Math4121 Exam 2 Review

Range: Chapter 2-4 of Bressoud’s A Radical Approach to Lebesgue’s Theory of Integration

Chapter 2

The Riemann-Stieltjes Integral

Definition of the Riemann-Stieltjes Integral

Let $f$ be a bounded function on $[a,b]$ and $\alpha$ be a bounded function on $[a,b]$.

We say that $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a,b]$ if there exists a number $I$ such that for every $\epsilon > 0$, there exists a $\delta > 0$ such that for every partition $P = \{a = x_0, x_1, \ldots, x_n = b\}$ of $[a,b]$ with $||P|| < \delta$, we have

If $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a,b]$, we write

Darboux Sums

Let $P = \{a = x_0, x_1, \ldots, x_n = b\}$ be a partition of $[a,b]$.

The upper Darboux sum of $f$ with respect to $\alpha$ is

where $M_i = \sup_{x \in [x_{i-1}, x_i]} f(x)$ and $\alpha_i = \sup_{x \in [x_{i-1}, x_i]} \alpha(x)$.

The lower Darboux sum of $f$ with respect to $\alpha$ is

where $m_i = \inf_{x \in [x_{i-1}, x_i]} f(x)$ and $\alpha_i = \inf_{x \in [x_{i-1}, x_i]} \alpha(x)$.

Fail of Riemann-Stieltjes Integration

Consider the function

We define

(i) The series converges uniformly over $x\in[0,1]$.

As a consequence, $f(x)\in \mathscr{R}$.

(ii) $f$ has a discontinuity at every rational number with even denominator.

Some integrable functions are not differentiable (violates the fundamental theorem of calculus)

Solve:

Define the oscilation of $f$ on $[x_{i-1}, x_i]$ as

And define continuous functions as those functions that have oscilation 0 on every subinterval of their domain.

that is, the function $f$ is continuous at $c$ if $\omega(f,c) = 0$.

And we claim that the function is integrable on $[a,b]$ if and only if the outer measure of the set of discontinuities of $f$ is 0.

Finite cover:

Given a set $S$, an finite cover of $S$ is a finite collection of open/ or closed/ or half-open intervals $\{I_1, I_2, \ldots, I_n\}$ such that $S \subseteq \bigcup_{i=1}^n I_i$. The set of all finite covers of $S$ is denoted by $\mathcal{C}_S$.

Length of a cover:

The length of a cover $\ell(C)$ is the sum of the lengths of the intervals in the cover. (open/closed/half-open doesn’t matter.)

Outer content:

The outer content of a set $S$ is the infimum of the lengths of all finite covers of $S$. $c_e(S) = \inf_{C\in \mathcal{C}_S}\ell(C)$. (e denotes “exterior”)

Homework question: You cannot cover an interval $[a,b]$ with length $k$ with a finite cover of length strictly less than $k$.

Proceed by counting the intervals $I_i = [l_i, r_i]$ in the cover, and $r_n-l_0$ is less than or equal to $c_e(S)$ and $l_0\leq a$ and $r_n\leq b$.

Theorem 2.5

Given a bounded function $f$ defined on the interval $[a,b]$, let $S_\sigma$ be the points in $[a,b]$ with oscilation greater than $\sigma$.

The function $f$ is Riemann-Stieltjes integrable over $[a,b]$ if and only if $\lim_{\sigma \to 0} |S_\sigma| = 0$. That is, for every $\sigma > 0$, the outer content of $S_\sigma$ is 0.

Extra terminology:

Dense:

A set $S$ is dense in the interval $I$ is every open subinterval of $I$ contains a point of $S$.

This is equivalent to saying that $S$ is dense in $I$ if every point of $I$ is a limit point of $S$ or a point of $S$. (proved in homework)

Totally discontinuous:

A discontinuous function is totally discontinuous in an interval if the set of points of continuity is not dense in that interval.

In other words, there exists an open interval $I$ such that the set of points of continuity of $f$ in $I$ is empty.

Pointwise discontinuity:

A discontinuous function is pointwise discontinuous if the set of points of discontinuity is dense in the domain of $f$.

Accumulation point (limit point):

A point $p$ is an accumulation point of a set $S$ if every neighborhood of $p$ contains a point of $S$ other than $p$ itself. (That is, there exists a convergent sequence $\{p_n\}_{n=1}^\infty$ in $S$ such that $\lim_{n\to\infty} p_n = p$ and $p_n \neq p$ for all $n \in \mathbb{N}$. Proved in Rudin)

Derived set:

The derived set of a set $S$ is the set of all accumulation points of $S$. $S' = \{p \in \mathbb{R} \mid \forall \epsilon > 0, \exists x \in S \text{ s.t. } 0 < |x-p| < \epsilon\}$.

Type 1 set:

A set $S$ is a type 1 set if $S'\neq \emptyset$ and $S''=\emptyset$.

Type $n$ set:

A set $S$ is a type $n$ set if $S'$ is a type $n-1$ set.

First species:

A set $S$ is of first species if it is type $n$ for some $n\geq 0$, otherwise it is of second species.

$\mathbb{Q}$ is not first species since it is dense in $\mathbb{R}$ and $\mathbb{Q}' = \mathbb{R}$.

$\mathbb{R}$ is not first species.

Chapter 3

Topology of $\mathbb{R}$

Open set:

A set $S$ is open if for every $x \in S$, there exists an $\epsilon > 0$ such that $B_\epsilon(x) \subseteq S$.

Closed set:

A set $S$ is closed if its complement is open.

Equivalently, a set $S$ is closed if it contains all of its limit points. That is $S' \subseteq S$.

Interior of a set:

The interior of a set $S$ is the set of all points in $S$ such that there exists an $\epsilon > 0$ such that $B_\epsilon(x) \subseteq S$. $S^\circ = \{x \in S \mid \exists \epsilon > 0 \text{ s.t. } B_\epsilon(x) \subseteq S\}$. (It is also the union of all open sets contained in $S$.)

Closure of a set:

The closure of a set $S$ is the set of all points that for every $\epsilon > 0$, $B_\epsilon(x) \cap S \neq \emptyset$. $\overline{S} = \{x \in \mathbb{R} \mid \forall \epsilon > 0, B_\epsilon(x) \cap S \neq \emptyset\}$.

Boundary of a set:

The boundary of a set $S$ is the set of all points in $S$ that are not in the interior of $S$. $\partial S = \overline{S} \setminus S^\circ$.

Theorem 3.4

Bolzano-Weierstrass Theorem:

Every bounded infinite set has an accumulation point.

Proof:

Let $S$ be a bounded infinite set. Cut the interval $[a,b]$ into two halves, and let $I_1$ be one with infinitely many points of $S$. (such set exists since $S$ is infinite.)

Let $I_2$ be the one half with infinitely many points of $I_1$.

By induction, we can cut the interval into two halves, and let $I_{n+1}$ be the one half with infinitely many points of $I_n$.

By the nested interval property, there exists a point $c$ that is in all $I_n$.

$c$ is an accumulation point of $S$.

QED

Theorem 3.6 (Heine-Borel Theorem)

For any open cover of a compact set, there exists a finite subcover.

Compact set:

A set $S$ is compact if every open cover of $S$ has a finite subcover. In $\mathbb{R}$, this is equivalent to being closed and bounded.

Cardinality:

The cardinality of $\mathbb{R}$ is $\mathfrak{c}$.

The cardinality of $\mathbb{N}$, $\mathbb{Z}$, and $\mathbb{Q}$ is $\aleph_0$.

Chapter 4

Nowhere Dense set

A set $S$ is nowhere dense if there are no open intervals in which $S$ is dense.

That is equivalent to $S'$ contains no open intervals.

Note: If $S$ is nowhere dense, then $S^c$ is dense. But if $S$ is dense, $S^c$ is not necessarily nowhere dense. (Consider $\mathbb{Q}$)

Perfect Set

A set $S$ is perfect if $S'=S$.

Example: open intervals, Cantor set.

Cantor set

The Cantor set ($SVC(3)$) is the set of all real numbers in $[0,1]$ that can be represented in base 3 using only the digits 0 and 2.

The outer content of the Cantor set is 0.

Generalized Cantor set (SVC(n))

The outer content of $SVC(n)$ is $\frac{n-3}{n-2}$.

Lemma 4.4

Osgood’s Lemma:

Let $G$ be a closed, bounded set and Let $G_1\subseteq G_2\subseteq \ldots$ and $G=\bigcup_{n=1}^{\infty} G_n$. Then $\lim_{n\to\infty} c_e(G_n)=c_e(G)$.

Key: Using Heine-Borel Theorem.

Theorem 4.5

Arzela-Osgood Theorem:

Let $\{f_n\}_{n=1}^{\infty}$ be a sequence of continuous, uniformly bounded functions on $[0,1]$ that converges pointwise to $0$. It follows that

Key: Using Osgood’s Lemma and do case analysis on bounded and unbounded parts of the Riemann-Stieltjes integral.

Theorem 4.7

Baire Category Theorem:

An open interval cannot be covered by a countable union of nowhere dense sets.