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^{∗}
[a,b]
and f is continuous at x ∈
[a,b]
. Show G
(y)
≡∫_{a}^{y}f
(t)
dt is
differentiable at x and G^{′}
(x)
= f
(x)
.
Suppose f has n + 1 derivatives on an open interval containing c. Show using induction
and integration by parts that
∫
∑n fk-(c) k 1- x (n+1) n
f (x ) = f (c)+ k! (x − c) + n! c f (t)(x− t) dt.
k=1
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)
= x^{−1∕2} for x > 0 and 0
for x = 0. Find ∫_{0}^{1}x^{−1∕2}dx. Hint: Let f_{n}
(x)
= 0 for x ∈
[0,1∕n]
and x^{−1∕2} for
x > 1∕n. Now consider each of these functions and use the monotone convergence
theorem.
Can you establish a version of the monotone convergence theorem which has a
decreasing sequence of functions,
{fk}
rather than an increasing sequence?
For E a subset of ℝ, let X_{E}
(x )
be defined by
{
1 if x ∈ E
XE (x) = 0 if x ∕∈ E
For F an integrator function, define E to be measurable if for all n ∈ ℕ, X_{E}∈ R^{∗}
[− n,n]
and in this case, let
{∫ n }
μ (E) ≡ sup XE (t)dt : n ∈ ℕ
−n
Show that if each E_{k} is measurable, then so is ∪_{k=1}^{∞}E_{k} 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
∫ ∫
∞ −tx−1 R −t x− 1
Γ (x ) ≡ 0 e t dt ≡ Rli→m∞ 0 e t dt
Show this limit exists. Also show that
Γ (x+ 1) = x Γ (x),Γ (1) = 1.
How does Γ
(n)
for n an integer compare with
(n − 1)
!?
Appendix A Construction Of Real Numbers
The purpose of this chapter is to give a construction of the real numbers from the
rationals.
Definition A.0.1Let R denote the set of Cauchy sequences of rationalnumbers. If
{xn}
_{n=1}^{∞}is such a Cauchy sequence, this will be denoted by x for thesake of simplicity of notation. A Cauchy sequence x will be called a null sequenceif
lim xn = 0.
n→∞
Denote the collection of such null Cauchy sequences asN.Then definex ∼ yif and onlyif
x− y ∈ N
Herex − ysignifies the Cauchy sequence
{xn − yn}
_{n=1}^{∞}. Also, for simplicity of notation, letQ denote the collection of constant Cauchy sequences: for a ∈ ℚ, leta ≡
{an}
_{n=1}^{∞}wherea_{n} = 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.0.3Define ℝ as the set of equivalence classes of ℝ. For
[x]
,
[y]
,
[z]
∈ ℝ, define
[x][y] ≡ [xy ]
wherexyis the sequence
{xnyn}
_{n=1}^{∞}. Also define
[x]+ [y] ≡ [x+ y ].
Theorem A.0.4With 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
Cauchy sequence.
for some constant which is independent of n,m. This follows because x,y being Cauchy
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
where C 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 →∞ and so
{x′ny′n − xnyn}
_{n=1}^{∞}∈ N which verifies that
[x′y′]
=
[xy ]
as hoped.
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
[0]
. Similarly
[1]
is a multiplicative identity. For
[x ]
≠
[0]
, let y_{n} = 1 if x_{n} = 0 and y_{n} = x_{n}^{−1} if x_{n}≠0. Is
y ∈ R? Since
[x]
≠
[0]
, there exists a subsequence of x,
{xnk}
and a positive number δ > 0
such that
|xnk| > 2δ
for all k. Therefore, since x is a Cauchy sequence, it must be the case that for some
N,
|xn| ≥ δ
for all n > N. Now for m,n > N,
| |
|y − y | = ||-1-−-1-||= |xn −-xm-|≤-1 |x − x |
n m |xn xm | |xn||xm| δ2 n m
which shows that
{yn}
_{n=1}^{∞}∈ R. Then clearly
[y ]
=
[x]
^{−1} because
[y][x] = [yx ]
and yx is a Cauchy sequence which equals 1 for all n large enough. Therefore,
[xy]
=
[1]
as
required. It is obvious that an additive inverse
[− x]
≡−
[x]
exists for each
[x]
∈ ℝ. Thus ℝ is
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
The sum of positive numbers is positive.
The product of positive numbers is positive.
[x ] is either positive [0], or − [x] is positive.
Definition A.0.5Define
[x]
>
[0]
if there exists δ > 0 and a subsequence of x
{xnk}
_{k=1}^{∞}with the property that
xnk > δ for all k.
First, it is convenient to present a simple lemma.
Lemma A.0.6
[x]
>
[0]
if and only if for every 0 < r < 1 there exists δ > 0 andm such that x_{n}> rδ for all n > m.
Proof:⇐= is obvious. Suppose then that
[x]
>
[0]
. Then there exists δ > 0 and a
subsequence
{xnk}
such that x_{nk}> δ. Let m be large enough that
|xn − xl|
<
(1 − r)
δ for
all l,n > m. Then pick k > m. It follows that n_{k}> m and so for n > m,
− (1 − r)δ < x − x < (1− r)δ
n nk
and so
rδ = δ− (1− r)δ < xnk − δ < xn ■
Theorem A.0.7The above definition makes ℝ into an ordered field.
Proof:First, why is the above definition even well defined? Suppose
[x]
is positive with δ
as in the definition. Is it true that if x^{′}∼ x, then x^{′} is also positive? By the above lemma,
there exists an N such that x_{n}> 2δ∕3 for all n > N. Since x − x^{′}∈N, it follows that there is
a larger N_{1} such that if n > N_{1}, then x_{n}^{′}> δ∕2. Thus the concept of being positive is well
defined.
Suppose both
[x]
,
[y]
are positive. Then by the above lemma, there exists an N,δ > 0
such that if n > N, then
xn > δ, yn > δ
Therefore, if n > N,
xnyn > δ2
and so
[xy]
=
[x ]
[y]
> 0. It is obvious that the sum of positive numbers is positive.
Next suppose that
[x]
≠
[0]
. Then there exists a subsequence
{xn }
k
such that
|x | > δ
nk
It follows that either x_{nk}> δ infinitely often or x_{nk}< −δ infinitely often but not both. Thus
either
[x ]
> 0 or
[x ]
< 0. ■
For a rational number r, denote by r the real number
[r]
,r being the constant sequence
equalling r. Then also r
[x]
=
[rx]
.
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
way.
(
{ [x] if [x ] > [0]
|[x]| ≡ ( [0] if [x] = [0] .
[− x] ≡ − [x] if [x ] < 0
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,
Therefore, simply pick k sufficiently large and it follows that for n > K_{ε},
|[xn]− [y]| < ε.
Hence the given Cauchy sequence converges to
[y]
. ■
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
study.
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 [32] 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.
d(x,y) = d (y,x), ∞ > d (x,y) ≥ 0
d(x,y)+ d(y,z) ≥ d(x,z)
d(x,y) = 0 if and only if x = y.
Appendix B 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