final update on 4121

pages/Math4121/Exam_reviews/Math4121_E1.md

@@ -0,0 +1,264 @@
+# Math4121 Exam 1 Review
+
+Range:  Chapter 5 and 6 of Rudin. We skipped (and so you will not be tested on)
+
+- Differentiation of Vector Valued Functions (pp. 111-113)
+- Integration of Vector-Valued Function and Rectifiable Curves (pp.135-137)
+
+You will also not be tested on Uniform Convergence and Integration, which we cover in class on Monday 2/10.
+
+## Chapter 5: Differentiation
+
+### Definition of the Derivative
+
+Let $f$ be a real function defined on an closed interval $[a,b]$. We say that $f$ is differentiable at a point $x \in [a,b]$ if the following limit exists:
+
+$$
+f'(x) = \lim_{t\to x} \frac{f(t) - f(x)}{t - x}
+$$
+
+If the limit exists, we call it the derivative of $f$ at $x$ and denote it by $f'(x)$.
+
+#### Theorem 5.2
+
+Every differentiable function is [continuous](https://notenextra.trance-0.com/Math4111/Math4111_L22#definition-45).
+
+The converse is not true, consider $f(x) = |x|$.
+
+#### Theorem 5.3
+
+If $f,g$ are differentiable at $x$, then
+
+1. $(f+g)'(x) = f'(x) + g'(x)$
+2. $(fg)'(x) = f'(x)g(x) + f(x)g'(x)$
+3. If $g(x) \neq 0$, then $(f/g)'(x) = (f'(x)g(x) - f(x)g'(x))/g(x)^2$
+
+#### Theorem 5.4
+
+Constant function is differentiable and its derivative is $0$.
+
+#### Theorem 5.5
+
+Chain rule: If $f$ is differentiable at $x$ and $g$ is differentiable at $f(x)$, then the composite function $g\circ f$ is differentiable at $x$ and
+
+$$
+(g\circ f)'(x) = g'(f(x))f'(x)
+$$
+
+#### Theorem 5.8
+
+The derivative of local extremum ($\exists \delta > 0$ s.t. $f(x)\geq f(y)$ or $f(x)\leq f(y)$ for all $y\in (x-\delta,x+\delta)$) is $0$.
+
+#### Theorem 5.9
+
+Generalized mean value theorem: If $f,g$ are differentiable on $(a,b)$, then there exists a point $x\in (a,b)$ such that
+
+$$
+(f(b)-f(a))g'(x) = (g(b)-g(a))f'(x)
+$$
+
+If we put $g(x) = x$, we get the mean value theorem.
+
+$$
+f(b)-f(a) = f'(x)(b-a)
+$$
+
+for some $x\in (a,b)$.
+
+#### Theorem 5.12
+
+Intermediate value theorem:
+
+If $f$ is differentiable on $[a,b]$, for all $\lambda$ between $f'(a)$ and $f'(b)$, there exists a $c\in (a,b)$ such that $f'(x) = \lambda$.
+
+#### Theorem 5.13
+
+L'Hôpital's rule: If $f,g$ are differentiable in $(a,b)$ and $g'(x) \neq 0$ for all $x\in (a,b)$, where $-\infty \leq a < b \leq \infty$,
+
+Suppose
+$$
+\frac{f'(x)}{g'(x)} \to A \text{ as } x\to a
+$$
+
+If 
+$$
+f(x) \to 0, g(x) \to 0 \text{ as } x\to a
+$$
+
+or if
+
+$$
+g(x) \to \infty \text{ as } x\to a
+$$
+
+then
+
+$$
+\lim_{x\to a} \frac{f(x)}{g(x)} = A
+$$
+
+#### Theorem 5.15
+
+Taylor's theorem: If $f$ is $n$ times differentiable on $[a,b]$, $f^{(n-1)}$ is continuous on $[a,b]$, and $f^{(n)}$ exists on $(a,b)$, for any distinct points $\alpha, \beta \in [a,b]$, there exists a point $x\in (\alpha, \beta)$ such that
+
+$$
+f(\beta) =\left(\sum_{k=0}^{n-1} \frac{f^{(k)}(\alpha)}{k!}(\beta-\alpha)^k\right) + \frac{f^{(n)}(x)}{n!}(\beta-\alpha)^n
+$$
+
+## Chapter 6: Riemann-Stieltjes Integration
+
+### Definition of the Integral
+
+Let $\alpha$ be a monotonically increasing function on $[a,b]$.
+
+A partition of $[a,b]$ is a set of points $P = \{x_0, x_1, \cdots, x_n\}$ such that
+
+$$
+a = x_0 < x_1 < \cdots < x_n = b
+$$
+
+Let $\Delta \alpha_i = \alpha(x_{i}) - \alpha(x_{i-1})$ for $i = 1, \cdots, n$.
+
+Let $m_i = \inf \{f(x) : x_{i-1} \leq x \leq x_{i}\}$ and $M_i = \sup \{f(x) : x_{i-1} \leq x \leq x_{i}\}$ for $i = 1, \cdots, n$.
+
+The lower sum of $f$ with respect to $\alpha$ is
+
+$$L(f,P,\alpha) = \sum_{i=1}^{n} m_i \Delta \alpha_i$$
+
+The upper sum of $f$ with respect to $\alpha$ is
+
+$$U(f,P,\alpha) = \sum_{i=1}^{n} M_i \Delta \alpha_i$$
+
+Let $\overline{\int_a^b} f(x) d\alpha(x)=\sup_P L(f,P,\alpha)$ and $\underline{\int_a^b} f(x) d\alpha(x)=\inf_P U(f,P,\alpha)$.
+
+If $\overline{\int_a^b} f(x) d\alpha(x) = \underline{\int_a^b} f(x) d\alpha(x)$, we say that $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a,b]$ and we write
+
+$$
+\int_a^b f(x) d\alpha(x) = \overline{\int_a^b} f(x) d\alpha(x) = \underline{\int_a^b} f(x) d\alpha(x)
+$$
+
+#### Theorem 6.4
+
+Refinement of partition will never make the lower sum smaller or the upper sum larger.
+
+$$
+L(f,P,\alpha) \leq L(f,P^*,\alpha) \leq U(f,P^*,\alpha) \leq U(f,P,\alpha)
+$$
+
+#### Theorem 6.5
+
+$\underline{\int_a^b} f(x) d\alpha(x) \leq \overline{\int_a^b} f(x) d\alpha(x)$
+
+#### Theorem 6.6
+
+$f\in \mathscr{R}(\alpha)$ on $[a,b]$ if and only if for every $\epsilon > 0$, there exists a partition $P$ of $[a,b]$ such that
+
+$$
+U(f,P,\alpha) - L(f,P,\alpha) < \epsilon
+$$
+
+#### Theorem 6.8
+
+Every continuous function on a closed interval is Riemann-Stieltjes integrable with respect to any monotonically increasing function.
+
+#### Theorem 6.9
+
+If $f$ is monotonically increasing on $[a,b]$ and **$\alpha$ is continuous on $[a,b]$**, then $f\in \mathscr{R}(\alpha)$ on $[a,b]$.
+
+Key: We can repartition the interval $[a,b]$ using $f$.
+
+#### Theorem 6.10
+
+If $f$ is bounded on $[a,b]$ and has only **finitely many discontinuities** on $[a,b]$, then $f\in \mathscr{R}(\alpha)$ on $[a,b]$.
+
+Key: We can use the bound and partition around the points of discontinuity to make the error arbitrary small.
+
+#### Theorem 6.11
+
+If $f\in \mathscr{R}(\alpha)$ on $[a,b]$, and $m\leq f(x) \leq M$ for all $x\in [a,b]$, and $\phi$ is a continuous function on $[m,M]$, then $\phi\circ f\in \mathscr{R}(\alpha)$ on $[a,b]$.
+
+_Composition of bounded integrable functions and continuous functions is integrable._
+
+#### Theorem 6.12
+
+Properties of the integral:
+
+Let $f,g\in \mathscr{R}(\alpha)$ on $[a,b]$, and $c$ be a constant. Then
+
+1. $f+g\in \mathscr{R}(\alpha)$ on $[a,b]$ and $\int_a^b (f(x) + g(x)) d\alpha(x) = \int_a^b f(x) d\alpha(x) + \int_a^b g(x) d\alpha(x)$
+2. $cf\in \mathscr{R}(\alpha)$ on $[a,b]$ and $\int_a^b cf(x) d\alpha(x) = c\int_a^b f(x) d\alpha(x)$
+3. $f\in \mathscr{R}(\alpha)$ on $[a,b]$ and $c\in [a,b]$, then $\int_a^b f(x) d\alpha(x) = \int_a^c f(x) d\alpha(x) + \int_c^b f(x) d\alpha(x)$.
+4. **Favorite Estimate**: If $|f(x)| \leq M$ for all $x\in [a,b]$, then $\left|\int_a^b f(x) d\alpha(x)\right| \leq M(\alpha(b)-\alpha(a))$.
+5. If $f\in \mathscr{R}(\beta)$ on $[a,b]$, then $\int_a^b f(x) d(\alpha+\beta) = \int_a^b f(x) d\alpha + \int_a^b f(x) d\beta$.
+
+#### Theorem 6.13
+
+If $f,g\in \mathscr{R}(\alpha)$ on $[a,b]$, then
+
+1. $fg\in \mathscr{R}(\alpha)$ on $[a,b]$
+2. $|f|\in \mathscr{R}(\alpha)$ on $[a,b]$ and $\left|\int_a^b f(x) d\alpha(x)\right| \leq \int_a^b |f(x)| d\alpha(x)$
+
+Key: (1), use Theorem 6.12, 6.11 to build up $fg$ from $(f+g)^2-f^2-g^2$. (2), take $\phi(x) = |x|$ in Theorem 6.11.
+
+#### Theorem 6.14
+
+Integration over indicator functions:
+
+If $a<s<b$, $f$ is bounded on $[a,b]$, and $f$ is continuous at $s$, and $\alpha(x)=I(x-s)$, then
+
+$$
+\int_a^b f(x) d\alpha(x) = f(s)
+$$
+
+Key: Note the max difference can be made only occurs at $s$.
+
+#### Theorem 6.15
+
+Integration over step functions:
+
+If $\alpha(x) = \sum_{i=1}^{n} c_i I(x-x_i)$ for $x\in [a,b]$, then
+
+$$
+\int_a^b f(x) d\alpha(x) = \sum_{i=1}^{n} c_i f(x_i)
+$$
+
+#### Theorem 6.21
+
+Fundamental theorem of calculus:
+
+Let $f\in \mathscr{R}(\alpha)$ on $[a,b]$, and $F(x) = \int_a^x f(t) d\alpha(t)$. Then
+
+1. $F$ is continuous on $[a,b]$
+2. If $f$ is continuous at $x\in [a,b]$, then $F$ is differentiable at $x$ and $F'(x) = f(x)$
+
+## Chapter 7: Sequence and Series of Functions
+
+### Example of non-Riemann integrable function
+
+$\lim_{m\to \infty} \lim_{n\to \infty} (\cos(m!\pi x))^{2n}=\begin{cases} 1 & x\in \mathbb{Q} \\ 0 & x\notin \mathbb{Q} \end{cases}$
+
+This function is everywhere discontinuous and not Riemann integrable.
+
+### Uniform Convergence
+
+#### Definition 7.7
+
+A sequence of functions $\{f_n\}$ converges uniformly to $f$ on $E$ if for every $\epsilon > 0$, there exists a positive integer $N$ such that
+
+$$
+|f_n(x) - f(x)| < \epsilon \text{ for all } x\in E \text{ and } n\geq N
+$$
+
+If $E$ is a point, then that's the common definition of convergence.
+
+If we have uniform convergence, then we can swap the order of limits.
+
+#### Theorem 7.16
+
+If $\{f_n\}\in \mathscr{R}(\alpha)$ on $[a,b]$, and $\{f_n\}$ converges uniformly to $f$ on $[a,b]$, then
+
+$$
+\int_a^b f(x) d\alpha(x) = \lim_{n\to \infty} \int_a^b f_n(x) d\alpha(x)
+$$
+
+Key: Use the definition of uniform convergence to bound the difference between the integral of the limit and the limit of the integral. $\int_a^b (f-f_n)d\alpha \leq |f-f_n| \int_a^b d\alpha = |f-f_n| (\alpha(b)-\alpha(a))$.

pages/Math4121/Exam_reviews/Math4121_E2.md

@@ -44,6 +44,45 @@ where $m_i = \inf_{x \in [x_{i-1}, x_i]} f(x)$ and $\alpha_i = \inf_{x \in [x_{i
 
 ### 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:
@@ -70,7 +109,7 @@ And we claim that the function is integrable on $[a,b]$ if and only if the outer
 
 > 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")
+> 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 $k$.
 
@@ -150,7 +189,7 @@ $\mathbb{R}$ is not first species.
 >
 > 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
+#### Theorem 3.4
 
 Bolzano-Weierstrass Theorem:
 
@@ -190,12 +229,51 @@ For any open cover of a compact set, there exists a finite subcover.
 
 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.
+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.
 
 
 

pages/Math4121/Exam_reviews/Math4121_Final.md

@@ -0,0 +1,362 @@
+# Math4121 Final Review
+
+## Guidelines
+
+There is one question from Exam 2 material.
+
+3 T/F from Exam 1 material.
+
+The remaining questions cover the material since Exam 2 (Chapters 5 and 6 of Bressoud and my lecture notes for the final week). 
+
+The format of the exam is quite similar to Exam 2, maybe a tad longer (but not twice as long, don't worry).
+
+## Chapter 5: Measure Theory
+
+### Jordan Measure
+
+> Content
+>
+> Let $\mathcal{C}_S^e$ be the set of all finite covers of $S$ by closed intervals ($S\subset C$, where $C$ is a finite union of closed intervals).
+>
+> Let $\mathcal{C}_S^i$ be the set of disjoint intervals that contained in $S$ ($\bigcup_{i=1}^n I_i\subset S$, where $I_i$ are disjoint intervals).
+>
+> Let $c_e(S)=\sup_{C\in\mathcal{C}_S^e} \sum_{i=1}^n |I_i|$ be the outer content of $S$.
+>
+> Let $c_i(S)=\inf_{I\in\mathcal{C}_S^i} \sum_{i=1}^n |I_i|$ be the inner content of $S$.
+>
+> _Here we use $|I|$ to denote the length of the interval $I$, in book we use volume but that's not important here._
+>
+> The content of $S$ is defined if $c(S)=c_e(S)=c_i(S)$
+
+Note that from this definition, **for any pairwise disjoint collection of sets** $S_1, S_2, \cdots, S_N$, we have
+
+$$
+\sum_{i=1}^N c_i(S_i)\leq c_i(\bigcup_{i=1}^N S_i)\leq c_e(\bigcup_{i=1}^N S_i)\leq \sum_{i=1}^N c_e(S_i)
+$$
+
+by $\sup$ and $\inf$ in the definition of $c_e(S)$ and $c_i(S)$.
+
+#### Proposition 5.1
+
+$$
+c_e(S)=c_i(S)+c_e(\partial S)
+$$
+
+Note the boundary of $S$ is defined as $\partial S=\overline{S}\setminus S$.
+
+Equivalently, $\forall x\in \partial S$, $\forall \epsilon>0$, $\exists p\notin S$ and $q\notin S$ s.t. $d(x,p)<\epsilon$ and $d(x,q)<\epsilon$.
+
+So the content of $S$ is defined if and only if $c_e(\partial S)=0$.
+
+> Jordan Measurable
+>
+> A set $S$ is Jordan measurable if and only if $c_e(\partial S)=0$, ($c(S)=c_e(S)=c_i(S)$)
+
+#### Proposition 5.2
+
+Finite additivity of content:
+
+Let $S_1, S_2, \cdots, S_N$ be a finite collection of pairwise disjoint Jordan measurable sets.
+
+$$
+c(\bigcup_{i=1}^N S_i)=\sum_{i=1}^N c(S_i)
+$$
+
+Example for Jordan measure of sets
+
+| Set | Inner Content | Outer Content | Content |
+| --- | --- | --- | --- |
+| $\emptyset$ | 0 | 0 | 0 |
+| $\{q\},q\in \mathbb{R}$ | 0 | 0 | 0 |
+| $\{\frac{1}{n}\}_{n=1}^\infty$ | 0 | 0 | 0 |
+| $\{[n,n+\frac{1}{2^n}]\}_{n=1}^\infty$ | 1 | 1 | 1 |
+| $SVC(3)$ | 0 | 1 | Undefined |
+| $SVC(4)$ | 0 | $\frac{1}{2}$ | Undefined |
+| $Q\cap [0,1]$ | 0 | 1 | Undefined |
+| $[0,1]\setminus Q$ | 0 | 1 | Undefined |
+| $[a,b], a<b\in \mathbb{R}$ | $b-a$ | $b-a$ | $b-a$ |
+| $[a,b),a<b\in \mathbb{R}$ | $b-a$ | $b-a$ | $b-a$ |
+| $(a,b],a<b\in \mathbb{R}$ | $b-a$ | $b-a$ | $b-a$ |
+| $(a,b),a<b\in \mathbb{R}$ | $b-a$ | $b-a$ | $b-a$ |
+
+### Borel Measure
+
+Our desired property of measures:
+
+1. Measure of interval is the length of the interval. $m([a,b])=m((a,b))=m([a,b))=m((a,b])=b-a$
+
+2. Countable additivity: If $S_1, S_2, \cdots, S_N$ are pairwise disjoint Borel measurable sets, then $m(\bigcup_{i=1}^N S_i)=\sum_{i=1}^N m(S_i)$
+
+3. Closure under set minus: If $S$ is Borel measurable and $T$ is Borel measurable, then $S\setminus T$ is Borel measurable with $m(S\setminus T)=m(S)-m(T)$
+
+#### Borel Measurable Sets
+
+$\mathcal{B}$ is the smallest $\sigma$-algebra that contains all closed intervals.
+
+> Sigma algebra: A $\sigma$-algebra is a collection of sets that is closed under **countable** union, intersection, and complement.
+>
+> That is:
+>
+> 1. $\emptyset\in \mathcal{B}$
+> 2. If $A\in \mathcal{B}$, then $A^c\in \mathcal{B}$
+> 3. If $A_1, A_2, \cdots, A_N\in \mathcal{B}$, then $\bigcup_{i=1}^N A_i\in \mathcal{B}$
+
+#### Proposition 5.3
+
+Borel measurable sets does not contain all Jordan measurable sets.
+
+Proof by cardinality of sets.
+
+Example for Borel measure of sets
+
+| Set | Borel Measure |
+| --- | --- |
+| $\emptyset$ | 0 |
+| $\{q\},q\in \mathbb{R}$ | 0 |
+| $\{\frac{1}{n}\}_{n=1}^\infty$ | 0 |
+| $\{[n,n+\frac{1}{2^n}]\}_{n=1}^\infty$ | 1 |
+| $SVC(3)$ | 0 |
+| $SVC(4)$ | 0 |
+| $Q\cap [0,1]$ | 0 |
+| $[0,1]\setminus Q$ | 1 |
+| $[a,b], a<b\in \mathbb{R}$ | $b-a$ |
+| $[a,b),a<b\in \mathbb{R}$ | $b-a$ |
+| $(a,b],a<b\in \mathbb{R}$ | $b-a$ |
+| $(a,b),a<b\in \mathbb{R}$ | $b-a$ |
+
+
+### Lebesgue Measure
+
+> Lebesgue measure
+>
+> Let $\mathcal{C}$ be the set of all countable covers of $S$.
+>
+> The Lebesgue outer measure of $S$ is defined as:
+>
+> $$m_e(S)=\inf_{C\in\mathcal{C}} \sum_{i=1}^\infty |I_i|$$
+>
+> If $S\subset[a,b]$, then the inner measure of $S$ is defined as:
+>
+> $$m_i(S)=(b-a)-m_e([a,b]\setminus S)$$
+>
+> If $m_i(S)=m_e(S)$, then $S$ is Lebesgue measurable.
+
+#### Proposition 5.4
+
+Subadditivity of Lebesgue outer measure:
+
+For any collection of sets $S_1, S_2, \cdots, S_N$,
+
+$$m_e(\bigcup_{i=1}^N S_i)\leq \sum_{i=1}^N m_e(S_i)$$
+
+#### Theorem 5.5
+
+If $S$ is bounded, then any of the following conditions imply that $S$ is Lebesgue measurable:
+
+1. $m_e(S)=0$
+2. $S$ is countable (measure of countable set is 0)
+3. $S$ is an interval
+
+> Alternative definition of Lebesgue measure
+>
+> The outer measure of $S$ is defined as the infimum of all the open sets that contain $S$.
+>
+> The inner measure of $S$ is defined as the supremum of all the closed sets that are contained in $S$.
+
+#### Theorem 5.6
+
+Caratheodory's criterion:
+
+A set $S$ is Lebesgue measurable if and only if for any set $X$ with finite outer measure,
+
+$$m_e(X-S)=m_e(X)-m_e(X\cap S)$$
+
+#### Lemma 5.7
+
+Local additivity of Lebesgue outer measure:
+
+If $I_1, I_2, \cdots, I_N$ are any countable collection of **pairwise disjoint intervals** and $S$ is a bounded set, then
+
+$$
+m_e\left(S\cup \bigcup_{i=1}^N I_i\right)=\sum_{i=1}^N m_e(S\cap I_i)
+$$
+
+#### Theorem 5.8
+
+Countable additivity of Lebesgue outer measure:
+
+If $S_1, S_2, \cdots, S_N$ are any countable collection of pairwise disjoint Lebesgue measurable sets, **whose union has a finite outer measure,** then
+
+$$
+m_e\left(\bigcup_{i=1}^N S_i\right)=\sum_{i=1}^N m_e(S_i)
+$$
+
+#### Theorem 5.9
+
+Any finite union or intersection of Lebesgue measurable sets is Lebesgue measurable.
+
+#### Theorem 5.10
+
+Any countable union or intersection of Lebesgue measurable sets is Lebesgue measurable.
+
+#### Corollary 5.12
+
+Limit of a monotone sequence of Lebesgue measurable sets is Lebesgue measurable.
+
+If $S_1\subseteq S_2\subseteq S_3\subseteq \cdots$ are Lebesgue measurable sets, then $\bigcup_{i=1}^\infty S_i$ is Lebesgue measurable. And $m(\bigcup_{i=1}^\infty S_i)=\lim_{i\to\infty} m(S_i)$
+
+If $S_1\supseteq S_2\supseteq S_3\supseteq \cdots$ are Lebesgue measurable sets, **and $S_1$ has finite measure**, then $\bigcap_{i=1}^\infty S_i$ is Lebesgue measurable. And $m(\bigcap_{i=1}^\infty S_i)=\lim_{i\to\infty} m(S_i)$
+
+#### Theorem 5.13
+
+Non-measurable sets (under axiom of choice)
+
+Note that $(0,1)\subseteq \bigcup_{q\in \mathbb{Q}\cap (-1,1)}(\mathcal{N}+q)\subseteq (-1,2)$
+
+$$
+\bigcup_{q\in \mathbb{Q}\cap (-1,1)}(\mathcal{N}+q)
+$$
+
+is not Lebesgue measurable.
+
+## Chapter 6: Lebesgue Integration
+
+### Lebesgue Integral
+
+Let the partition on y-axis be $l=l_0<l_1<\cdots<l_n=L$, and $S_i=\{x|l_i<f(x)<l_{i+1}\}$
+
+The Lebesgue integral of $f$ over $[a,b]$ is bounded by:
+
+$$
+\sum_{i=0}^{n-1} l_i m(S_i)\leq \int_a^b f(x) \, dx\leq \sum_{i=0}^{n-1} l_{i+1} m(S_i)
+$$
+
+> Definition of measurable function:
+>
+> A function $f$ is measurable if for all $c\in \mathbb{R}$, the set $\{x\in [a,b]|f(x)>c\}$ is Lebesgue measurable.
+>
+> Equivalently, a function $f$ is measurable if any of the following conditions hold:
+>
+> 1. For all $c\in \mathbb{R}$, the set $\{x\in [a,b]|f(x)>c\}$ is Lebesgue measurable.
+> 2. For all $c\in \mathbb{R}$, the set $\{x\in [a,b]|f(x)\geq c\}$ is Lebesgue measurable.
+> 3. For all $c\in \mathbb{R}$, the set $\{x\in [a,b]|f(x)<c\}$ is Lebesgue measurable.
+> 4. For all $c\in \mathbb{R}$, the set $\{x\in [a,b]|f(x)\leq c\}$ is Lebesgue measurable.
+> 5. For all $c<d\in \mathbb{R}$, the set $\{x\in [a,b]|c\leq f(x)<d\}$ is Lebesgue measurable.
+>
+> Prove by using the fact$\{x\in [a,b]|f(x)\geq c\}=\bigcap_{n=1}^\infty \{x\in [a,b]|f(x)>c-\frac{1}{n}\}$
+
+#### Proposition 6.3
+
+If $f,g$ is a measurable function, and $k\in \mathbb{R}$, then $f+g,kf,f^2,fg,|f|$ is measurable.
+
+> Definition of almost everywhere:
+>
+> A property holds almost everywhere if it holds everywhere except for a set of Lebesgue measure 0.
+
+#### Proposition 6.4
+
+If $f_n$ is a sequence of measurable functions, then $\limsup_{n\to\infty} f_n, \liminf_{n\to\infty} f_n$ is measurable.
+
+#### Theorem 6.5
+
+Limit of measurable functions is measurable.
+
+> Definition of simple function:
+>
+> A simple function is a linear combination of indicator functions of Lebesgue measurable sets.
+
+#### Theorem 6.6
+
+Measurable function as limit of simple functions.
+
+$f$ is a measurable function if and only if ffthere exists a sequence of simple functions $f_n$ s.t. $f_n\to f$ almost everywhere.
+
+### Integration
+
+#### Proposition 6.10
+
+Let $\phi,\psi$ be simple functions, $c\in \mathbb{R}$ and $E=E_1\cup E_2$ where $E_1\cap E_2=\emptyset$.
+
+Then
+
+1. $\int_E \phi(x) \, dx=\int_{E_1} \phi(x) \, dx+\int_{E_2} \phi(x) \, dx$
+2. $\int_E (c\phi)(x) \, dx=c\int_E \phi(x) \, dx$
+3. $\int_E (\phi+\psi)(x) \, dx=\int_E \phi(x) \, dx+\int_E \psi(x) \, dx$
+4. If $\phi\leq \psi$ for all $x\in E$, then $\int_E \phi(x) \, dx\leq \int_E \psi(x) \, dx$
+
+> Definition of Lebesgue integral of simple function:
+>
+> Let $\phi$ be a simple function, $\phi=\sum_{i=1}^n l_i \chi_{S_i}$
+>
+> $$\int_E \phi(x) \, dx=\sum_{i=1}^n l_i m(S_i\cap E)$$
+
+> Definition of Lebesgue integral of measurable function:
+>
+> Let $f$ be a nonnegative measurable function, then
+>
+> $$\int_E f(x) \, dx=\sup_{\phi\leq f} \int_E \phi(x) \, dx$$
+>
+> If $f$ is not nonnegative, then
+>
+> $$\int_E f(x) \, dx=\int_E f^+(x) \, dx-\int_E f^-(x) \, dx$$
+>
+> where $f^+(x)=\max(f(x),0)$ and $f^-(x)=\max(-f(x),0)$
+
+#### Proposition 6.12
+
+Integral over a set of measure 0 is 0.
+
+#### Theorem 6.13
+
+If a nonnegative measurable function $f$ has integral 0 on a set $E$, then $f(x)=0$ almost everywhere on $E$.
+
+#### Theorem 6.14
+
+Monotone convergence theorem:
+
+If $f_n$ is a sequence of monotone increasing measurable functions and $f_n\to f$ almost everywhere, and $\exists A>0$ s.t. $|\int_E f_n(x) \, dx|\leq A$ for all $n$, then $f(x)=\lim_{n\to\infty} f_n(x)$ exists almost everywhere and it's integrable on $E$ with
+
+$$
+\int_E f(x) \, dx=\lim_{n\to\infty} \int_E f_n(x) \, dx
+$$
+
+#### Theorem 6.19
+
+Dominated convergence theorem:
+
+If $f_n$ is a sequence of integrable functions and $f_n\to f$ almost everywhere, and there exists a nonnegative integrable function $g$ s.t. $|f_n(x)|\leq g(x)$ for all $x\in E$ and all $n$, then $f(x)=\lim_{n\to\infty} f_n(x)$ exists almost everywhere and it's integrable on $E$ with
+
+$$
+\int_E f(x) \, dx=\lim_{n\to\infty} \int_E f_n(x) \, dx
+$$
+
+#### Theorem 6.20
+
+Fatou's lemma:
+
+If $f_n$ is a sequence of nonnegative integrable functions, then
+
+$$
+\int_E \liminf_{n\to\infty} f_n(x) \, dx\leq \liminf_{n\to\infty} \int_E f_n(x) \, dx
+$$
+
+> Definition of Hardy-Littlewood maximal function
+>
+> Given integrable $f$m and an interval $I$, look at the averaging operator $A_I f(x)=\frac{\chi_I(x)}{m(I)}\int_I f(y)dy$.
+>
+> The maximal function is defined as
+>
+> $$f^*(x)=\sup_{I \text{ is an open interval}} A_I f(x)$$
+
+### Lebesgue's Fundamental theorem of calculus
+
+If $f$ is Lebesgue integrable on $[a,b]$, then $F(x) = \int_a^x f(t)dt$ is differentiable **almost everywhere** and $F'(x) = f(x)$ **almost everywhere**.
+
+Outline:
+
+Let $\lambda,\epsilon > 0$. Find $g$ continuous such that $\int_{\mathbb{R}}|f-g|dm < \frac{\lambda \epsilon}{5}$.
+
+To control $A_I f(x)-f(x)=(A_I(f-g)(x))+(A_I g(x)-g(x))+(g(x)-f(x))$, we need to estimate the three terms separately.
+
+Our goal is to show that $\lim_{r\to 0^+}\sup_{I\text{ is open interval}, m(I)<r, x\in I}|A_I f(x)-f(x)|=0$. For $x$ almost every $x\in[a,b]$.
+

pages/Math4121/Math4121_L13.md

@@ -89,12 +89,16 @@ x-\lfloor x \rfloor & x \in [\lfloor x \rfloor, \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 $x\in[0,1]$.
 
 $$

pages/Math4121/Math4121_L19.md

@@ -22,7 +22,23 @@ $$
 
 However, the quarter cantor set removes $\frac{3^{n-1}}{4^n}$ of the total "content", and the total length removed after infinitely many steps is:
 
-_skip this part, some error occurred._
+Every time we remove $\frac{1}{4^n}$ of the remaining intervals. So on each layer, we remove $\frac{2^{n-1}}{4^n}$ of the total "content".
+
+So the total length removed is:
+
+$$
+\begin{aligned}
+1-\frac{1}{4}-\frac{2}{4^2}-\frac{2^2}{4^3}-\cdots&=1-\frac{1}{4}\sum_{n=0}^{\infty} \left(\frac{2}{4}\right)^n\\
+&=1-\frac{1}{4}\cdot\frac{1}{1-\frac{2}{4}}\\
+&=1-\frac{1}{4}\cdot\frac{4}{2}\\
+&=1-\frac{1}{2}\\
+&=\frac{1}{2}
+\end{aligned}
+$$
+
+#### Generalized Cantor set (SVC(n))
+
+The outer content of $SVC(n)$ is $\frac{n-3}{n-2}$.
 
 #### Monotonicity of outer content
 

pages/Math4121/Math4121_L24.md

@@ -22,13 +22,13 @@ Since $c_e(S)=0$ and $\partial S=[0,1]$, $c_i(S)=1$.
 
 So $c_e(\partial S)=1\neq 0$.
 
-2. SVC(3) is Jordan measurable.
+2. $SVC(3)$ is Jordan measurable.
 
 Since $c_e(S)=0$ and $\partial S=0$, $c_i(S)=0$. The outer content of the cantor set is $0$.
 
 > Any set or subset of a set with $c_e(S)=0$ is Jordan measurable.
 
-3. SVC(4)
+3. $SVC(4)$
 
 At each step, we remove $2^n$ intervals of length $\frac{1}{4^n}$.
 

pages/Math4121/Math4121_L9.md

@@ -75,30 +75,9 @@ Let $f,g\in \mathscr{R}(\alpha)$ on $[a, b]$.
 
 Proof:
 
-Property (aa), (b), (e) holds for Riemann Sums themselves.
+**Property (aa), (b), (e) holds for Riemann Sums themselves.**
 
-$$
-\sup cf(x) = c\sup f(x)\quad \forall c\in \mathbb{R}
-$$
-
-$$
-U(P,cf, \alpha) = cU(P,f,\alpha)
-$$
-
-For (b), notice that if $f(x)\leq g(x)$, then $\sup f(x)\leq \sup g(x)$, $U(P,f,\alpha)\leq U(P,g,\alpha)$. and $L(P,f,\alpha)\leq L(P,g,\alpha)$.
-
-For (e), notice that
-
-$$
-\begin{aligned}
-\Delta (\alpha+\beta)_i &= \alpha(x_i)-\alpha(x_{i-1})+\beta(x_i)-\beta(x_{i-1}) \\
-&= \Delta \alpha_i + \Delta \beta_i
-\end{aligned}
-$$
-
-(c),(d) are left as homework.
-
-For (a), Set $h(x)=f(x)+g(x)$. Then $h\in \mathscr{R}(\alpha)$ on $[a, b]$ and we will show $\int_a^b h d\alpha \leq \int_a^b f d\alpha + \int_a^b g d\alpha$.
+**For (a)**, Set $h(x)=f(x)+g(x)$. Then $h\in \mathscr{R}(\alpha)$ on $[a, b]$ and we will show $\int_a^b h d\alpha \leq \int_a^b f d\alpha + \int_a^b g d\alpha$.
 
 Since $f,g\in \mathscr{R}(\alpha)$ on $[a, b]$, for any $\epsilon > 0$, there exists a partition $P_1,P_2$ of $[a, b]$ such that $U(f,P_1,\alpha)-L(f,P_1,\alpha) < \epsilon$ and $U(g,P_2,\alpha)-L(g,P_2,\alpha) < \epsilon$.
 
@@ -108,4 +87,76 @@ So $U(P,h,\alpha)\leq U(P,f,\alpha)+U(P,g,\alpha)\leq \int_a^b f d\alpha + \int_
 
 Since $\epsilon$ is arbitrary, $\int_a^b h d\alpha \leq \int_a^b f d\alpha + \int_a^b g d\alpha$.
 
+
+$$
+\sup cf(x) = c\sup f(x)\quad \forall c\in \mathbb{R}
+$$
+
+$$
+U(P,cf, \alpha) = cU(P,f,\alpha)
+$$
+
+**For (b)**, notice that if $f(x)\leq g(x)$, then $\sup f(x)\leq \sup g(x)$, $U(P,f,\alpha)\leq U(P,g,\alpha)$. and $L(P,f,\alpha)\leq L(P,g,\alpha)$.
+
+**For (c)**, if $f\in \mathscr{R}(\alpha)$ on $[a,b]$, and if $a<c<b$, then $f\in \mathscr{R}$ on $[a,c]$ and $[c,b]$, and 
+
+$$
+\int_a^c f d\alpha + \int_c^b f d\alpha=\int_a^b f d\alpha
+$$
+
+For every partition $P=\{x_0,x_1,\cdots,x_n\}$ of $[a,b]$, we have a refinement $P^*=P\cup\{c\}$ of $[a,b]$. Let $P_1=\{x_0,x_1,\cdots,x_j,c\}$ and $P_2=\{c,x_{j+1},\cdots,x_n\}$ be the partitions of $[a,c]$ and $[c,b]$ respectively. So 
+$$
+\begin{aligned}
+U(P^*,f,\alpha)&=\sum_{i=0}^{n}M_i(x_i-x_{i+1})\\
+&=M_c(c-x_j)+\sum_{i=0}^{j-1}M_i(x_i-x_{i+1})+\sum_{i=j+1}^{n}M_i(x_i-x_{i+1})\\
+&=U(P_1,f,\alpha)+U(P_2,f,\alpha) 
+\end{aligned}
+$$
+and
+$$
+\begin{aligned}
+L(P^*,f,\alpha)&=\sum_{i=0}^{n}m_i(x_i-x_{i+1})\\
+&=m_c(x_j-c)+\sum_{i=0}^{j-1}m_i(x_i-x_{i+1})+\sum_{i=j+1}^{n-1}m_i(x_i-x_{i+1})\\
+&=L(P_1,f,\alpha)+L(P_2,f,\alpha)
+\end{aligned}
+$$
+
+Since $P^*$ is a refinement of $P$, by \textbf{Theorem 6.4}, we have $U(P^*,f,\alpha)\leq U(P,f,\alpha)$ and $L(P^*,f,\alpha)\geq L(P,f,\alpha)$.
+
+So $\int_a^c f d\alpha+\int_c^b f d\alpha\leq U(P^*,f,\alpha)\leq U(P,f,\alpha)=\int_a^b f d\alpha$.
+
+Similarly, we have $\int_a^c f d\alpha+\int_c^b f d\alpha\geq L(P^*,f,\alpha)\geq L(P,f,\alpha)=\int_a^b f d\alpha$.
+
+Therefore, $\int_a^c f d\alpha+\int_c^b f d\alpha=\int_a^b f d\alpha$.
+
+**For (d)**, if $f\in \mathscr{R}(\alpha)$ on $[a,b]$, and if $|f(x)| \leq M$ on $[a,b]$, then 
+$$
+\left|\int_a^b f d\alpha\right| \leq M(\alpha(b)-\alpha(a))
+$$
+
+Since $|f(x)|\leq M$ on $[a,b]$, $\forall x\in [a,b]$, we have $f(x)\in [-M,M]$ on $[a,b]$. So $\sup|f(x)|\leq M$ and $\inf|f(x)|\leq M$. Since $L(P,f,\alpha)\leq \int_a^b f d\alpha\leq U(P,f,\alpha)$, we have
+
+So
+
+$$
+\begin{aligned}
+    \left|\int_a^b f d\alpha\right|&\leq \max\left\{|L(P,f,\alpha)|,|U(P,f,\alpha)|\right\}\\
+    &=\max\left\{\sum_{i=0}^{n-1}|M_i|\Delta x_i,\sum_{i=0}^{n-1}|m_i|\Delta x_i\right\}\\
+    &\leq \sum_{i=0}^{n-1}\max\{|M_i|,|m_i|\}\Delta x_i\\
+    &\leq \sum_{i=0}^{n-1}M\Delta x_i\\
+    &=M(\alpha(b)-\alpha(a))
+\end{aligned}
+$$
+
+Therefore, $\left|\int_a^b f d\alpha\right| \leq M(\alpha(b)-\alpha(a))$.
+
+**For (e)**, notice that
+
+$$
+\begin{aligned}
+\Delta (\alpha+\beta)_i &= \alpha(x_i)-\alpha(x_{i-1})+\beta(x_i)-\beta(x_{i-1}) \\
+&= \Delta \alpha_i + \Delta \beta_i
+\end{aligned}
+$$
+
 QED