Question. Let {fn} be a sequence of integrable functions that converges at every point of a cell K ⊂ Rp to a function f. Is f integrable on K? Suppose that f is integrable on K, is it true that R
Kf = limR
Kfn K?
Examples.
(a) Let Q ∩ [0, 1] = {xn}∞n=1 and fn be a monotone sequence of integrable functions on [0, 1] defined by fn(x) =
(1 if x ∈ {x1, x2, . . . , xn}, 0 otherwise.
Then the limit function f is defined by f (x) = lim fn(x) = (
1 if x ∈ Q ∩ [0, 1], 0 if x ∈ [0, 1] \ Q.
Note that the convergence of fn to f is not uniform on [0, 1], f is not integrable on [0, 1], and 0 = lim
Z 1
0
fn 6=
Z 1
0
lim fn since R1
0 f does not exist.
(b) Define (discontinuous) fn and (continuous) f on K = [0, 1] by n ≥ 1 by fn(x) =
(
n if x ∈ (0,n1),
0 otherwise, and f (x) = 0.
Note that the convergence of fn to f is not uniform on [0, 1], f is Riemann integrable on K, and 1 = limR1
0 fn 6=R1
0 lim fn=R1
0 f = 0.
(c) Let K = [0, 1], and (continuous function) fn be defined for n ≥ 2 by fn(x) =
n2x if x ∈ [0,n1],
−n2(x −n2) if x ∈ [n1,n2], 0 if x ∈ [n2, 1],
and (continuous) f (x) = lim fn(x) = 0 for all x ∈ K. Note that the convergence of fn to f is not uniform on [0, 1], f is integrable on K, and 1 = limR1
0 fn6=R1
0 lim fn=R1
0 f = 0.
These examples indicate that a convergence theorem for the Riemann integral will require some con- dition in addition to pointwise convergence.
Theorem. Let {fn} be a sequence of integrable functions that converges uniformly on a closed cell K ⊂ Rp to a function f. Then f is integrable and R
Kf = limR
Kfn.
Proof. Let ² > 0 and N be such that kfN − f kK < ². Now let PN be a partition of K such that if P, Q are refinements of PN, then |SP(fN, K) − SQ(fN, K)| < ², for any choice of the intermedi- ate points. If we use the same intermediate points for f and fN, then |SP(fN, K) − SP(f, K)| ≤ kfN−f kKc(K) < ²c(K). Since a similar estimate holds for the partition Q, then for refinements P, Q of PN and corresponding Riemann sums, we have |SP(f, K) − SQ(f, K)| ≤ |SP(f, K) − SP(fN, K)| +
|SP(fN, K) − SQ(fN, K)| + |SQ(fN, K) − SQ(f, K)| ≤ ²(1 + 2c(K)). This implies that f is integrable on K.
Since |R
Kf −R
Kfn| = |R
K(f − fn)| ≤ kf − fnkKc(K), we have R
Kf = limR
Kfn. Example. Let K = [0, 1], and fn be defined by
fn(x) = (
sin(nπx) if x ∈ [0,n1], 0 if x ∈ (n1, 1].
Note that fnconverges to the zero function on [0, 1], and the convergence is not uniform on K. How- ever, limR1
0 fn = lim 2
nπ = 0 = R1
0 lim fn. This example demonstrates that the uniform convergence is not a necessary condition in the theorem.
Bounded Convergence Theorem. Let {fn} be a sequence of integrable functions on a closed cell K ⊂ Rp. Suppose that there exists B > 0 such that kfn(x)k ≤ B for all n ∈ N, x ∈ K. If the function f (x) = lim fn(x), x ∈ K, exists and is integrable, then R
Kf = limR
Kfn.
Remark. This theorem has replaced the uniform convergence of fn by the uniform boundedness of fn and the integrability of f.
Outline of the Proof. Since f (x) = lim fn(x) for x ∈ K and kfnkK ≤ B for all n ∈ N, there exists M such that |f (x)| ≤ M and |fn(x)| ≤ M for all x ∈ K and all n ≥ 1. Since |f − fn| is integrable on K, there exists a subset A ⊆ K such that c(K \ A) = 0 and |f − fn| converges uniformly to 0 on A. This implies thatR
Kf = limR
Kfn.
To find A, we observe that the convergence of fn to f is not uniform on K if there exists ² > 0 such that the set An= {x ∈ K | ∃ j ≥ n such that |fj(x) − f (x)| ≥ ²} 6= ∅. Note that
(a) {An} is a nested sequence, i.e. A1 ⊇ A2 ⊇ · · · An⊇ An+1 ⊇ · · · ,
(b) lim
n→∞An =
\∞ n=1
An = ∅ and lim c(An) = 0,
(c) An6= ∅ for any δ < ²,
(d) fn converges uniformly to f on each K \ Aj, j ∈ N.
Examples. Use suitable convergence theorem to prove the following.
(a) If a > 0, then lim
n
Z a
0
e−nxdx = 0.
(b) If 0 < a < 2, then lim
n
Z 2
a
e−nx2dx = 0. What happens if a = 0?
(c) If a > 0, then lim
n
Z π
a
sin nx
nx dx = 0. What happens if a = 0?
(d) Let fn(x) = nx
1 + nx for x ∈ [0, 1], and let f (x) =
(0 if x = 0, 1 if x ∈ (0, 1].
Then lim fn(x) = f (x) for all x ∈ [0, 1] and that limR1
0 fn(x) dx = R1
0 f (x) dx.
(e) Let hn(x) = nxe−nx2 for x ∈ [0, 1] and let h(x) = 0. Then 0 =R1
0 h(x) dx 6= limR1
0 hn(x) dx = 1 2. Monotone Convergence Theorem. Let {fn} be a monotone sequence of integrable functions on a closed cell K ⊂ Rp. If the function f (x) = lim fn(x), x ∈ K, exists and is integrable, then R
Kf = limR
Kfn.
Proof. Suppose that f1(x) ≤ f2(x) ≤ · · · ≤ f (x) for all x ∈ K, then fn(x) ∈ [f1(x), f (x)] for all n ∈ N and kfn(x)k ≤ |f1(x)| + |f (x)| ≤ sup
x∈K
|f1(x)| + sup
x∈K
|f (x)| = B for all x ∈ K and for all n ∈ N, so we can apply the Bounded Convergence Theorem to establish that R
Kf = limR
Kfn. Remark. Note that the convergence theorem may fail if K in not compact.
Example. Let fn(x) =
1
x if x ∈ [1, n]
0 if x > n. Then each fn is integrable on [1, ∞), and {fn} is a bounded, monotone sequence that converges uniformly on [1, ∞) to a continuous function f (x) = 1/x.
Note that limR∞
1 fn6=R∞
1 lim fn since f is not integrable on [1, ∞).
Example. Let gn(x) =
1
n if x ∈ [0, n2]
0 if x > n2. Then each gn is integrable on [1, ∞), and {gn} is a bounded, monotone sequence that converges [1, ∞) to an integrable function g(x) ≡ 0. Note that limR∞
1 gn6=R∞
1 lim gn.
Definition. If {fn} is a sequence of functions defined on a subset D of Rp with values in Rq, the sequence of partial sums (sn) of the series P
fn is defined for x in D by sn(x) = Xn
j=1
fj(x). In case the sequence {sn} converges on D to a function f, we say that the infinite series of functionsP
fn
converges to f on D. If the sequence {sn}converges uniformly on D to a function f, we say that the infinite series of functionsP
fn converges uniformly to f on D.
Remark (Cauchy Criterion). It is easy to see that P
fk converges uniformly on D if and only if for each ² > 0, there exists M = M(²) ∈ N such that for any n, m ≥ M and any x ∈ D, we have ksn(x) − sm(x)k < ².
Dirichlet’s Test Let {fn} be a sequence of functions on D ⊆ Rp to Rq such that the partial sums sn=
Xn j=1
fj, n ∈ N, are all bounded. Let {φn} be a decreasing sequence of functions on D to R which
converges uniformly on D to zero. Then the series X∞ n=1
(fnφn) converges uniformly on D.
Abel’s Test Let X∞ n=1
fn be a series of functions on D ⊆ Rp to Rq which is uniformly convergent on
D. Let {φn} be a sequence of functions on D to R which is bounded on D. Then the series X∞ n=1
(fnφn) converges uniformly on D.
Outline of the proofs. For Dirichlet’s test, observe that | Xm j=n
φjfj| = |φm+1sm− φnsn−1+ Xm j=n
(φj−
φj+1sj|. For Abel’s test, if |φj(x)| ≤ B for all j ∈ N and for all x ∈ D, then | Xm j=n
φjfj| ≤ B Xm j=n
|fj|.
Theorem. If fn is continuous on D ⊆ Rp to Rq for each n ∈ N and if P
fn converges to f uniformly on D, then f is continuous on D.
Term-by-Term Integration Theorem. Suppose that the real-valued functions fn, n ∈ N, are integrable on K = [a, b]. If the series P
fn converges to f uniformly on K, then f is integrable on K and
Z
K
f = X∞
j=1
Z
K
fn.
Term-by-Term Differentiation Theorem. For each n ∈ N, let fn be a real-valued function on K = [a, b] which has a derivative fn0 on K. Suppose that the infinite series P
fn converges for at least one point of K and that the series of derivatives P
fn0 converges uniformly on K. Then there exists a real-valued function f on K such that P
fn converges uniformly on K to f. In addition, f has a derivative on K and f0 =P
fn0.
Proof. Suppose that the partial sum snofP
fnconverges at x0 ∈ K. For each x ∈ K and any m, n ∈ N, by the Mean Value Theorem, the equality sm(x)−sn(x) = sm(x0)−sn(x0)+(x−x0)(s0m(y)−s0n(y)) holds for some y lying between x and x0. The uniform convergence of P
fn0 and the convergence of Pfn(x0) lead to the uniform convergence of P
fn on K.
Suppose thatP
fn0 converges uniformly to g on K. For each x, c ∈ K and any m, n ∈ N, by the Mean Value Theorem, the equality sm(x) − sn(x) = sm(c) − sn(c) + (x − c)(s0m(y) − s0n(y)) holds for some y lying between x and c. We infer that, when x 6= c, then |sm(x) − sm(c)
x − c −sn(x) − sn(c)
x − c | ≤ ks0m−s0nkK. Given ² > 0, by the uniform convergence ofP
fn0, there exists a M(²) such that if m, n ≥ M(²) then ks0m− s0nkK < ². Taking the limit with respect to m, we get |f (x) − f (c)
x − c −sn(x) − sn(c)
x − c | ≤ ² when n ≥ M(²). Since g(c) = lim s0n(c), there exists an N(²) such that if n ≥ N(²), then |s0n(c) − g(c)| < ².
Now let L = max{M(²), N(²)}. In view of the existence of s0L(c), if 0 < |x − c| < δL(²), then
|sL(x) − sL(c)
x − c −s0L(c)| < ². Therefore, it follows that if 0 < |x−c| < δL(²), then |f (x) − f (c)
x − c −g(c)| <
3². This shows that f0(c) exists and equals g(c).
Example (a). For each k ∈ N and for each x ∈ [−1, 1], define fk(x) = xk
k2. Then X∞ k=1
fk converges
uniformly on [−1, 1], and X∞
k=1
fk0 = X∞ k=1
xk−1
k converges uniformly on any [−r, r], where 0 ≤ r < 1.
Example (b). For each k ≥ 0 and for each x ∈ (−1, 1), define fk(x) = (−1)kxk. Then X∞ k=0
fk converges to f (x) = 1
1 + x uniformly on any [−r, r], where 0 ≤ r < 1.
Example (c). Using (a), one observes that X∞
k=0
fk0 = X∞ k=0
(−1)kx2k converges to 1
1 + x2 on (−1, 1) and it is not convergent at x = ±1, while
X∞ k=0
fk = X∞ k=0
(−1)k
2k + 1x2k+1 converges to tan−1x on (−1, 1) uniformly on [−1, 1].
Example (d). Note that X∞ k=1
fk = X∞ k=1
sin kx
k2 converges uniformly on R by Cauchy Criterion and the criterion is not applicable for
X∞ k=1
fk0 = X∞ k=1
cos kx k since
X∞ k=1
1
k diverges.
Definition. Let f be defined on [a, ∞) × [α, β] to R. Suppose that for each t ∈ J = [α, β] the infinite integral F (t) = R∞
a f (x, t)dx = limc→∞Rc
af (x, t)dx exists. We say that this convergence is uniform on J if for every ² > 0 there exists a number M(²) such that if c ≥ M(²) and t ∈ J, then
|F (t) −Rc
a f (x, t)dx| < ².
Dominated Convergence Theorem. Suppose that f is integrable over [a, c] for all c ≥ a and all t ∈ J = [α, β]. Suppose that there exists a positive function φ defined for x ≥ a such that
|f (x, t)| ≤ φ(x) for x ≥ a, t ∈ J, and such that the integral R∞
a φ(x)dx exists. Then, for each t ∈ J, the integral F (t) =R∞
a f (x, t)dx is (absolutely) convergent and the convergence is uniform on J.
Dirichlet’s Test Let f be continuous on [a, ∞) × [α, β] and suppose that there exists a constant A such that |Rc
af (x, t)dx| ≤ A for c ≥ a, t ∈ J = [α, β]. Suppose that for each t ∈ J, the function φ(x, t) is monotone decreasing for x ≥ a, and converges to 0 as x → ∞ uniformly for t ∈ J. Then the integral F (t) =R∞
a f (x, t)φ(x, t)dx converges uniformly on J.
Examples.
(a) R∞
0
dx
x2+ t2 converges uniformly for |t| ≥ a > 0.
(b) R∞
0
dx
x2+ t converges uniformly for t ≥ a > 0 and diverges when t ≤ 0.
(c) R∞
0 e−xcos txdx converges uniformly for t ∈ R by the dominated convergence theorem.
(d) R∞
0 e−x2−t2/x2dx converges uniformly for t ∈ R by the dominated convergence theorem.
Theorem. Let f be continuous on K = [a, b] × [c, d] to R and F : [c, d] → R be defined by F (t) =Rb
af (x, t)dx. Then F is continuous on [c, d] to R.
Proof. Let ² > 0, since f is uniformly continuous on K, there exists a δ(²) > 0 such that if t and t0 belong to [c, d] and |t − t0| < δ(²), then |f (x, t) − f (x, t0)| < ², for all x ∈ [a, b]. It follows that
|F (t) − F (t0)| = |Rb
a
¡f (x, t) − f (x, t0)¢
dx| ≤ Rb
a|f (x, t) − f (x, t0)|dx ≤ ²(b − a), which establishes the continuity of F.
Remark. Suppose that f is continuous on [a, ∞) × [c, d] to R and F (t) = R∞
a f (x, t)dx converges uniformly on [c, d], we let Fn(t) =Ra+n
a f (x, t)dx. Then Fnis continuous on [c, d] and F is continuous on [c, d] since Fn converges to F uniformly on [c, d].
Theorem. Let f and its partial derivative ft be continuous on K = [a, b] × [c, d] to R. Then the function F (t) =Rb
a f (x, t)dx is differentiable on (c, d) and F0(t) =Rb
a ft(x, t)dx for t ∈ (c, d).
Proof. From the uniform continuity of ft on K we infer that if ² > 0, then there is a δ(²) > 0 such
that if |t − t0| < δ(²), then |ft(x, t) − ft(x, t0)| < ² for all x ∈ [a, b]. Let t, t0 satisfy this condition and apply the Mean Value Theorem to obtain a t1 ( which may depend on x and lies between t and t0) such that f (x, t) − f (x, t0) = (t − t0)ft(x, t1). Combining these two relations, we infer that if 0 < |t − t0| < δ(²), then |f (x, t) − f (x, t0)
t − t0 − ft(x, t0)| < ² for all x ∈ [a, b]. Thus, we obtain
|F (t) − F (t0) t − t0 −Rb
a ft(x, t0)dx| ≤ Rb
a |f (x, t) − f (x, t0)
t − t0 − ft(x, t0)|dx ≤ ²(b − a), which establishes F0(t) =Rb
a ft(x, t)dx.
Generalization. Let S be a measurable subset of Rn and T a subset of Rm. Suppose f (x, y) is a function on T × S that is integrable as a function of y ∈ S for each x ∈ T, and let F be defined on T by F (x) =R
Sf (x, y)dy for x ∈ T.
(a) If f (x, y) is continuous as a function of x ∈ T for each y ∈ S, and there exists a constant C such that |f (x, y)| ≤ C for all x ∈ T and y ∈ S, then F is continuous on T.
(b) Suppose T is open. If f (x, y) is of class C1 as a function of x ∈ T for each y ∈ S, and there is a constant C such that |∇xf (x, y)| ≤ C for all x ∈ T and y ∈ S, then F is of class C1 on T and
∂F
∂xj
(x) =R
S
∂f
∂xj
(x, y)dy for x ∈ T.
Proof. Let {xj} be a sequence in T converging to x ∈ T. For each j ∈ N and y ∈ S, let fj(y) = f (xj, y) and let f (y) = f (x, y). Then each fj and f are integrable on S, and |fj(y)| ≤ C and fj(y) → f (y) for all j and all y ∈ S. The bounded convergence theorem implies that lim F (xj) = limR
Sf (xj, y)dy = limR
Sfj = R
Slim fj = R
Slim f (xj, y) = R
Sf (x, y) = F (x). Hence, F is continu- ous and this proves (a).
Part (b) is proved by applying the bounded convergence theorem to the sequence of difference quo- tients f (x + hjei, y) − f (x, y)
hj , where ei denotes the unit vector in the xi−coordinate and {hj} is a sequence of numbers tending to zero. The uniform bound on these quotients is obtained by applying the mean value theorem .
Examples.
(a) Let f (x, t) = cos tx
1 + x2 for x ∈ [0, ∞) and t ∈ (−∞, ∞). Then R∞
0 f (x, t) converges uniformly for t ∈ R by Dominated Convergence Theorem.
(b) Let f (x, t) = e−xxt for x ∈ [0, ∞) and t ∈ [0, ∞). For any β > 0, the integral R∞
0 f (x, t) converges uniformly for t ∈ [0, β] by Dominated Convergence Theorem. Similarly, the Laplace transform of xn, n = 0, 1, 2, . . . , defined by L {xn}(t) = R∞
0 xne−txdx also converges uniformly for t ≥ γ > 0 to n!
tn+1. For t ≥ 1, define the gamma function Γ by Γ(t) = R∞
0 xt−1e−xdx. Then it is uniformly convergent on an interval containing t. Note that Γ(t + 1) = tΓ(t) and hence Γ(n + 1) = n! for any n ∈ N.
(c) Let f (x, t) = e−txsin x for x ∈ [0, ∞) and t ≥ γ > 0. Then the integral F (t) = R∞
0 e−txsin xdx is converges uniformly for t ≥ γ > 0 by Dominated Convergence Theorem and it is called the laplace transform of sin x, denoted by L {sin x}(t). Note that an elementary calculation shows that L {sin x}(t) = 1
1 + t2. (d) Let f (x, t, u) = e−txsin ux
x for x ∈ [0, ∞) and t, u ∈ [0, ∞). By taking φ = e−tx/x and by applying the Dirichlet’s test, one can show that R∞
γ f (x, t, u) converges uniformly for t ≥ γ ≥ 0. Note that if F (t, u) = L {sin ux
x }(t) =R∞
0 e−txsin ux
x dx, then ∂F
∂u(t, u) =R∞
0 e−txcos uxdx = t t2+ u2 and F (t, u) = tan−1 u
t. By setting u = 1 and by letting t → 0+, we obtain thatR∞
0
sin x
x dx = π 2.
(e) Let G(t) =R∞
0 e−x2−t2/x2dx for t > 0. Then G0(t) = −2G(t) and G(t) =
√π 2 e−2t. (f) Let F (t) =R∞
0 e−x2cos txdx for t ∈ R. Then F0(t) = −t
2F (t) and F (t) =
√π 2 e−t2/4.
Leibiniz’s Formula. Let f and its partial derivative ft be continuous on K = [a, b] × [c, d] to R and α and β be differentiable functions on [c, d] and have values in [a, b]. Then the function φ(t) = Rβ(t)
α(t) f (x, t)dx is differentiable on (c, d) and φ0(t) = f (β(t), t)β0(t) − f (α(t), t)α0(t) +Rβ(t)
α(t) ft(x, t)dx for t ∈ (c, d).
Proof. Let H be defined for (u, v, t) by H(u, v, t) = Ru
v f (x, t)dx, where u, v belong to [a, b] and t belongs to [c, d]. Then φ(t) = H(β(t), α(t), t). Applying the Chain Rule to obtain the result.
Interchange Theorem. Let f be continuous on K = [a, b] × [c, d] to R.
Then Rd
c
nRb
a f (x, t)dx o
dt =Rb
a
nRd
c f (x, t)dt o
dx.
Proof. Since f is uniformly continuous on K, if ² > 0 there exists a δ(²) > 0 such that if |x−x0| < δ(²) and |t − t0| < δ(²), then |f (x, t) − f (x0, t0)| < ². Let n be chosen so large that (b − a)/n < δ(²) and (d − c)/n < δ(²) and divide K into n2 equal rectangles by dividing [a, b] and [c, d] each into n equal parts. For j = 0, 1, . . . , n, we let xj = a + (b − a)j/n, tj = c + (d − c)j/n.
Then Z d
c
½Z b
a
f (x, t)dx
¾ dt =
Xn i=1
Xn j=1
Z ti
ti−1
(Z xj
xj−1
f (x, t)dx )
dt = Xn
i=1
Xn j=1
f (x0j, t0i)(xj − xj−1)(ti − ti−1).
Similarly, Z b
a
½Z d
c
f (x, t)dt
¾ dx =
Xn i=1
Xn j=1
f (x00j, t00i)(xj− xj−1)(ti− ti−1).
The uniform continuity of f implies that two iterated integrals differ by at most ²(b − a)(d − c). Since
² > 0 is arbitrary, the equality of these integrals is confirmed.
Example. Let A ⊆ R2 be the set consisting of all pairs (i/p, j/p) where p is a prime number, and i, j = 1, 2, . . . , p − 1. (a) Show that each horizontal and each vertical line in R2 intersects A in a finite number (often zero) of points and that A does not have content. Let f be defined on K = [0, 1]×[0, 1]
by f (x, y) = 1 for (x, y) ∈ A and f (x, y) = 0 otherwise. (b) Show that f is not integrable on K.
However, the iterated integrals exist and satisfyR1
0
nR1
0 f (x, y)dx o
dy =R1
0
nR1
0 f (x, y)dy o
dx.
Example. 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 thatR
Kf =R1
0
nR1
0 f (x, y)dxo
dy = 0, but that R1
0 f (x, y)dy does not exist for rational x.
Fubini’s Theorem. Let f be continuous on K = [a, b]×[c, d] to R. ThenR
Kf =Rd
c
nRb
af (x, y)dx o
dy = Rb
a
nRd
c f (x, y)dy o
dx.
Proof. Let F be defined for y ∈ [c, d] by F (y) = Rb
af (x, y)dx. Let c = y0 < y1 < · · · < yr = d be a partition of [c, d], let a = x0 < x1 < · · · < xs = b be a partition of [a, b], and let P denote the partition of K obtained by using the cells [xi−1, xi] × [yj−1, yj]. Let yj∗ be any point in [yj−1, yj] and note that F (y∗j) =
Z b
a
f (x, yj∗)dx = Xs
i=1
Z xi
xi−1
f (x, y∗j)dx. The Mean Value Theorem implies that
for each j, there exists a x∗ji ∈ [xi−1, xi] such that F (yj∗) = Xs
i=1
f (x∗ji, y∗j)(xi− xi−1). We multiply by
(yj− yj−1) and add to obtain Xr
j=1
F (y∗j)(yj− yj−1) = Xr
j=1
Xs i=1
f (x∗ji, yj∗)(xi− xi−1)(yj− yj−1). 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.
Generalization Theorem. Let f be integrable on K = [a, b] × [c, d] to R and suppose that 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 R
Kf = Rd
c
nRb
a f (x, y)dx o
dy.
Corollary. Let A ⊆ R2 be given by A = {(x, y) : α(y) ≤ x ≤ β(y), c ≤ y ≤ d}, where α and β are continuous functions on [c, d] with values in [a, b]. If f is continuous on A 7→ R, then f is integrable on A andR
Af =Rd
c
nRβ(y)
α(y) f (x, y)dx o
dy.
Proof. Let K ⊇ A be a closed cell and fK be the extension of f to K. Since ∂A has content zero, fK is integrable on K. Now for each y ∈ [c, d] the function x 7→ fK(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 R
Af =R
KfK =Rd
c
nRb
afK(x, y)dx o
dy =Rd
c
nRβ(y)
α(y) f (x, y)dx o
dy.
Example. Let R denote the triangular region in the first quadrant bounded by the lines y = x, y = 0, and x = 1. ThenR1
0
R1
y
sin x
x dxdy =R1
0
Rx
0
sin x x dydx.
Example.
Z 2
0
Z 1
y/2
ye−x3 dxdy = Z 1
0
Z 2x
0
ye−x3 dydx = Z 1
0
y2
2 e−x3|2x0 dx
= Z 1
0
2x2e−x3dx = −2e−x3 3 |10 = 2
3(1 − e−1).
Example. For β > α > 0, let R = [0, ∞) × [α, β] and f (x, t) = e−tx. Then log β
α = Rβ
α
1 tdt = Rβ
α
R∞
0 e−txdxdt =R∞
0
Rβ
α e−txdtdx =R∞
0
e−αx− e−βx
x dx.
Lemma. Let Ω ⊆ Rp be open, φ : Ω → Rp belong to Class C1(Ω), and A be a bounded set with Cl(A) = ¯A ⊂ Ω.
Then there exists a bounded open set Ω1 with
A ⊂ Ω¯ 1 ⊂ ¯Ω1 ⊆ Ω
and a constant M > 0 such that if A is contained in the union of a finite number of closed cells in Ω, with total content at most α, then φ(A) is contained in the union of a finite number of closed cells in Ω, with total content at most (√
pM )pα.
Proof. If Ω = Rp, let δ = 1; otherwise let δ = 1
2inf{ka − xk : a ∈ ¯A, x /∈ Ω} > 0.
Let
Ω1 = {y ∈ Rp : ky − ak < δ for some a ∈ A}
M = sup{kdφ(x)kpp= sup
06=v∈Rp
kdφx(v)k/kvk : x ∈ Ω1} < ∞.
If A ⊆ I1∪ · · · ∪ Im, where the Ij are closed cells contained in Ω1, then it follows that for x, y ∈ Ij we have
kφ(x) − φ(y)k ≤ Mkx − yk.
Suppose the side length of Ij is 2rj and take x to be the center of Ij; then if y ∈ Ij, we have kx − yk ≤√
prj. Thus φ(Ij) is contained in a closed cell of side length 2√
pMrj, and φ(A) is contained in the union of a finite number closed cells with total content at most (√
pM )pα.
Theorem. Let Ω ⊆ Rp be open, φ : Ω → Rp belong to Class C1(Ω). If A ⊂ Ω has content zero and if ¯A ⊂ Ω, then φ(A) has content zero.
Proof. Apply the lemma for arbitrary α > 0.
Corollary. Let r < p, Ω ⊆ Rr be open, and ψ : Ω → Rp belong to Class C1(Ω). If A ⊂ Ω is a bounded set with ¯A ⊂ Ω, then ψ(A) has content zero in Rp.
Proof. Let Ω0 = Ω × Rp−r. Then Ω0 is open in Rp. Define φ : Ω0 → Rp by
φ(x1, . . . , xr, xr+1, . . . , xp) = ψ(x1, . . . , xr).
Thus φ ∈ C1(Ω0).
Let A0 = A × {(0, . . . , 0)}. Then ¯A0 ⊂ Ω0 and A0 has content zero in Rp. It follows that ψ(A) = φ(A0) has content zero in Rp.
Theorem. Let Ω ⊆ Rp be open, φ : Ω → Rp belong to Class C1(Ω). Suppose that A has content, A ⊂ Ω, and the Jacobian determinant J¯ φ(x) = det(dφ)(x) 6= 0 for all x ∈ Int(A). Then φ(A) has content.
Proof. Since φ( ¯A) is compact and φ(A) ⊆ φ( ¯A), φ(A) is bounded in Rp, and φ(A) ⊆ φ( ¯A).
Now ∂φ(A) ∪ Int(φ(A)) = φ(A) ⊆ φ( ¯A) = φ(Int(A) ∪ ∂A),
and since φ ∈ C1(Ω) and Jφ(x) 6= 0 for all x ∈ Int(A), we conclude that φ(Int(A)) ⊆ Int(φ(A)) by the inverse function theorem.
Hence, we infer that ∂φ(A) ⊆ φ(∂A), and c(∂φ(A)) = 0.
Thus φ(A) has content.
Corollary. Let Ω ⊆ Rp be open, φ : Ω → Rp belong to Class C1(Ω) and be injective on Ω.
If A has content, ¯A ⊂ Ω, and Jφ(x) 6= 0 for all x ∈ Int(A).
Then ∂φ(A) = φ(∂A).
Proof. The proof of the inclusion φ(∂A) ⊇ ∂φ(A) is given in the proof of the proceeding theorem.
To establish the identity ∂φ(A) = φ(∂A), we only need to show the that φ(∂A) ⊆ ∂φ(A).
Let x ∈ ∂A, then there exists a sequence {xn} in A and a sequence {yn} in Ω \ A, such that
n→∞lim xn= x = lim
n→∞yn.
Since φ is continuous, we have lim
n→∞φ(xn) = φ(x) = lim
n→∞φ(yn).
On the other hand, since φ is injective on Ω, φ(yn) /∈ φ(A).
Thus φ(x) ∈ ∂φ(A) which implies that φ(∂A) ⊆ ∂φ(A).
Theorem. Let
L ∈ L (Rp) = {L : Rp → Rp | L = (lij) is an p × p matrix over R}
= the space of linear mappings on Rp, and let A ∈ D(Rp).
Then c(L(A)) = | det L| c(A).
Outline of the Proof. If det L = 0, then L(Rp) = Rr for some r < p, and that c(L(A)) = 0 for all A ∈ D(Rp).
If det L 6= 0, and if A ∈ D(Rp), then L(A) ∈ D(Rp).
Also, note that
(1) if A ∈ D(Rp), then L(A) has content and c(L(A)) ≥ 0.
(2) suppose A, B ∈ D(Rp) and A∩B = ∅, then L(A)∩L(B) = ∅ and c(L(A∪B)) = c(L(A)∪L(B)) = c(L(A)) + c(L(B)).
(3) if x ∈ Rp and A ∈ D(Rp), then L(x + A) = L(x) + L(A) and c(L(x + A)) = c(L(A)).
These imply that
(i) there exists a constant mL ≥ 0 such that c(L(A)) = mLc(A) for all A ∈ D(Rp).
(ii) if L, M ∈ L (Rp) are nonsingular maps and if A ∈ D(Rp), since mL◦Mc(A) = c(L ◦ M(A)) = mLc(M(A)) = mLmMc(A), we have mL◦M = mLmM.
(iii) one can show that mL= | det L| since every nonsingular L ∈ L (Rp) is the composition of linear maps of the following three forms:
(a) L1(x1, . . . , xp) = (αx1, x2, . . . , xp) for some α 6= 0;
(b) L2(x1, . . . , xi, xi+1, . . . , xp) = (x1, . . . , xi+1, xi, . . . , xp);
(c) L3(x1, . . . , xp) = (x1+ x2, x2, . . . , xp).
If K0 = [0, 1) × · · · × [0, 1) in Rp, then one can show that
mL1 = mL1c(K0) = c(L1(K0)) = |α| = | det L1|, mL2 = mL2c(K0) = c(L2(K0)) = 1 = | det L2|, mL3 = mL3c(K0) = c(L3(K0)) = 1 = | det L3|.
Hence, we have mL= | det L|.
Lemma. Let K ⊂ Rp be a closed cell with center 0, Ω be an open set containing K and ψ : Ω → Rp belong to Class C1(Ω) and be injective. Suppose further that Jψ(x) 6= 0 for x ∈ K and that kψ(x) − xk ≤ αkxk for x ∈ K, and some constant 0 < α < 1
√p. Then
(1 − α√
p)p ≤ c(ψ(K))
c(K) ≤ (1 + α√ p)p. Proof. If the side length of K is 2r and if x ∈ ∂K, then we have
r ≤ kxk ≤ r√ p.
This implies that
kψ(x) − xk ≤ αkxk ≤ αr√ p, i.e. ψ(x) is within distance αr√
p of x ∈ ∂K.
Note that Rp\ ∂K is a disjoint union of two nonempty open sets. Let Ci be an open cell with center 0 and side length 2(1 − α√
p)r, Co be a closed cell with center 0 and side length 2(1 + α√
p)r.
Then we have Ci ⊂ ψ(K) ⊂ Co, which implies that (1−α√
p)pc(K) = (1−α√
p)p(2r)p = c(Ci) ≤ c(ψ(K)) ≤ c(Co) = (1+α√
p)p(2r)p = (1+α√
p)pc(K).
The Jacobian Theorem. Let Ω ⊆ Rp be open, φ : Ω → Rp belong to Class C1(Ω), and be injective on Ω with Jφ(x) 6= 0 for x ∈ Ω. Let A have content and satisfy that ¯A ⊂ Ω. If ² > 0 is given, then there exists γ > 0 such that if K is a closed cell with center x ∈ A and side length less than 2γ, then
|Jφ(x)|(1 − ²)p ≤ c(φ(K))
c(K) ≤ |Jφ(x)|(1 + ²)p. Proof. For each x ∈ Ω, let Lx = (dφ(x))−1,
then 1 = det(Lx◦ dφ(x)) = (det Lx) (det dφ(x)), it follows that det Lx = 1
det dφ(x) = 1 Jφ(x). Let Ω1 be a bounded open subset of Ω such that
A ⊂ Ω¯ 1 ⊂ ¯Ω1 ⊆ Ω
and dist(A, ∂Ω1) = 2δ > 0.
Since φ ∈ C1(Ω), the entries in the standard matrix for Lx are continuous There exists a constant M > 0 such that kLxkpp ≤ M for all x ∈ Ω1. Let 0 < ² < 1 be a constant.
Since dφ is uniformly continuous on Ω1, there exists β with 0 < β < δ such that if x1, x2 ∈ Ω1 and kx1− x2k ≤ β, then kdφ(x1) − dφ(x2)kpp ≤ ²/M√
p.
Now let x ∈ A be given, hence if kzk ≤ β, then x, x + z ∈ Ω1. Hence,
kφ(x + z) − φ(x) − dφ(x)(z)k ≤ kzk sup
0≤t≤1kdφ(x + tz) − dφ(x)kpp
≤ ²
M√ pkzk.
This implies that for a fixed x ∈ A and for each kzk ≤ β if we set ψ(z) = Lx[φ(x + z) − φ(x)], Then we have
kψ(z) − zk = kLx[φ(x + z) − φ(x) − dφ(x)(z)]k
≤ Mkzk sup
0≤t≤1
kdφ(x + tz) − dφ(x)kpp
≤ ²
√pkzk for kzk ≤ β.
If K1 is any closed cell with center 0 and contained in the open ball with radius β, then (1 − ²)p ≤ c(ψ(K1))
c(K1) ≤ (1 + ²)p.
It follows that if K = x + K1 then K is a closed cell with center x and that c(K) = c(K1) and c(ψ(K1)) = | det Lx|c(φ(x + K1) − φ(x)) = 1
Jφ(x)c(φ(K)).
This establishes the inequality
|Jφ(x)|(1 − ²)p ≤ c(φ(K))
c(K) ≤ |Jφ(x)|(1 + ²)p.
for those closed cell K with center x ∈ A and side length less than 2γ = 2β/√ p.
Change of Variables Theorem. Let Ω ⊆ Rp be open, φ : Ω → Rp belong to Class C1(Ω), and be injective on Ω, and Jφ(x) 6= 0 for x ∈ Ω. Suppose that A has content, ¯A ⊂ Ω, and f : φ(A) → R is bounded and continuous.
Then R
φ(A)f =R
A(f ◦ φ)|Jφ|.
Example. Find RR
S
¡x2 + y2¢
dA if S be the region in the first quadrant bounded by the curves xy = 1, xy = 3, x2− y2 = 1, and x2− y2 = 4. Setting u = x2− y2, v = xy, we haveRR
S
¡x2+ y2¢ Rv=3 dA =
v=1
Ru=4
u=1
1
2 dudv = 3.
Example. ³R∞
0 e−x2dx´2
=R∞
0
R∞
0 e−x2−y2dxdy =Rπ/2
0
R∞
0 re−r2drdθ = π 2. Summary.
(a) Let A ⊂ Rn. Define what it means to say that A has content (or A is Jordan measurable).
(b) Let A ⊂ Rn. Define the content c(A) of A when A has content (or A is Jordan measurable).
(c) Let A ⊂ Rn. Define what it means to say that A has content (or measure) c(A) zero.
(d) Let A be a bounded subset of Rn, and let f be a bounded function defined on A to R. Define what it means to say that f is integrable on A. Give a class of A and a class of f from which R
Af exists.
(e) Let A be a bounded subset of Rnand let f be an integrable function defined on A to R. Discuss the continuity of f on A.
(f) Let A be a bounded subset of Rn and let f be a continuous function defined on A to R. Discuss the integrability of f on A?
(g) Let A be a bounded subset of Rn, let f, g be integrable functions defined on A to R, and let a, b ∈ R. Discuss the integrability of af + bg and f g on A.
(h) Let A be a bounded subset of Rn and let {fn} be a sequence of integrable function defined on A to R. Assume that f (x) = lim fn(x) exists for each x ∈ A. Discuss the integrability of f on A, and conditions on which the equality limR
Afn =R
Alim fn holds.
(i) Let Ω ⊆ Rp be open and let φ : Ω → Rp belong to Class C1(Ω). Suppose that A has content (or A is measurable), ¯A ⊂ Ω. Discuss the measurability of φ(A).
(j) Let Ω ⊆ Rp be open and let φ : Ω → Rp belong to Class C1(Ω). Suppose that A has content zero and if ¯A ⊂ Ω, discuss the measurability of φ(A).
(k) Let L ∈ L (Rp) = {B : Rp → Rp | B = (bij) is an p × p matrix over R} = the space of linear mappings on Rp, and let A ∈ D(Rp). Find c(L(A)).
(l) Let Ω ⊆ Rp be open and let φ : Ω → Rp belong to Class C1(Ω). Then dφ(x) : Rp → Rp is a linear mapping in L (Rp). If Jφ(x) 6= 0 for all x ∈ Ω and suppose that A has content (or A is measurable), ¯A ⊂ Ω. Discuss the geometric meaning of dφ(x)(v) for each v ∈ Rp, and the influence of dφ(x) on the volume element of A at x.
