formula holds for invertible linear maps because the determinant is the signed volume-scaling factor of a linear map: for every rectangle $R$ , $| T (R) | = | \det T | | R |,$ and the constant-function reduction then gives the integral formula.

We now prove the theorem by induction on $n$ . The case $n = 1$ is the ordinary one-dimensional substitution formula. Assume the theorem is known in dimension $n - 1$ , and let $a \in A$ . By the local reduction it is enough to prove the formula on some open neighborhood of $a$ . Composing with the inverse of the linear map $D g (a)$ , and using the linear case and the composition step, we may assume ${[D g (a)]}_{E} = I_{n} .$

Write $g = (g^{1}, \dots, g^{n})$ . Define $h (x^{1}, \dots, x^{n}) = (g^{1} (x), \dots, g^{n - 1} (x), x^{n}) .$ Since ${[D h (a)]}_{E} = I_{n}$ , the inverse function theorem gives an open neighborhood $U^{'}$ of $a$ on which $h$ is one-to-one and $\det {[D h]}_{E} \neq 0$ . On $h (U^{'})$ , define $k (y^{1}, \dots, y^{n}) = (y^{1}, \dots, y^{n - 1}, g^{n} (h^{- 1} (y))) .$ Then $g = k \circ h$ on $U^{'}$ . Shrinking $U^{'}$ if needed, the inverse function theorem also gives $\det {[D k]}_{E} \neq 0$ on $h (U^{'})$ . By the composition step, it remains to prove the formula for maps of the form $h$ and $k$ , each of which changes only one coordinate after the other coordinates are fixed.

We prove it for $h$ ; the proof for $k$ is the same. Let $W = D \times [a_{n}, b_{n}] \subseteq U^{'}$ , where $D \subseteq ℝ^{n - 1}$ is a rectangle. For each fixed $x^{n}$ , define $h_{x^{n}} (x^{1}, \dots, x^{n - 1}) = (g^{1} (x), \dots, g^{n - 1} (x)) .$ Each $h_{x^{n}}$ is a one-to-one continuously differentiable map in $ℝ^{n - 1}$ , and $\det {[D h_{x^{n}} (x^{1}, \dots, x^{n - 1})]}_{E} = \det {[D h (x^{1}, \dots, x^{n})]}_{E} .$ Applying the induction hypothesis to each slice and then applying Fubini's theorem gives $\int_{h (W)} 1 = \int_{a_{n}}^{b_{n}} \int_{D} | \det {[D h (x^{1}, \dots, x^{n})]}_{E} | d x^{1} \dots d x^{n - 1} d x^{n} = \int_{W} | \det {[D h]}_{E} | .$ By the constant-function reduction, the change-of-variables formula holds for $h$ on $W$ . These rectangles form a local cover of $U^{'}$ , so the local reduction proves the formula near $a$ . Since $a$ was arbitrary, the theorem follows on all of $A$ .

Polar Coordinate Change of Variables

Let

W = {(u, v) \in ℝ^{2} : u^{2} + v^{2} < R^{2}}

with

R > 0

. If

q : W \to ℝ

is integrable, then

\int_{W} q (u, v) d u d v = \int_{0}^{2 π} \int_{0}^{R} q (r \cos θ, r \sin θ) r d r d θ .

Let

ϕ (r, θ) = (r \cos θ, r \sin θ)

(0, R) \times (0,2 π)

. This map is one-to-one onto

W

with the origin and one radius removed. Its Jacobian matrix is

ϕ^{'} (r, θ) = [\begin{matrix} \cos θ & - r \sin θ \\ \sin θ & r \cos θ \end{matrix}],

| \det ϕ^{'} (r, θ) | = r .

The omitted origin and radius have measure zero, so by removing a measure-zero set, the integral over

W

equals the integral over the image of

ϕ

. Applying the change of variables theorem gives

\int_{W} q (u, v) d u d v = \int_{(0, R) \times (0,2 π)} q (r \cos θ, r \sin θ) r d r d θ .

Writing this integral as an iterated integral gives the stated formula.

Sard's Theorem for the Equal-Dimension

C^{1}

Case

Let

A \subseteq ℝ^{n}

be open and let

g : A \to ℝ^{n}

be continuously differentiable. If

B = {x \in A : \det {[D g (x)]}_{E} = 0},

then

g (B)

has measure zero.