- Prove that if fn ∈ R∗ and
converges uniformly to
f, then f ∈ R∗
Ifn = ∫
- In Example 12.2.5 there is the function given
It equals the derivative of a function as explained in this example. Thus g is generalized
Riemann integrable on . What about the function
h = max
- Let f ∈ R∗ and consider the function
dt. Is this function continuous?
Explain. Hint: Let ε > 0 be given and let a gauge, δ be such that if P is δ fine
Now pick h < δ for some
x ∈ such that
x + h < b. Then consider
the single tagged interval, where
x is the tag. By Corollary
Now you finish the argument and show f is continuous from the right. A similar
argument will work for continuity from the left.
- Generalize Problem 3 to the case where the integrator function is more general. You
need to consider two cases, one when the integrator function is continuous at x and the
other when it is not continuous.
- Suppose f ∈ R∗ and
f is continuous at x ∈
. Show G
differentiable at x and G′ =
- Suppose f has n + 1 derivatives on an open interval containing c. Show using induction
and integration by parts that
Would this technique work with the ordinary Riemann or Darboux integrals?
- The ordinary Riemann integral is only applicable to bounded functions. However, the
Generalized Riemann integral has no such restriction. Let f =
x−1∕2 for x > 0 and 0
for x = 0. Find ∫
01x−1∕2dx. Hint: Let fn = 0 for
x ∈ and
x > 1∕n. Now consider each of these functions and use the monotone convergence
- Can you establish a version of the monotone convergence theorem which has a
decreasing sequence of functions, rather than an increasing sequence?
- For E a subset of ℝ, let XE be defined by
For F an integrator function, define E to be measurable if for all n ∈ ℕ, XE ∈ R∗
and in this case, let
Show that if each Ek is measurable, then so is ∪k=1∞Ek and if E is measurable, then so
is ℝ ∖ E. Hint: This will involve the monotone convergence theorem.
- The gamma function is defined for x > 0 as
Show this limit exists. Also show that
How does Γ for
n an integer compare with !?
Construction Of Real Numbers
The purpose of this chapter is to give a construction of the real numbers from the
Definition A.0.1 Let R denote the set of Cauchy sequences of rational
n=1∞ is such a Cauchy sequence, this will be denoted by x for the
sake of simplicity of notation. A Cauchy sequence x will be called a null sequence
Denote the collection of such null Cauchy sequences as N.Then define x ∼ y if and only
Here x − y signifies the Cauchy sequence
n=1∞. Also, for simplicity of notation, let
Q denote the collection of constant Cauchy sequences: for a ∈ ℚ, let a ≡
= a for all n. Thus Q ⊆ R.
Then the following proposition is very easy to establish and is left to the reader.
Proposition A.0.2 ∼ is an equivalence relation on R.
Definition A.0.3 Define ℝ as the set of equivalence classes of ℝ. For
∈ ℝ, define
where xy is the sequence
n=1∞. Also define
Theorem A.0.4 With the two operations defined above, ℝ is a field.
Proof: First it is necessary to verify that the two operations are binary operations on ℝ.
This is obvious for addition. The question for multiplication reduces to whether xy is a
for some constant which is independent of n,m
. This follows because x,y
implies that these sequences are both bounded.
Now why are these operations well defined? Consider multiplication because it is fairly
obvious that addition is well defined. If x ∼ x′ and y ∼ y′, is it true that
n=1∞ ∈ N
is a constant which bounds all terms of all four given Cauchy sequences, the constant
existing because these are all Cauchy sequences. By assumption, the last expression converges
to 0 as n →∞
which verifies that
The case for addition is similar but easier.
Commutative and associative laws for addition and multiplication are all obvious. So is
the distributive law. The existence of an additive identity is clear. You just use
is a multiplicative identity. For
= 1 if xn
= 0 and yn
y ∈ R
there exists a subsequence of x,
and a positive number
for all k. Therefore, since x is a Cauchy sequence, it must be the case that for some
for all n > N. Now for m,n > N,
which shows that
. Then clearly
and yx is a Cauchy sequence which equals 1 for all n large enough. Therefore,
required. It is obvious that an additive inverse
exists for each
. Thus ℝ
a field as claimed. ■
Of course there are two other properties which need to be considered. Is ℝ ordered? Is ℝ
complete? First recall what it means for ℝ to be ordered. There is a subset of ℝ called the
positive numbers, such that
Definition A.0.5 Define
if there exists δ >
0 and a subsequence of x
k=1∞ with the property that
First, it is convenient to present a simple lemma.
if and only if for every
0 < r <
1 there exists δ >
m such that xn > rδ for all n > m.
Proof: ⇐= is obvious. Suppose then that
Then there exists δ >
0 and a
xnk > δ
. Let m
be large enough that
all l,n > m.
Then pick k > m.
It follows that nk > m
and so for n > m,
Theorem A.0.7 The above definition makes ℝ into an ordered field.
Proof: First, why is the above definition even well defined? Suppose
is positive with
as in the definition. Is it true that if x′∼ x,
is also positive? By the above lemma,
there exists an N
such that xn >
3 for all n > N.
Since x − x′∈ N,
it follows that there is
a larger N1
such that if n > N1,
then xn′ > δ∕
Thus the concept of being positive is well
are positive. Then by the above lemma, there exists an
such that if n > N,
Therefore, if n > N,
It is obvious that the sum of positive numbers is positive.
Next suppose that
. Then there exists a subsequence
It follows that either xnk > δ infinitely often or xnk < −δ infinitely often but not both. Thus
For a rational number r, denote by r the real number
being the constant sequence
. Then also r
It remains to verify that ℝ is complete. This requires a measure of distance between two
elements of ℝ. So how do you do this? Simply define the absolute value the usual
Then it is routine to verify that this satisfies the usual properties of absolute value. The main
difficulty is, as usual, the triangle inequality. Note that from the definition,
it is a similar argument. Thus the triangle inequality holds. What
? You can simply check the various cases. Consider the case that
Lemma A.0.8 If r > 1, then there exists mr such that for all n > mr,
Proof: Consider the last claim first. If
then there exists δ <
0 such that for all
large enough xn < δ <
so these are equal. The
0 is similar. Consider the other assertion.
and so, it follows as in Lemma A.0.6 that for all n large enough, the above conclusion holds.
Finally, here is a proof that ℝ is complete.
Theorem A.0.9 ℝ as defined above is a complete ordered field.
is a Cauchy sequence in ℝ
. Then there exists a subsequence,
still denoted as
I will show that a subsequence of this one converges. Then it will follow that the original
Cauchy sequence converges.
Since each xn is a Cauchy sequence, there exists N
that if k,l ≥ N
By making N
larger if necessary, it can also be assumed that for each
k ≥ N
Then define a sequence y as
the second term on the right of the top line being less that 2−k
is a Cauchy sequence.
It remains to verify if possible that limk→∞
= 0. First consider the
Let l ≥ N
Now both l,N
and so the middle term above is less than 2−k. Consider the last term. If l ≥ i ≥ k, then
Therefore, in the case that k≠l, the last term is dominated by
Next consider the first term. For k ≤ i ≤ N
Therefore, supposing that N
(There is nothing to show if this is not so.)
It follows that for l > N
The given sequence is a Cauchy sequence, so if ε > 0 is given, there exists Kε such that if
m,n > Kε, then
Pick k > Kε. Then for n > Kε,
Therefore, simply pick k
sufficiently large and it follows that for n > Kε,
Hence the given Cauchy sequence converges to
It follows you can consider each real number in ℝ as an equivalence class of
Cauchy sequences. It seems to me that it is easier to consider it simply as a point on
the number line and begin with the real number line as the fundamental thing to
There are other ways to construct the real numbers from the rational numbers. The
technique of Dedekind cuts is often used for example. See the advanced calculus book by
Rudin  for this. It might be a little shorter and easier to understand. However, the above
technique of the consideration of equivalence classes of Cauchy sequences can also be used to
complete any metric space. The technique of Dedekind cuts cannot. A metric space is just a
nonempty set X on which is defined a distance function (metric) which satisfies the following
axioms for x,y,z ∈ X.
Theory of Functions of Many Variables∗
After consideration of functions of one variable, one then goes to functions of many variables.
This chapter contains the basic theory which will allow this generalization. This book is
limited mostly to functions of one variable.
It is often the case that one desires to consider the complex numbers denoted as ℂ. It is
convenient to have the th