From 2ed300e35e1cd4fb9dc2703ea48a32f7fdc31ef8 Mon Sep 17 00:00:00 2001
From: Trance-0 <60459821+Trance-0@users.noreply.github.com>
Date: Tue, 25 Mar 2025 22:30:10 -0500
Subject: [PATCH] update

---
 pages/Math4121/Exam_reviews/Math4121_E2.md | 192 +++++++++++++++++++++
 pages/Math416/Exam_reviews/Math416_E1.md   |   7 +
 2 files changed, 199 insertions(+)

diff --git a/pages/Math4121/Exam_reviews/Math4121_E2.md b/pages/Math4121/Exam_reviews/Math4121_E2.md
index 600d76c..7a4eeaf 100644
--- a/pages/Math4121/Exam_reviews/Math4121_E2.md
+++ b/pages/Math4121/Exam_reviews/Math4121_E2.md
@@ -4,6 +4,198 @@ Range: Chapter 2-4 of Bressoud's A Radical Approach to Lebesgue's Theory of Inte
 
 ## 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
+
+#### 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 $S$. $\partial S = \overline{S} \setminus S^\circ$.
+
+#### Missing Thoerem 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.
+
+### Perfect Set
+
+A set $S$ is **perfect** if $S'=S$.
+
+
+
+
diff --git a/pages/Math416/Exam_reviews/Math416_E1.md b/pages/Math416/Exam_reviews/Math416_E1.md
index 653d5ad..3bcb4ae 100644
--- a/pages/Math416/Exam_reviews/Math416_E1.md
+++ b/pages/Math416/Exam_reviews/Math416_E1.md
@@ -208,6 +208,13 @@ $$
 \int \frac{1}{x^2+a^2} dx=\frac{1}{a}\arctan(\frac{x}{a})
 $$
 
+$$
+\int \frac{1}{\sqrt{x^2-a^2}} dx=\ln|x+\sqrt{x^2-a^2}|
+$$
+
+$$
+\int \frac{1}{\sqrt{x^2+a^2}} dx=\ln|x+\sqrt{x^2+a^2}|
+$$
 
 ## Chapter 1 Complex Numbers