To aide in our proofs, let’s create an equivalent definition of Riemann Integrable.

Definition 1. A function f : A → R for A ⊂ Rⁿ is Riemann Integrable if for all real ε > 0 there exists a partition P of A such that

Note this is simply Theorem 3-3 of Spivak.

1 Problem 3 from Slides

Solution:
We will make use of Definition 1 for an easier proof. So let ε > 0. Define η > 0 so that

(1.1)

Because A is compact and f continuous on A, then f is uniformly continuous on A. Hence there is a δ > 0 such that |x - y| < δ implies

(1.2)

for all x,y ∈ A. Now define a partition P = (P₁,…,P_n) of A by P_i = {a_i,a_i + k_i,a_i + 2k_i} where k_i = and m is an integer chosen so that b_i -a_i < m for all i. Defining P in this way means that any two points x,y contained in the same rectangle S ∈ P will have

Hence, by equation 1.2 we have

(1.3)

due to f attaining its maximum and minimum value on the compact set S. Finally, through the use of equations 1.1 and 1.3 we obtain

2 Problem 4 from Slides

Let A ⊂ Rⁿ and denote it by A = [a₁,b₁] ×× [a_n,b_n]. Show that if f : A → R has only finitely many points of discontinuity, then it is Riemann integrable.

Solution:

Let ε > 0. We will show the existence of a partition P of A such that U(f,P) -L(f,P) < ε. To ease notation and therefore our proof, we define M = |f(x)| and define η > 0 so that

(2.4)

both of which we will use shortly.

Secluding Discontinuities. Let E = {x₁,…,x_k} be the points of discontinuity of f denoting each x_i by x_i = (x_i1,x_i2,…,x_in). Then define k closed squares = {A₁,…,A_k} by setting the length of the sides of each square to ℓ where ℓ is chosen so that

and

(2.5)

and then putting each A_i = [x_i1 -,x_i1 + ] ×× [x_in -,x_in + ]. By defining A_i this way, the first restriction on ℓ above ensures A₁,…,A_k are mutually exclusive. The second restriction above simply allows an important bound we’ll see shortly.

Partitioning A. Since E is covered by and each point of E in the interior of an element of , then K = A-∪_iA_i^∘, where A_i^∘ is the interior of A_i, is compact and has no intersection with E. Since f is continuous on A, it is continuous on K, and thus uniformly continuous on compact K. Hence we can find a δ > 0 such that

(2.6)

whenever |s - t| < δ for any s,t ∈ K. Now define a partition P = (P₁,…,P_n) of A by

where r_i = and z is an integer chosen so that < δ for all i. Defining P_i in this manner ensures that no point of , for any j, is contained in P_i and that all points outside of those intervals are at most within a distance of δ of each other. This restriction on the distance yields

(2.7)

for any S ∈ P - due to equation 2.6.

Conclusion. Given our definition of P and because |M_S(f) - m_S(f)|≤ 2M for any S ∈ P, we have the following sequence of equations allotted to us by equations 2.4, 2.5, and 2.7

U(f,P) - L(f,P)	= ∑ S∈P|MS(f) - mS(f)|v(S)
	= ∑ S∈P-|MS(f) - mS(f)|v(S) + ∑ S∈|MS(f) - mS(f)|v(S)
	< ∑ S∈P-η v(S) + ∑ S∈|MS(f) - mS(f)|v(S)
	≤∑ S∈P-η v(S) + 2M ∑ S∈v(S)
	≤∑ S∈P-η v(S) + 2M ∑ S∈ℓn
	= η ∑ S∈P-v(S) + 2Mkℓn
	≤ η v(A) + 2Mkℓn
	< ε∕2 + ε∕2
	= ε

3 Problem 5 from Slides

(a) Show that f + g is Riemann Integrable

(3.8)

and

(3.9)

implying that

(3.10)

This we state for later use.

So now let ε > 0. Since f and g are Riemann Integrable we can then find partitions P₁ and P₂ of A such that U(f,P₁) - L(f,P₁) < ε∕2 and U(g,P₂) - L(g,P₂) < ε∕2. Putting P = P₁ ∪ P₂ refines both P₁ and P₂ simultaneously, thus yielding U(f,P) - L(f,P) < ε∕2 and U(g,P) - L(g,P) < ε∕2. Adding these inequalities gives us

so that through application of equation 3.10 we get

ε	> +
	= ∑ S∈Pv(S) + ∑ S∈Pv(S)
	= ∑ S∈Pv(S)
	≥∑ S∈Pv(S)
	= U(f + g,P) - L(f + g,P)

(b) Show that ∫ _A = ∫ _Af + ∫ _Ag

for f and g. Hence for any partition P of A,

which implies

Since f and g are Riemann Integrable, then L(f,P) + L(g,P) and U(f,P) + U(g,P) can be brought arbitrarily close to each other. Thus the above inequality implies that all of L(f + g,P), L(f,P) + L(g,P), U(f,P) + U(g,P), and U(f + g,P) can be made arbitrarily close to each other by choosing an appropriate partition. Hence

(c) For constant c, show that cf is Riemann Integrable

Since U(cf,P) = cU(f,P) and L(cf,P) = cL(f,P), then

so that cf is Riemann Integrable.

(d) For constant c, show that ∫ _Acf = c∫ _Af

as desired.

4 Problem 6 from Slides

Show that we can use open rectangles instead of closed rectangles in the definition of “measure zero” and the sets that have measure zero will remain unchanged

Solution:

There is nothing really to prove to show that the open rectangle definition implies the closed rectangle definition since a countable set of open rectangles is a subset of set of those rectangles’ closures, but have the same volume.

To prove that the closed rectangle definition implies the open rectangle definition, let A ⊂ Rⁿ be a set of measure zero using the closed rectangle definition. Let ε > 0. Then we can find a countable collection of closed sets {V _i} that covers A such that

and denote each V _i by [a_i1,b_i1] ×× [a_in,b_in]. Choose r > 0 so that (1 + r)ⁿ < 2 and define a countable collection of open sets {U_i} by

for each i. Then with this definition we have V _i ⊂ U_i indicating that {U_i} is a cover of A by open rectangles. But furthermore, the volume of each open rectangle is

implying that the volume of the entire collection is

which reveals that A has measure zero according to the open rectangle definition, as well.

5 Problem 8 from Slides

Show that if a < b, then the closed interval [a,b] ⊂ R does not have content zero.

Solution:

Since [a,b] is a closed rectangle of volume b - a, then any covering of it by closed rectangles must have a total volume of at least b - a. Hence by setting ε = b - a there will never be a cover (finite or countably infinite) of [a,b] by closed rectangles with total volume less than this ε. Hence [a,b] must not be a set of content zero.

6 Problem 9 from Slides

Show that a compact set A ⊂ Rⁿ has measure zero if and only if it has content zero.

Solution:

Let A be a compact set with measure zero. Let ε > 0, then in light of problem 4 there is an uncountable collection of open rectangles = {U_i} with v( ) < ε. However, since A is compact there is a finite subcollection of , say U_n1,U_n2,…,U_nk, which covers A. Furthermore, this subcollection has the property

which implies that A has content zero.

The converse is trivial as content zero implies measure zero.

7 Problem 10 from Slides

Show that the set A of rational numbers between 0 and 1 does not have content zero.

Solution:

Let = {V _i} be a finite collection of n closed rectangles with total volume less than 1∕2. Denote each V _i by [a_i,b_i] and define j = argmin_i{a_i}. Note we know j exists due to the finite cardinality of . We then have two possible scenarios.

1. a_j > 0: If this is the case, then (0,a_j) would be uncovered by . Since A is dense in [0,1] then there would be a point of A contained in (0,a_j) and hence not be covered by .
2. a_j ≤ 0: If this is the case, we may repeat our process developed here to determine if A ∩ (b_j,1) is covered by - V _j. Given that is finite, we will have two eventualities; either we will come across the previous case, or we will hit this current case for at most n times, ending when is empty. If the former, we know does not cover A, but if the latter, then will cover [0,x] and not (x,∞) for some x ≥ 0. However, given that v( ) < 1∕2, then x < 1∕2, i.e. does not cover (1∕2,1). Since A is dense in [0,1], then A ∩ (1∕2,1) is nonempty and, furthermore, not covered by .

Because all cases result in some subset of A remaining uncovered by then there must be no finite set of closed rectangles that cover A and have volume less than 1∕2. Hence A does not have content zero.

8 Problem 11 from Slides

Let f : [a,b] → R be an increasing function. Show that the set of all points x ∈ [a,b] where f is discontinuous has measure zero.

Solution:

By Rudin’s Theorem 4.30, the set of points E in [a,b] where f is discontinuous is countable. So denote the points of E by x₁,x₂,x₃,…. Thus for any ε > 0 we can cover E with closed rectangles A₁,A₂,A₃,… by

so that

as desired.

9 Problem 12 from Slides

Show that the bounded function f : A → R is continuous at a ∈ A if and only if

Solution:
First assume that f is continuous. Let ε > 0. Then we can find a δ > 0 such that |x-a| < δ implies |f(x) -f(a)| < for all x ∈ A. Hence if r < δ then

In other words lim_r→0 = 0

Conversely, assume that lim_r→0 = 0. Let ε > 0. Then we can find δ > 0 such that r < δ implies that |M(a,f,r) - m(a,f,r)| < ε. Hence for any x ∈ A with |x - a| < δ we have

so that f is continuous.

10 Problem 13 from Slides

Let A ⊂ Rⁿ be a closed set and f : A → R a bounded function. Show that the set {x ∈ A|o(x,f) ≥ ε} is closed for any ε > 0.

Solution:
Let ε > 0 and put B = {x ∈ A|o(x,f) ≥ ε}. We show B is closed by showing it’s complement is open. Let b ∈ B^c, in which case either b ⁄∈ A or both b ∈ A and o(f,b) < ε. If the former then since A is closed there’s a neighborhood of b contained in A^c which is contained in B^c implying B is closed in this case. So assume the latter. Then lim_r→0 = ℓ for some ℓ with 0 ≤ ℓ < ε. Hence there exists a δ > 0 such that r < δ implies M(b,f,r) - m(b,f,r) - ℓ < ϵ - ℓ, i.e.

(10.11)

whenever r < δ. Let y ∈ B_δ∕2(b) where B_δ∕2(b) is the open ball around b of radius δ∕2. Then we have B_δ∕4(y) ⊂ B_δ∕2(b) which together with equation 10.11 implies

whenever r < δ∕4. In other words,

Hence y ∈ B^c. Since y ∈ B_δ∕2(b) was arbitrary, then B_δ∕2(b) ⊂ B^c, implying that B^c is open and it’s complement B is closed, as desired.

11 Problem 14 from Slides

Let A ⊂ Rⁿ be a closed rectangle and f : A → R a bounded function such that for all x ∈ A, o(f,x) < ε for a fixed ε > 0. Show that there is a partition P of A such that U(f,P) - L(f,P) < εv(A).

Solution:
Since o(f,x) < ε for all x ∈ A, then putting

for each x ∈ A yields ℓ_x < ε, i.e. ε-ℓ_x > 0 for each x ∈ A. The above equation then implies that for each x ∈ A there is a δ_x > 0 such that M(x,f,r) -m(x,f,r) -ℓ_x < ε-ℓ_x whenever r < δ_x, in other words M(x,f,r) -m(x,f,r) < ε whenever r < δ_x. Thus by setting δ = inf _x{δ_x} we have

(11.12)

for all x ∈ A whenever r < δ. For later ease of notation, put η = δ∕2, noting that therefore η < δ and so equation 11.12 applies for r = η.

Now denote A by [a₁,b₁] ×× [a_n,b_n] and define a partition P = (P₁,…,P_n) of A by setting

where we define k_i = , and m is chosen so that b_i - a_i < m for all i. Defining P in this way ensures that each rectangle S ∈ P has sides of length less than . This implies that for each such rectangle there’s an x_S ∈ A with S ⊂ B_η(x_S) where B_η(x_S) is the open ball of radius η centered at x_S. Hence equation 11.12 implies

U(f,P) - L(f,P)	= ∑ S∈Pv(S)
	≤∑ S∈Pv(S)
	< ∑ S∈Pεv(S)
	= ε∑ S∈P v(S)
	= εv(A)

12 Problem 20 from Slides

and define g : R → R by

for some real numbers a < b.

(a) Prove that f is of class C^∞

(12.13)

for some polynomial p(x). Since f′(x) = 2x^-3f(x), then

so that f⁽ⁿ⁺¹⁾(x) is the product of f and a polynomial. The inductive hypothesis thus tells us that 12.13 holds for all positive n, as desired.

(b) Prove that g is of class C^∞ and positive on (a,b) but zero elsewhere

(12.14)

for some polynomial p(x). Since g′(x) = (2(x - a)^-3 - 2(b - x)^-3)g(x), then

g(n+1)(x)	= p′(x)g(x) + p(x)g′(x)
	= p′(x)g(x) + p(x)(2(x - a)-3 - 2(b - x)-3)g(x)
	= g(x)

Furthermore, for any x₀ ≤ a we have x₀ -a ≤ 0 so that f(x₀ -a) = 0 which in turn means g(x₀) = 0. Likewise, when x₀ ≥ b then b-x₀ ≤ 0 so that f(b-x₀) = 0 implying g(x₀) = 0. Finally, whenever x₀ ∈ (a,b) we have both 0 < b-x₀ and 0 < x₀ - a implying that g(x₀) = f(x₀ - a)f(b - x₀) = e^{-(x₀-a)-2-(b-x ₀)-2} > 0 so that g(x₀) is positive in this case.

(c)

and then define h : R → R by

Show h is of class C^∞

Since g ∈ C^∞, then the fact that

implies that h is of class C^∞.

Show h(x) = 0 for x ≤ a

According to the previous part of this problem, g(x) = 0 for x ≤ a, so that for some c ∈ R

whenever x ≤ a.

Show 0 < h(x) < 1 for a < x < b

According to the previous part of this problem, g(x) > 0 for all x ∈ (a,b) and zero elsewhere. Hence

(12.15)

and

(12.16)

Inequality 12.15 allows us to add the value on its left to both sides of inequality 12.16. Doing so yields our desired upper bound.

∫ -∞xg(x)dx + ∫ x∞g(x)dx	> ∫ -∞xg(x)dx
∫ -∞∞g(x)dx	> ∫ -∞xg(x)dx
M	> ∫ -∞xg(x)dx
1	> ∫ -∞xg(x)dx
1	> h(x)

so that M > 0. Combining this with inequality 12.15 gives us the lower bound we desire.

∫ -∞xg(x)dx	> (0)
∫ -∞xg(x)dx	> 0
h(x)	> 0

Show h(x) = 1 for x ≥ b

Since g(x) = 0 whenever x ≥ b, then for x ≥ b we have

∫ -∞xg(x)dx	= ∫ -∞∞g(x)dx
	= 1
∫ -∞xg(x)dx	= 1
h(x)	= 1

(d)

Then given the properties of h we proved above, we obtain the following properties of k:

• When |x|≤ a h(|x|) = 0 so that k(x) = 1 - 0 = 1
• When a < |x|b 0 < h(|x|) < 1 so that

  0 >	- h(x) > -1
  1 >	1 - h(x) > 0
  1 >	k(x) > 0

• When |x|≥ b h(|x|) = 1 so that k(x) = 1 - 1 = 0

Furthermore, since the partial derivative of k with respect to x_i is

then the fact that h is of class C^∞ implies that k is also of the class C^∞, as desired.