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# 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

$$
\left| \int_a^b f \, d\alpha - I \right| < \epsilon
$$

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

$$
\int_a^b f \, d\alpha = I
$$

#### 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

$$
U(f, \alpha, P) = \sum_{i=1}^n M_i (x_i - x_{i-1})
$$

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

$$
L(f, \alpha, P) = \sum_{i=1}^n m_i (x_i - x_{i-1})
$$

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

$$
((x)) = \begin{cases} 
x-\lfloor x \rfloor & x \in [\lfloor x \rfloor, \lfloor x \rfloor + \frac{1}{2}) \\
0 & x=\lfloor x \rfloor + \frac{1}{2}\\
x-\lfloor x \rfloor - 1 & x \in (\lfloor x \rfloor + \frac{1}{2}, \lfloor x \rfloor + 1] \end{cases}
$$

![Graph of y=((x))](https://notenextra.trance-0.com/Math4121/y=((x)).png)

We define

$$
f(x) = \sum_{n=1}^{\infty} \frac{((nx))}{n^2}=\lim_{N\to\infty}\sum_{n=1}^{N} \frac{((nx))}{n^2}
$$

![Graph of y=f(x)](https://notenextra.trance-0.com/Math4121/sum_y=((x)).png)

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

$$
\left|f(x)-\sum_{n=1}^{N} \frac{((nx))}{n^2}\right|\leq \sum_{n=N+1}^{\infty}\frac{|((nx))|}{n^2}\leq \sum_{n=N+1}^{\infty} \frac{1}{n^2}<\epsilon
$$

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

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

$$
\begin{aligned}
\lim_{h\to 0^+}f(\frac{a}{2b}+h)-f(\frac{a}{2b})&=\lim_{h\to 0^+}\sum_{n=1}^{\infty}\frac{((\frac{na}{2b}+h))}{n^2}-\sum_{n=1}^{\infty}\frac{((\frac{na}{2b}))}{n^2}\\
&=\lim_{h\to 0^+}\sum_{n=1}^{\infty}\frac{((\frac{na}{2b}+h))-((\frac{na}{2b}))}{n^2}\\
&=\sum_{n=1}^{\infty}\lim_{h\to 0^+}\frac{((\frac{na}{2b}+h))-((\frac{na}{2b}))}{n^2}\\
&>0
\end{aligned}
$$


#### 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

$$
\omega(f, [x_{i-1}, x_i]) = \sup_{x,y \in [x_{i-1}, x_i]} |f(x) - f(y)|-\inf_{x,y \in [x_{i-1}, x_i]} |f(x) - f(y)|
$$

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 

$$
\lim_{n\to\infty}\int_0^1 f_n(x) \, dx = \int_0^1 \lim_{n\to\infty} f_n(x) \, dx=0
$$

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.