if A = Yp j=1 [aj, bj] is a closed cell in Rp, one can define its content to be c(A

Let A ⊆ R^p and f be a function defined over A. To define R

Af, the integral of f over A, it is reasonable to require that

(a) the content (volume) of A is measurable, e.g. if A = Yp j=1

[a_j, b_j] is a closed cell in R^p, one can

define its content to be c(A) = Yp j=1

(b_j − a_j).

(b) the function f is summable, i.e. integrable, over A.

In the first part of this handout, we shall discuss how to define the integral of a function on cells in R^p. In the second part of the handout, we shall extend the definition to a function on more general (measurable) sets in R^p.

Part 1: Integrable Functions on Cells:

Definitions:

(a) K is called a cell in R^p (or a p−cell, or a p−dimensional rectangle) if K = I₁× · · · × I_p, where I_j = [a_j, b_j] ⊂ R for j = 1, . . . , p.

(b) The (p−)content c(K) of K is defined to be c(K) = (b1− a1) × · · · × (bp− ap) = Yp j=1

(bj− aj).

(c) A set Z ⊂ R^p has p−content zero if ∀ ² > 0, ∃ a finite set C = {K_j}^m_j=1 of p−cells such that (a) Z ⊂

[m j=1

K_j,

(b) Xm

j=1

c(Kj) < ².

Remark 1. Note that the definition implies that if K is a cell (not necessarily closed) in R^p, then the boundary ∂K of K is a set of p−content zero.

Remark 2. Note that the definition of the content for a cell is well defined since it is easy to see that the following properties are satisfied.

(a) Let K be a cell in R^p) and K is a finite disjoint union of cells in R^p), i.e. K = [l i=1

K_i, then

c(K) = Xl

i=1

c(K_i).

(b) Let K₁, K₂ be cells in R^p). Then c(K₁ ∪ K₂) = c(K₁\ K₂) + c(K₁∩ K₂) + c(K₂\ K₁).

(c) Let x ∈ R^p, K be a cell in R^p) and x + K = {x + z | z ∈ K}. Then x + K is a cell in R^p) with c(x + K) = c(K), i.e. the definition of content for cells is invariant under translations.

Remark 3. By taking ²/2 > 0, if it is necessary, one may also assume that Z ⊂ Int¡[^m

j=1

K_j¢ . Example (1). Let Z = {x_j ∈ R | lim

j→∞x_j = x}, a (0-dim’l) subset of R. Then (1 − dim’l) c(Z) = 0 since ∀² > 0, ∃ a 1-d cell K_xsuch that x ∈ Int(K_x), c(K_x) < ²/2, and x_j ∈ K_x∀j ≥ L. For each j = 1, . . . , L − 1, let Kj be a 1-d cell such that xj ∈ Kj, and c(Kj) ≤ ²/(2L).

Example (2). 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 that satisfies (a) will have Xm

j=1

c(K_j) ≥ 1.

Example (3). Let Z = {(x, y) | |x| + |y| = 1}, a (1-dim’l) subset of R². Then the (2-d) content c(Z) = 0.

Definition (4). A collection of sets C = {K_j}^m_j=1 in Rⁿ is called a partition of a set K in Rⁿ if (a)

[m i=1

Ki = K, and

(b) IntKi∩ IntKj = ∅ holds for each 1 ≤ i 6= j ≤ m.

Example. Let K = [a₁, b₁] × · · · × [a_n, b_n] = I₁× · · · × I_n, and P_j = {[x^j_i, x^j_i+1] | a_j = x^j₀ < · · · <

x^j_m(j) = b_j}, for j = 1, . . . , n, be a partition of I_j into m(j) (a finite number of) closed cells in R.

Then the set P = {Q_n

j=1[x^j_ij, x^j_ij+1] | 0 ≤ i_j ≤ m(j) − 1} (induced by P_j’s) partitions K into m(1) × · · · × m(n) (finite number of) parallel closed n-cells.

Example. Let K = [0, 1]×[2, 4] ⊂ R², and for each n = 1, 2, . . . , let Pn = {[ⁱ⁻¹_n ,_nⁱ]×[2+²⁽ⁱ⁻¹⁾_n , 2+²ⁱ_n] | 1 ≤ i ≤ n} be a partition of K that divides each side of K into n equal length subintervals. One can define the norm of a partition P_n to be the kP_nk = max

Kj∈Pn

diam(K_j). In this example, the norm of the partition is kP_nk = ^√_n⁵.

Definition(5). Let P = {Ii}^l_i=1 and Q = {Kj}^m_j=1 be partitions of (an n-cell) K. We say that P is a refinement of Q, denoted Q ⊂ P, if each cell in P is contained in some cell in Q, i.e. for each I_i ∈ P ∃ K_j ∈ Q such that I_i ⊂ K_j.

Note that if P , Q are partitions of K, then P ∩ Q is a (common) refinement of P and Q, and, in general, P ∪ Q is Not a partition,

Example. Let K = [0, 1]×[2, 4] ⊂ R², and for each n = 1, 2, . . . , let P_n= {[ⁱ⁻¹_n ,_nⁱ]×[2+²⁽ⁱ⁻¹⁾_n , 2+²ⁱ_n] | 1 ≤ i ≤ n}. Then I = [0, 1/2] × [2, 3] ∈ P2 ⊂ P2 ∪ P3, J = [0, 1/3] × [2, 8/3] ∈ P3 ⊂ P2∪ P3, but (0, 1/2) × (2, 3) ∩ (0, 1/3) × (2, 8/3) 6= ∅. Therefore, P₂∪ P₃ is not a partition of K.

Definition (6). Let f be a bounded function defined on a closed n−cell K with values in R. A Riemann sum SP(f, K) corresponding to a partition P = {Kj}^m_j=1 of K is given by S_P(f, K) =

Xm i=1

f (x_i)c(K_i), where x_i is any point in K_i, and c(K_i) denotes the (n-dim’l) content of K_i.

Remark. Note that Xm

i=1

m_ic(K_i) = L_P(f ) ≤ S_P(f ) = S_P(f, K) ≤ U_P(f ) = Xm

i=1

M_ic(K_i), where m_i = inf

Ki

f ≤ f (x_i) ≤ M_i = sup

Ki

f, and L_P(f ), and U_P(f ) are called the lower sum and upper sum, respectively, of f with respect to the partition P of K.

Remark. (Monotonicity of lower and upper sums) If P, Q are partitions of K, and P ⊂ Q i.e.

Q is finer than P, then we have

L_P(f ) ≤ L_Q(f ) ≤ S_Q(f ) ≤ U_Q(f ) ≤ U_P(f ).

Since the set {L_P(f ) | P is a partition of K} is nonempty, and bounded from above by¡

sup_Kf¢ c(K), the L(f, K) = sup_P L_P(f ) = sup{L_P(f ) | P is a partition of K} exists.

Analogously, the U(f, K) = inf_P U_P(f ) = inf{U_P(f ) | P is a partition of K} exists.

If {P_k} be any sequence of partitions of K such that P_j ⊂ P_j+1 for each j = 1, 2 . . . , and kP_jk ≥ kP_j+1k → 0, then lim

kPjk→0L_Pj(f ) = L(f, K), and lim

kPjk→0U_Pj(f ) = U(f, K).

Definition (of integrability on cells). A bounded function f is called Riemann integrable on K if L(f, K) = U(f, K) and this common value, denoted

Z

K

f, is called the (Riemann) integral of f on K.

Remark. f is integrable on K if and only if there exists a unique number L such that for each partition P of K we have L_P(f ) ≤ L ≤ U_P(f )

Proof: (⇒) Since L_P(f ) ≤ L(f, K) = U(f, K) ≤ U_P(f ) holds for each partition P of K, by setting L = L(f, K), the inequality LP(f ) ≤ L ≤ UP(f ) holds for each partition P of K. Suppose that L₁ is a number such that the inequality L_P(f ) ≤ L₁ ≤ U_P(f ) holds for each partition P of K.

Then L₁ is an upper (resp. a lower) bound of the set {L_P(f ) | P is a partition of K} (resp.

{UP(f ) | P is a partition of K}) which implies that L = L(f, K) ≤ L1 (resp. L1 ≤ U(f, K) = L)).

Hence, L₁ = L is the unique number such that the inequality L_P(f ) ≤ L ≤ U_P(f ) holds for each partition P of K.

(⇐) In order to show that L(f, K) = U(f, K), we show that L(f, K) = L, and U(f, K) = L. Since the inequality L_P(f ) ≤ L ≤ U_P(f ) holds for each partition P of K, we have L(f, K) = sup_PL_P(f ) ≤ L ≤ inf_P U_P(f ) = U(f, K). Suppose that L(f, K) < L (resp. L < U (f, K)), then ∃ ²₀ > 0 such that L(f, K) < L − ²0 (resp. L + ²0 < U (f, K)). Thus, for each partition P of K, we have L_P(f ) ≤ L(f, K) < L − ²₀ < L ≤ U_P(f ) (resp. L_P(f ) ≤ L < L + ²₀ < U(f, K) ≤ U_P(f )) which contradicts to the uniqueness of L. This implies that L(f, K) = L (resp. U(f, K) = L), and L(f, K) = L = U(f, K), i.e. f is integrable on K.

Criterion of integrability: Let f be a bounded function defined on K. Then the following are equivalent.

(1) f is integrable on K, i.e. L(f, K) = U(f, K), with integral L =R

Kf = L(f, K)

(2) (Riemann Criterion for Integrability) ∀ ² > 0, ∃ partition P_², of K, such that if P is a refinement of P_², then |U_P(f ) − L_P(f )| < ².

(3) (Cauchy Criterion for Integrability) ∀ ² > 0, ∃ partition P², of K, such that if P and Q are any refinements of P_², and S_P(f, K) and S_Q(f, K) are any corresponding Riemann sums, then |S_P(f, K)−

S_Q(f, K)| < ².

(4) ∀ ² > 0, ∃ partition P², of K, such that if P is any refinement of P², and SP(f, K) is any corresponding Riemann sum, then |S_P(f, K) − L| < ².

Proof Since L_P(f ) ≤ L(f, K) ≤ U(f, K) ≤ U_P(f ), (by drawing a picture) one notes that

|U(f, K) − L(f, K)| ≤^(∗) |UP(f ) − LP(f )| ≤^(†) |UP(f ) − L| + |LP(f ) − L|.

(1) ⇒ (2) : Given ² > 0, since L(f, K) = U(f, K) (and by the definitions that L(f, K) being the smallest number that satisfies L(f, K) ≥ L_P(f ), and U(f, K) being the largest number that satisfies U(f, K) ≤ UP(f ) for all P ), there exists a partition P² such that

L(f, K) − ²/2 < L_P²(f ) ≤ L(f, K), and

L(f, K) = U(f, K) ≤ U_P²(f ) < U (f, K) + ²/2 = L(f, K) + ²/2.

Thus, if P is any refinement of P_², then

L(f, K) − ²/2 < L_P²(f ) ≤ L_P(f ) ≤ U_P(f ) ≤ U_P²(f ) < L(f, K) + ²/2.

Setting L = L(f, K) in the (second) inequality (†), we get that

|U_P(f ) − L_P(f )| ≤ |U_P(f ) − L| + |L_P(f ) − L| < ²/2 + ²/2 = ².

Thus, the conclusion of (2) holds.

(2) ⇒ (1) : For each ² > 0, since the (first) inequality (∗), and (2) hold, there exists a partition P_² such that if P is any refinement of P², then

|U(f, K) − L(f, K)| ≤ |U_P(f ) − L_P(f )| < ².

Letting ² → 0, we get 0 ≤ lim

²→0|U(f, K) − L(f, K)| ≤ lim

²→0² = 0 ⇒ U(f, K) = L(f, K)

and f is integrable.

(2) ⇔ (3) : For any refinements P, Q of P_², we have

LP²(f ) ≤ LP(f ) ≤ SP(f, K) ≤ UP(f ) ≤ UP²(f ) LP²(f ) ≤ LQ(f ) ≤ SQ(f, K) ≤ UQ(f ) ≤ UP²(f ) Thus, we have

|S_P(f, K) − S_Q(f, K)| ≤ |U_P²(f ) − L_P²(f )|

and (2) ⇒ (3).

Conversely, for any refinements P, Q of P_², if |S_P(f, K) − S_Q(f, K)| < ²/2 then, since

|U_P(f ) − L_Q(f )| = sup |S_P(f, K) − S_Q(f, K)|,

where the supremum is taken on all possible Riemann sum SP(f, K) and SQ(f, K) corresponding to the given (fixed) partitions P and Q, respectively, we have

|U_P(f ) − L_Q(f )| ≤ sup |S_P(f, K) − S_Q(f, K)| ≤ ²/2 and (3) ⇒ (2).

(3) ⇔ (4) : Let {Q_j} be a sequence of refinements of P_² such that Q_j ⊂ Q_j+1 and lim

j→∞kQ_jk = 0.

then

|S_P(f, K) − L| = lim

j→∞|S_P(f, K) − S_Qj(f, K)| ≤ ² and (3) ⇒ (4). Conversely, since

|S_P(f, K) − S_Q(f, K)| ≤ |S_P(f, K) − L| + |S_Q(f, K) − L|

holds, we have (4) ⇒ (3).

Example (1). For a < b, let f (x) =

(a if x ∈ Q ∩ [0, 1], b if x ∈ [0, 1] \ Q.

Then f is not continuous at each x ∈ [a, b] and f is not integrable on [0, 1] since LP(f, [0, 1]) = a 6=

b = U_P(f, [a, b]) for any partition P of [a, b].

Example (2). Let f (x) =







1

n if x = ^m_n ∈ Q ∩ [0, 1], where m, n ∈ N = {1, 2, . . .} and gcd(m, n) = 1, 1 if x = 0,

0 if x ∈ [0, 1] \ Q.

Then f is integrable on [0, 1] and f is continuous at every irrational and discontinuous at every ra- tional.

Observations. (1) There are finitely many rational numbers ^p_q ∈ [0, 1] such that q < n. In fact, for fixed q < n, the number of ^p_q ∈ [0, 1] is at most q + 1, which is at most n. Moreover, there are less than n positive q such that q < n. Thus, the set A_n = {^p_q ∈ [0, 1] : q < n} contains no more than n² element and note that if ^m_n ∈ (0, 1) is in lowest terms (m and n have no common factors except one), then min{|^p_q − ^m_n| : ^p_q ∈ A_n} > _n¹2.

(2) Fix 2 ≤ n ∈ N, and let ˆP = {^p_q ±_n¹3 : q < n, q ∈ N, p = 0, 1, . . . , q − 1}. Since the set ˆP is finite, it yields a partition P =¡P ∩ [0, 1]ˆ ¢

∪ {1} of [0, 1]. Define a step function s_n(x) =

(1 if ∃ q ∈ N, p ∈ {0, 1, . . . , q − 1} with q < n such that^p_q − _n¹3 < x < ^p_q + _n¹3, 0 if there do not exist such p and q.

Also, define f_n(x) =

(f (x) if ∃ q ∈ N, p ∈ {0, 1, . . . , q − 1} with q < n such that^p_q − _n¹3 < x < ^p_q +_n¹3, 0 if there do not exist such p and q.

For each n ≥ 2, since An contains no more than n² elements, there exist no more than n² intervals (^p_q−_n¹3,^p_q+_n¹3) in the interval [0, 1]. Thus, we have 0 ≤ U(f_n, [0, 1]) ≤ U_P(f_n, [0, 1]) ≤ U_P(s_n, [0, 1]) = R₁

0 sn(x) ≤ n²¡₂

n³

¢ = _n² for all n ≥ 2. By letting n go to infinity, we get 0 = U(f, [0, 1]) ≥ L(f, [0, 1]) = 0. This proves that f is integrable on [0, 1].

(3) Let x = ^m_n be a rational number in lowest terms. Note that if y ∈ [0, 1] satisfying that

|y − x| < _4n¹2 = _(2n)¹ 2 then either y ∈ [0, 1] \ Q or y = ^p_q with q < 2n (since any ^p_q ∈ A2n will have |x − ^p_q| ≥ _(2n)¹ 2). Assume f is continuous at x. Given ² = _2n¹ , there exists δ > 0 such that if y ∈ [0, 1] and |y − x| < δ then |f (x) − f (y)| < ². Let d = min{δ,_(2n)¹ 2} and note that if y ∈ [0, 1] and

|y − x| < d then |f (x) − f (y)| > _2n¹ (regardless that y is rational or irrational). This contradicts our assumption.

(4) Let α ∈ [0, 1] \ Q. Given ² > 0, choose n ∈ N such that ¹_n < ². Since there are finitely many rational numbers ^p_q ∈ [0, 1] such that q < n, the minimum distance δ between ^p_q and α for q < n, i.e. δ = min{|α − ^p_q| : q < n}, exists and it is positive since α is irrational. If |x − α| < δ, then either x ∈ [0, 1] \ Q, or x = ^p_q with p and q having no common factors except one and q ≥ n, since

|x − α| < δ. Thus, we have |f (x) − f (α)| =

(0 < ² if x ∈ (α − δ, α + δ) \ Q,

|f (^p_q)| = ¹_q ≤ ¹_n < ² if x = ^p_q ∈ (α − δ, α + δ).

Some basic properties of integrable functions on cells:

(1) Suppose that K, K₁, K₂ are closed n−cells such that K = K₁∪ K₂, and Int(K₁) ∩ Int(K₂) = ∅.

If f is integrable on K, then f is integrable on K₁, and K₂, and R

Kf =R

K1f +R

K2f.

Proof. Given ² > 0, since f is integrable, there is a partition P²of K such that if P is any refinement of P_², then |U_P(f, K) − L_P(f, K)| < ². For i = 1, 2, let P_²,i = P_²∩ K_i, then P_²,i is a partition of K_i, and P_²,1∪ P_²,2 is a refinement of P_². Thus, we have

² > U_{P²,1∪P²,2}(f, K₁∪ K₂) − L_{P²,1∪P²,2}(f, K₁∪ K₂)

= U_P²,1(f, K₁) − L_P²,1(f, K₁) + U_P²,2(f, K₂) − L_P²,2(f, K₂)

≥ UP²,i(f, Ki) − LP²,i(f, Ki)

≥ 0 Thus, for each i = 1, 2,

² > UP²,i(f, Ki) − LP²,i(f, Ki) ≥ 0 and if P_i is any refinement of P_²,i, then

² > UP²,i(f, Ki) − LP²,i(f, Ki) ≥ UPi(f, Ki) − LPi(f, Ki) ≥ 0.

This implies that f is Riemann integrable on Ki and Li =R

Kif exists, for i = 1, 2, and L = L(f, K) = sup{L_P(f ) | P is any refinement of P_²}

≤ sup{L_P1(f ) + L_P2(f ) | P_iis any refinement of P_²,ii = 1, 2}

≤ L₁+ L₂

≤ inf{U_P1(f ) + U_P2(f ) | P_iis any refinement of P_²,ii = 1, 2}

= inf{UP(f ) | P is any refinement of P²,1∪ P²,2}

= U(f, K) = L Thus, we have L = L₁+ L₂.

(2) If f and g are integrable on K, then, for any c ∈ R, cf + g is integrable on K, andR

K(cf + g) = cR

Kf +R

Kg.

Proof. Given ² > 0, since f, g are integrable on K, there exists a partition P_² of K such that if P is any refinement of P² then

|U_P(f ) − L_P(f )| < ²/2(1 + |c|)

and

|U_P(g) − L_P(g)| < ²/2.

Thus, we have

|U_P(cf + g) − L_P(cf + g)| ≤ |c||U_P(f ) − L_P(f )| + |U_P(g) − L_P(g)| < |c|²/2(1 + |c|) + ²/2 < ² which implies that cf + g is integrable on K. Let {P_j} be a sequence of partitions of K satisfying that P_j ⊂ P_j+1 for all j = 1, 2, . . ., and lim

j→∞kP_jk = 0. Since Z

K

f = lim

j→∞SPj(f, K) and Z

K

g = lim

j→∞SPj(g, K), we have

c Z

K

f + Z

K

g = c lim

j→∞S_Pj(f, K) + lim

j→∞S_Pj(g, K)

= lim

j→∞SPj(cf, K) + lim

j→∞SPj(g, K)

= lim

j→∞S_Pj(cf + g, K)

= Z

K

(cf + g).

(3) Suppose that f and g are integrable on K. If f (x) ≤ g(x) for each x ∈ K, then R

Kf ≤R

Kg.

Proof. Since −f + g ≥ 0 on K and, by (2), it is integrable on K, we have 0 ≤ L_P(−f + g) ≤

Z

K

(−f + g) = − Z

K

f + Z

K

g, where P is any partition of K. Since R

Kf ∈ R, by adding R

Kf on both sides of the inequality, we get R

Kf ≤R

Kg.

(4) If f is integrable on K, then |f | is integrable on K, and |R

Kf | ≤ R

K|f |.

Proof. Given ² > 0, since f is integrable on K, there exists a partition P_² of K such that if P = {K_j}^m_j=1 is any refinement of P_² then

|U_P(|f |)−L_P(|f |)| = | Xm

i=1

¡sup

Ki

|f |−inf

Ki

|f |¢

c(K_i)| ≤ | Xm

i=1

¡sup

Ki

f −inf

Ki

f¢

c(K_i)| = |U_P(f )−L_P(f )| < ².

Thus, |f | is integrable (by Riemann’s Criterion for integrability).

Since ±f, |f | are integrable, and ±f ≤ |f | on K, we have ±R

Kf ≤ R

K|f | ⇒ |R

Kf | ≤R

K|f |.

Examples of integrability.

(1) Let Z ⊂ Rⁿhave (n-)content zero, and f be a bounded function defined on Z. Then f is integrable on Z, andR

Zf = 0.

Proof. For each ² > 0, since Z has content zero, there exists a collection of cells {K_i}^m_i=1 such that Z ⊂ ∪^m_i=1K_i = K, and P_m

i=1c(K_i) < ²/2¡

sup_Z|f | + 1¢

. Define

f (x) =¯

(f (x) if x ∈ Z, 0 if x ∈ K \ Z.

Then,

| ¯f |(x) =

(|f |(x) ≥ 0 if x ∈ Z,

0 if x ∈ K \ Z,

and note that

| sup

Kj∩Zf − inf

Kj∩Zf | = | sup

Kj∩Z

f − inf¯

Kj∩Z

f |¯

≤ | sup

Kj

f − inf¯

Kj

f |¯

≤ | sup

Kj

| ¯f | − inf

Kj

¡− | ¯f |¢

|

= | sup

Kj

| ¯f | −¡

− sup

Kj

| ¯f |¢

|

= 2 sup

Kj

| ¯f | = 2 sup

Kj∩Z

|f |

≤ 2M.

Thus, if P is any partition of K = ∪^m_i=1K_i, we have

|UP(f, Z) − LP(f, Z)| ≤ 2 sup

Z |f | Xm

i=1

c(Ki) < ² which implies that f is integrable on Z with R

Zf = 0.

(2) Let I be a closed interval in R, and f be a bounded and monotonic function defined on I = [a, b]

Then f is integrable on I.

Proof. Since f is bounded on I, L(f, I) = sup

P LP(f ) and U(f, I) = inf

P UP(f ) exist.

Let P_n be the partition that divides I into 2ⁿ equal length subintervals. Thus,

n→∞lim L_Pn(f ) = L(f, I) and lim

n→∞U_Pn(f ) = U(f, I).

Since

n→∞lim |U_Pn(f ) − L_Pn(f )| = lim

n→∞

¡f (b) − f (a)¢¡

b − a¢

/2ⁿ = 0, we get L(f, I) = U(f, I), i.e. f is integrable on I.

(3) Let K be a closed n−cell, and f be a continuous function on K. Then f is integrable on K.

Proof. Since f is continuous on (compact set) K, f is uniformly continuous on K.

Hence, for any given ² > 0 there exists δ > 0 such that

if x, y ∈ K and kx − yk < δ then |f (x) − f (y)| < ²/¡

c(K) + 1¢ . Let P² be a partition of K such that kP²k = max

Kj∈P²

diam(Kj) = max

Kj∈P²

sup{kx − yk : x, y ∈ Kj} < δ.

If P is any refinement of P_² then |U_P(f ) − L_P(f )| < ²c(K)/¡

c(K) + 1¢

< ².

Therefore, f is integrable on K.

(4) Let K be a closed n−cell, and f be a bounded function defined on K. If there exists a (n-)content zero subset Z ⊂ K, such that f is continuous on K \ Z, i.e. f is continuous everywhere on K except at a content zero subset Z of K, then f is integrable on K.

Proof. For each ² > 0, since Z has content zero, there exists a collection of cells {I_i}^l_i=1 such that Z ⊂ Int¡[^l

i=1

I_i¢

, IntI_i∩ IntI_j = ∅, and Xl

i=1

c(I_i) < ²/4¡ sup

K

|f | + 1¢ .

Since K \ Int¡[^l

i=1

I_i¢

is compact, f is uniformly continuous there.

Thus, for the given ² > 0, there exists a δ > 0 such that if x, y ∈ Kj \ Int¡[^l

i=1

Ii

¢, then |f (x) − f (y)| < ²/2¡

c(K) + 1¢ .

Let P² = {Kj}^m_j=1 be a partition of K such that {Ii∩K}^l_i=1⊂ P²and kP²k = max

1≤j≤m sup

x,y∈Kj

kx−yk < δ.

If P is any refinement of P² then, by using (1) and (3), we have

|UP(f ) − LP(f )| ≤ Xl

i=1

¡sup

Z∩Ii

f − inf

Z∩Ii

f¢

c(Ii) + |UP(f, K \ Int¡

∪^l_i=1Ii

¢) − LP(f, K \ Int¡

∪^l_i=1Ii

¢)|

≤ 2 sup

Z |f | Xl

i=1

c(Ii) + c(K)²/2¡

c(K) + 1¢

< ²

This implies that f is integrable on K.

Theorems: (1) Suppose f and g are integrable on a closed n−cell K, and f = g everywhere on K except at a content zero subset Z of K, thenR

Kf =R

Kg.

Proof. Since f, g are integrable on K and Z has content zero, f − g is integrable on K and it is continuous with value 0 on K \ Z. Given ² > 0, let {Ii}^l_i=1 be a collection of cells such that

Z ⊂ Int¡[^l

i=1

Ii

¢ and Xl

i=1

c(Ii) < ²/4¡ sup

K |f − g| + 1¢ .

Let P_²= {K_j}^m_j=1 such that {I_i∩ K}^l_i=1⊂ P_². Thus, if P is any refinement of P_² then

|U_P(f − g) − L_P(f − g)| ≤ |U_P(f − g, K \ Int¡[^l

i=1

I_i¢

) − L_P(f − g, K \ Int¡[^l

i=1

I_i¢ )|

+ Xl

i=1

¡sup

Z∩Ii

(f − g) − inf

Z∩Ii

(f − g)¢ c(I_i)

< ².

This implies that R

K(f − g) = 0. Since R

Kg ∈ R, we have R

Kf =R

Kg.

(2) Fundamental Theorem of Calculus Let f be integrable on [a, b]. For each x ∈ [a, b], let F (x) =R

a,x]f =R_x

a f (t)dt. Then F is continuous on [a, b]; moreover, F⁰(x) exists and equals f (x) at every x at which f is continuous.

Remark. For each x ∈ [a, b], the existence of F (x) is due to [a, x] ⊆ [a, b] and the existence of R_b

a f.

Existence ofR_b

a f implies that for each ² > 0 ∃ a partition P² of [a, b] such that if P is any refinement of P_², then |U_P(f, [a, b]) − L_P(f, [a, b])| < ².

Let

P_²^l = P²∩ [a, x] and P_²^r = P²∩ [x, b].

Then P_²^l∪ P_²^r is a refinement of P_² and if P^l is any refinement of P_²^l, then

|U_P²^l(f, [a, x]) − L_P²^l(f, [a, x])| ≤ |U_P²^l(f, [a, x]) − L_P²^l(f, [a, x]) + UP_²^r(f, [x, b]) − LP_²^r(f, [x, b])|

= |U_P²^l_∪P²^r(f, [a, b]) − L_P²^l_∪P²^r(f, [a, b])|

< ².

Hence, f is integrable on [a, x], i.e. F (x) exists.

Proof of the Theorem. If x, y ∈ [a, b] ⇒ F (y) − F (x) =R_y

x f (t)dt.

Let c = sup{|f (t)| : t ∈ [a, b]}. (c exists since f is integrable on [a, b] ⇒ f is bounded on [a, b].) Then |F (y) − F (x)| ≤ |R_y

x |f (t)|dt| ≤ c|R_y

x dt| = c|y − x|

⇒ F is (Lipschitz, hence) continuous on [a, b].

Suppose that f is continuous at x; thus ∀ ² > 0, ∃ δ > 0 such that if t ∈ [a, b] and |t − x| < δ then |f (t) − f (x)| < ².

Since f (x) = f (x)_y−x¹ R_y

x dt = _y−x¹ R_y

x f (x)dt.

Hence, if y ∈ [a, b] and |y − x| < δ ⇒ t ∈ [a, b] and |t − x| < δ for all t between y and x.

⇒ |f (t) − f (x)| < ² and this implies that

⇒ |F (y) − F (x)

y − x − f (x)| ≤ 1

|y − x||R_y

x |f (t) − f (x)|dt| ≤ 1

|y − x||R_y

x ²dt| = ².

⇒ lim

y→x

F (y) − F (x)

y − x = f (x), i.e. F⁰(x) exists and equals f (x) at every x at which f is continuous.

(3) Let F be a continuous function on [a, b] that is differentiable except at finitely many points in [a, b], and let f be a function on [a, b] that agrees with F⁰ at all points where F⁰ is defined. If f is integrable on [a, b], then R_b

a f (t)dt = F (b) − F (a).

Proof. Let {z₁, . . . , z_m} ⊂ [a, b] be the set at which F⁰ does not exist, i.e. F is not differentiable at z_i, i = 1, . . . , m.

Let P = {a = x0 < x1 < · · · < xn = b} be a partition of [a, b] with zi, i = 1, . . . , m, being partition point, i.e. each z_i ∈ {x₀, x₁, . . . , x_n}.

⇒ F is continuous on each [x_j−1, x_j], j = 1, . . . , n, and differentiable on each (x_j−1, x_j).

By the Mean Value Theorem, F (xj) − F (xj−1) = F⁰(tj) (xj − xj−1) = f (tj) (xj − xj−1) for some t_j ∈ (x_j−1, x_j) and for each j = 1, . . . , n.

Thus, we have F (b) − F (a) = Xn

j=1

F (x_j) − F (x_j−1) = Xn

j=1

f (t_j) (x_j − x_j−1)

⇒ L_P(f, [a, b]) ≤ F (b) − F (a) ≤ U_P(f, [a, b])

⇒ sup

P LP(f, [a, b]) ≤ F (b) − F (a) ≤ inf

P UP(f, [a, b]) If f is integrable then R_b

a f (t)dt = sup_P LP(f, [a, b]) = infP UP(f, [a, b]) = F (b) − F (a).

Part 2: Integrable Functions on General Measurable Sets:

In the following we shall extend the concept of content of a cell in Rⁿ to more general mea- surable subsets of Rⁿ and extend the definition of integrability of a function to general subsets of Rⁿ.

Definition (of integrability on general Euclidaen bounded subsets.) Let A ⊂ Rⁿ be a bounded set and let f : A → R be a bounded function. Let K be a closed cell containing A and define f_K : K → R 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 R

Af =R

Kf_K.

Remark. If A = K is a closed cell in Rⁿ, then, since f_K = f on K, it is obvious the integrability of f on A agrees with the integrability of fK on K and R

Af is defined to be R

KfK.

Remark. Let I be any closed cell containing A. Then K ∩ I is a closed cell containing A ⇒ f_K = f_K∩I = f_I everywhere in K ∩ I, f_K = 0 on K \ (K ∩ I), and f_I = 0 on I \ (K ∩ I). Hence, R

KfK =R

K∩IfK∩I =R

IfI ⇒ the definition (of integrability of f ) only depends on f and A (and it is independent of the choice of K ⊇ A).

Basic properties of integrable functions on general sets:

(1). Let f and g be integrable functions defined on a bounded set A ⊂ Rⁿ and let α, β ∈ R. Then the function αf + βg is integrable on A and R

A(αf + βg) = αR

Af + βR

Ag.

Proof. For any partition P of a cell K ⊇ A, since S_P(αf_K+ βg_K, K) = αS_P(f_K, K) + βS_P(g_K, K) when the same intermediate points xj are used, the function αf + βg is integrable on A. Thus, by choosing the intermediate points from A whenever it is possible, we obtain that S_P(αf + βg, A) = αS_P(f, A) + βS_P(g, A) which implies thatR

A(αf + βg) = αR

Af + βR

Ag.

(2) Let A1 and A2 be bounded sets with no pints in common, and let f be a bounded function. If f is integrable on A₁ and on A₂, then f is integrable on A₁∪ A₂ and R

A1∪A2f = R

A1f +R

A2f.

Proof. Let K be a closed cell containing both A₁and A₂, and let f_K(x) = (

f (x) if x ∈ A₁ ∪ A₂, 0 if x ∈ K \ (A₁ ∪ A₂) and f_Kⁱ (x) =

(f (x) if x ∈ A_i,

0 if x ∈ K \ A_i for i = 1, 2. Since f is integrable on A_i, i = 1, 2, f_Kⁱ is integrable on K and, since f_K = f_K¹ + f_K², and f is integrable on A₁∪ A₂. Also, for any partition P of K, note that S_P(f_K, K) = S_P(f_K¹, K) + S_P(f_K², K) when the same intermediate points x_j are used. Thus, we haveR

A1∪A2f = R

A1f +R

A2f.

(3) If f : A → R is integrable on (bounded set) A and f (x) ≥ 0 for x ∈ A, then R

Af ≥ 0.

Proof. For any closed cell K ⊇ A and any partition P of K, note that S_P(f_K, K) ≥ 0 for any Riemann sum. Thus, R

Af ≥ 0.

Remark. This implies that if f and g are integrable on A and f (x) ≤ g(x) for x ∈ A, then (a) R

Af ≤R

Ag, and (b) |f | is integable on A, and |R

Af | ≤ R

A|f |.

(4) Let f : A → R be a bounded function and suppose that A has content zero. Then f is integrable on A andR

Af = 0.

Proof. LetK ⊇ A be a closed cell. If ² > 0 is given, let P_² be a partition of K such that those cells in P_² which contain points of A have total content less than ². Now if P is a refinement of P_², then those cells in P containing points of A will also have total content less than ². Hence if |f (x)| ≤ M for x ∈ A, we have |S_P(f_K, K) − 0| ≤ M² for any Riemann sum corresponding to P. Since ² is arbitrary, this implies that R

Af = 0.

(5) Let f, g : A → R be bounded functions and suppose that f is integrable on (bounded set) A. Let Z ⊆ A have content zero and suppose that f (x) = g(x) for all x ∈ A \ Z. Then g is integrable on A and R

Af =R

Ag.

Proof. Extend f and g to functions fK, gK defined on a closed cell K ⊇ A. Thus, the function h_K = f_K− g_K is bounded and equals 0 except on Z. Hence, h_K is integrable on K and R

Kh_K = 0.

Since f_K is also integrable on K, we have R

Af =R

Kf_K =R

K(f_K− h_K) =R

Kg_K =R

Ag.

(6) Let U be a connected, open subset of Rⁿ and let f : U ⊂ Rⁿ→ Rⁿ⁺¹ be a C¹ map on U. If K is any convex, compact subset of U, then f (K) has measure (or content) zero.

Definition. Let A ⊂ Rⁿ be a bounded set. The characteristic function of A is the function χ_A defined by χ_A(x) =

(1 if x ∈ A,

0 otherwise. Now, suppose A is a bounded subset of Rⁿand f is a bounded function on Rⁿ. Let K be a closed cell that contains A. We say that f is integrable on A if f χ_A is integrable on K, and define R

Af =R

Kf χ_A. (Note that f χ_A= 0 on K \ A, so it is independent of the choice of K ⊇ A.)

Question: Let f ≡ 1 on A ⊂ Rⁿ. What does it mean when we say that f is integrable on A?

Definition. A set A ⊂ Rⁿ is said to have content (or it is said to be (Jordan) measurable) if it is bounded and its boundary ∂A has content zero. Let D(Rⁿ) (= {A ⊂ Rⁿ | A has content } = {A ⊂ Rⁿ| A is measurable }) denote the set of all measurable subsets of Rⁿ.

Remark. If A ∈ D(Rⁿ) and if K is a closed cell containing A, then the function g_K defined by g_K(x) =

(

1 if x ∈ A

0 if x ∈ K \ A is continuous on K except possibly at points of ∂A (which has content zero). Thus, g_K is integrable on K and we define the content of A, denoted c(A), by c(A) =R

Kg_K =R

A1.

Remark. ∀ ² > 0, since c(A) =R

Kg_K, ∃ a partition P_² = {I_j}^m_j=1of K such that |S_P²(g_k, K)−c(A)| <

² for any Riemann sum S_P²(g_k, K). By choosing the intermediate points in S_P²(g_k, K) to belong to A whenever possible, we have P_m

j=1c(I_j) + ² ≥ S_P²(g_k, K) + ² > c(A) > S_P²(g_k, K) − ², where the first inequality holds since A ⊂ ∪^m_j=1Ij.

Thus, we have: A set A ⊂ Rⁿ has content zero if and only if A has content and R

A1 = c(A) = 0.