format updates

content/Math4121/Math4121_L17.md

@@ -24,7 +24,8 @@ Given a set $S$, the power set of $S$, denoted $\mathscr{P}(S)$ or $2^S$, is the
 
 Cardinality of $2^S$ is not equal to the cardinality of $S$.
 
-Proof:
+<details>
+<summary>Proof of Cantor's Theorem</summary>
 
 Assume they have the same cardinality, then $\exists \psi: S \to 2^X$ which is one-to-one and onto. (this function returns a subset of $S$)
 
@@ -38,7 +39,7 @@ If $b\in T$, then by definition of $T$, $b \notin \psi(b)$, but $\psi(b) = T$, w
 
 If $b \notin T$, then $b \in \psi(b)$, which is also a contradiction since $b\in T$. Therefore, $2^S$ cannot have the same cardinality as $S$.
 
-QED
+</details>
 
 ### Back to Hankel's Conjecture
 

content/Math4121/Math4121_L18.md

@@ -12,13 +12,16 @@ By modifying this example, we can find similar with any outer content between 0
 
 $S\subseteq[0,1]$ is perfect if $S=S'$.
 
-Example:
+<details>
+<summary>Examples of perfect set</summary>
 
 - $[0,1]$ is perfect
   - perfect sets are closed
 - Finite collection of points is not perfect because they do not have limit points.
   - perfect sets are uncountable (no countable sets can be perfect)
 
+</details>
+
 #### Middle third Cantor set
 
 We construct the set by removing the middle third of the interval.
@@ -49,7 +52,8 @@ $$
 
 $C$ is perfect and nowhere dense, and outer content is 0.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 (i) $c_e(C)=0$
 
@@ -70,3 +74,4 @@ It is sufficient to show $C$ contains no intervals.
 Any open intervals has a real number with 1 in it's base 3 decimal expansion (proof in homework)
 
 _take some interval in $(a,b)$ we can change the digits that is small enough and keep the element still in the set_
+</details>

content/Math4121/Math4121_L19.md

@@ -44,11 +44,12 @@ The outer content of $SVC(n)$ is $\frac{n-3}{n-2}$.
 
 If $S\subseteq T$, then $c_e(S)\leq c_e(T)$.
 
-Proof: 
+<details>
+<summary>Proof of Monotonicity of outer content</summary>
 
 If $C$ is cover of $T$, then $S\subseteq T\subseteq C$, so $C$ is a cover of $S$. Since $c_e(s)$ takes the inf over a larger set that $c_e(T)$, $c_e(S) \leq c_e(T)$.
 
-QED
+</details>
 
 #### Theorem Osgood's Lemma
 

content/Math4121/Math4121_L20.md

@@ -30,7 +30,8 @@ If $S=\bigcup_{n=1}^{\infty} I_n$, $T=\bigcup_{n=1}^{\infty} J_n$, where $I_n$ a
 
 Let $S$ be a closed, bounded set in $\mathbb{R}$, and $S_1\subseteq S_2\subseteq \ldots$, and $S=\bigcup_{n=1}^{\infty} S_n$. Then $\lim_{k\to\infty} c_e(S_k)=c_e(S)$.
 
-Proof:
+<details>
+<summary>Proof of Osgood's Lemma</summary>
 
 Trivial that $c_e(S_k)\leq c_e(S)$.
 
@@ -70,7 +71,7 @@ c_e(S)&\leq c_e(U)\\
 \end{aligned}
 $$
 
-QED
+</details>
 
 ### Convergence Theorems for sequences of functions
 
@@ -96,7 +97,8 @@ $$
 \lim_{n\to\infty}\int_a^b f_n(x)\ dx=\int_a^b f(x)\ dx
 $$
 
-Proof:
+<details>
+<summary>Proof of Arzela-Osgood Theorem (incomplete)</summary>
 
 Define $\Gamma_{\alpha}=\{x:\forall m\in \mathbb{N} \textup{ and }\forall \delta>0, \exists n\geq m \textup{ s.t. } |y-x|<\delta \textup{ and } |f_n(y)-f_m(y)|>\alpha\}$.
 
@@ -105,3 +107,4 @@ _$\Gamma_{\alpha}$ is the negation of $(\alpha,\delta)$ definition of limit._
 $\Gamma_{\alpha}$ is closed and nowhere dense.
 
 Continue on next lecture.
+</details>

content/Math4121/Math4121_L21.md

@@ -20,7 +20,8 @@ $$
 
 Fact: $\Gamma_{\alpha}$ is closed and nowhere dense.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Without loss of generality, we can assume $f=0$. Given any $\alpha > 0$, $\exists N$ such that
 
@@ -70,5 +71,5 @@ This implies $\ell(P_2)\leq \frac{\alpha}{4B}$.
 
 Continue on Friday.
 
-QED
+</details>
 

content/Math4121/Math4121_L22.md

@@ -2,7 +2,8 @@
 
 ## Continue on Arzela-Osgood Theorem
 
-Proof:
+<details>
+<summary>Proof continuation of Arzela-Osgood Theorem</summary>
 
 Part 2: Control the integral on $\mathcal{U}$
 
@@ -40,7 +41,7 @@ $$
 
 $\forall N\geq K$.
 
-QED
+</details>
 
 ### Baire Category Theorem
 
@@ -50,7 +51,8 @@ Nowhere dense sets can be large, but they canot cover an open (or closed) interv
 
 An open interval cannot be covered by a countable union of nowhere dense sets.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Suppose $(0,1)\subset \bigcup_{n=1}^\infty S_n$ where each $S_n$ is nowhere dense. In particular, $\exists I_1$ closed interval such that $I_1\subset (0,1)$ and $I_1\cap S_1=\emptyset$.
 
@@ -62,7 +64,7 @@ Then $x\in (0,1)$ and $x\notin \bigcup_{n=1}^\infty S_n$.
 
 Contradiction with the assumption that $(0,1)\subset \bigcup_{n=1}^\infty S_n$.
 
-QED
+</details>
 
 #### Definition First Category
 
@@ -72,13 +74,14 @@ A countable union of nowhere dense sets is called a set of **first category**.
 
 Complement of a set of first category in $\mathbb{R}$ is dense in $\mathbb{R}$.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 We need to show that for every interval $I$, $\exists x\in I\cap S^c$. ($\exists x\in I$ and $x\notin S$)
 
 This is equivalent to the Baire Category Theorem.
 
-QED
+</details>
 
 Recall a function is pointwise discontinuous if $\mathcal{C}=\{c\in [a,b]: f\text{ is continuous at } c\}$ is dense in $[a,b]$.
 
@@ -88,7 +91,8 @@ $\mathcal{D}=[a,b]\setminus \mathcal{C}$ is called the set of points of disconti
 
 $f$ is pointwise discontinuous if and only if $\mathcal{D}$ is of first category.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Part 1: If $\mathcal{D}$ is of first category, then $f$ is pointwise discontinuous.
 
@@ -104,13 +108,14 @@ Let $I\subseteq [a,b]$ so $\exists c\in \mathcal{C}\cap I$. So by definition of
 
 Thus, $P_k$ is nowhere dense.
 
-QED
+</details>
 
 #### Corollary 4.10
 
 Let $\{f_n\}$ be a sequence of pointwise discontinuous functions. The set of points at which all $f_n$ are simultaneously continuous is dense (it's also uncountable).
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 $$
 \bigcap_{n=1}^\infty \mathcal{C}_n=\left(\bigcup_{n=1}^\infty \mathcal{D}_n\right)^c
@@ -118,4 +123,4 @@ $$
 
 The complement of a set of first category is dense.
 
-QED
+</details>

content/Math4121/Math4121_L23.md

@@ -110,7 +110,8 @@ $$
 
 So $S$ is Jordan measurable if and only if $c_e(\partial S)=0$.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Let $\epsilon > 0$, and $\{R_j\}_{j=1}^N$ be an open cover of $\partial S$. such that $\sum_{j=1}^N \text{vol}(R_j) < c_e(\partial S)+\frac{\epsilon}{2}$.
 
@@ -136,4 +137,4 @@ If $\eta$ is small enough (depends on $\delta$), then $\mathcal{C}_\eta=\{Q\in K
 
 Suppose $\exists x\in S$ but not in $\mathcal{C}_\eta$. Then $x$ is closed to $\partial S$ so in some $Q_j$. (This proof is not rigorous, but you get the idea. Also not clear in book actually.)
 
-EOP
+</details>

content/Math4121/Math4121_L24.md

@@ -14,7 +14,8 @@ $$
 
 where $\partial S$ is the boundary of $S$ and $c_e(\partial S)=0$.
 
-Example:
+<details>
+<summary>Examples for Jordan measurable</summary>
 
 1. $S=\mathbb{Q}\cap [0,1]$ is not Jordan measurable.
 
@@ -56,6 +57,8 @@ So $c_e(SVC(4))=\frac{1}{2}$.
 
 > General formula for $c_e(SVC(n))=\frac{n-3}{n-2}$, and since $SVC(n)$ is nowhere dense, $c_i(SVC(n))=0$.
 
+</details>
+
 ### Additivity of Content
 
 Recall that outer content is sub-additive. Let $S,T\subseteq \mathbb{R}^n$ be disjoint.
@@ -80,7 +83,8 @@ $$
 c(\bigcup_{i=1}^N S_i)=\sum_{i=1}^N c(S_i)
 $$
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 $$
 \begin{aligned}
@@ -96,7 +100,7 @@ $$
 c(\bigcup_{i=1}^N S_i)=\sum_{i=1}^N c(S_i)
 $$
 
-QED
+</details>
 
 ##### Failure for countable additivity for Jordan content
 

content/Math4121/Math4121_L25.md

@@ -48,7 +48,8 @@ The Borel sets are Borel measurable.
 
 (proof in the following lectures)
 
-Examples:
+<details>
+<summary>Examples for Borel measurable</summary>
 
 1. Let $S=\{x\in [0,1]: x\in \mathbb{Q}\}$
 
@@ -62,6 +63,8 @@ Since $c_e(SVC(4))=\frac{1}{2}$ and $c_i(SVC(4))=0$, it is not Jordan measurable
 
 $S$ is Borel measurable with $m(S)=\frac{1}{2}$. (use setminus and union to show)
 
+</details>
+
 #### Proposition 5.3
 
 Let $\mathcal{B}$ be the Borel sets in $\mathbb{R}$. Then the cardinality of $\mathcal{B}$ is $2^{\aleph_0}=\mathfrak{c}$. But the cardinality of the set of Jordan measurable sets is $2^{\mathfrak{c}}$.

content/Math4121/Math4121_L27.md

@@ -14,7 +14,7 @@ where $I_j$ is an open interval
 
 1. $m_e(I)=\ell(I)$
 2. Countably sub-additive: $m_e\left(\bigcup_{n=1}^\infty S_n\right)\leq \sum_{n=1}^\infty m_e(S_n)$ (Prove today)
-3. does not repect complementation (Build in to Borel measure)
+3. does not respect complementation (Build in to Borel measure)
 
 Why does Jordan content respect complementation?
 
@@ -64,7 +64,8 @@ $$
 m_e\left(\bigcup_{n=1}^\infty S_n\right)\leq\sum_{n=1}^\infty m(S_n)
 $$
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Let $\epsilon>0$ and for each $j$, let $\{I_{i,j}\}_{i=1}^\infty$ be a cover of $S_j$ s.t.
 
@@ -84,21 +85,22 @@ $$
 m_e\left(\bigcup_{j=1}^\infty S_j\right)\leq\sum_{j=1}^\infty m_e(S_j)=\sum_{j=1}^\infty m(S_j)
 $$
 
-QED
+</details>
 
-#### Corollary
+#### Corollary: inner measure is always less than or equal to outer measure
 
 $$
 m_i(S)\leq m_e(S)
 $$
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 $$
 m_i(S)=m(I)-m_e(I\setminus S)\leq m(I)-m_i(I\setminus S)=m_e(S)
 $$
 
-QED
+</details>
 
 ### Caratheodory's Criterion
 
@@ -110,7 +112,8 @@ $$
 m_e\left(S\cap \left(\bigcup_{j=1}^\infty I_j\right)\right)=m_e\left(\bigcup_{j=1}^\infty (S\cap I_j)\right)=\sum_{j=1}^\infty m_e(S\cap I_j)
 $$
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 For each $j$, let $\{J_i\}_{i=1}^\infty$ be a cover of $S\cap \left(\bigcup_{j=1}^\infty I_j\right)$ such that $\sum_{i=1}^\infty \ell(J_i)<c_e(S\cap \left(\bigcup_{j=1}^\infty I_j\right))+\epsilon$. Since $\{I_j\}_{j=1}^\infty$ are pairwise disjoint, so is $\{J_i\cap I_j\}_{j=1}^\infty$ for each $i$.
 
@@ -132,7 +135,7 @@ $$
 m_e\left(S\cap \left(\bigcup_{j=1}^\infty I_j\right)\right)\leq \sum_{j=1}^\infty m_e(S\cap I_j)
 $$
 
-QED
+</details>
 
 #### Theorem 5.6 (Caratheodory's Criterion)
 

content/Math4121/Math4121_L28.md

@@ -31,7 +31,8 @@ $$
 > $$m_e\left(S\cap \bigcup_{j=1}^{\infty} I_j\right) = \sum_{j=1}^{\infty} m_e(S\cap I_j)$$
 > Proved on Friday
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 $\implies$ If Lebesgue criterion holds for $S$, then for any $X$ of finite outer measure,
 
@@ -72,7 +73,7 @@ m_e(X)&\leq m_e(X\cap S)+m_e(S^c\cap X)\\
 \end{aligned}
 $$
 
-QED
+</details>
 
 ### Revisit Borel's criterion
 
@@ -88,7 +89,8 @@ $$
 m_e(S)=\sum_{j=1}^{\infty} m_e(S_j)
 $$
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 First we prove $m_e(\bigcup_{j=1}^{\infty} S_j)=\sum_{j=1}^{\infty} m(S_j)$ by induction.
 
@@ -116,16 +118,17 @@ Therefore, $\sum_{j=1}^{\infty} m(S_j)\leq m_e(S)\leq \sum_{j=1}^{\infty} m(S_j)
 
 So $S$ is measurable.
 
-QED
+</details>
 
 #### Proposition 5.9 (Preview)
 
 Any finite union (and intersection) of measurable sets is measurable.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Let $S_1, S_2$ be measurable sets.
 
 We prove by verifying the Caratheodory's criteria for $S_1\cup S_2$.
 
-QED
+</details>

content/Math4121/Math4121_L29.md

@@ -34,7 +34,8 @@ Towards proving $\mathfrak{M}$ is closed under countable unions:
 
 Any finite union/intersection of Lebesgue measurable sets is Lebesgue measurable.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Suppose $S_1, S_2$ is a measurable, and we need to show that $S_1\cup S_2$ is measurable. Given $X$, need to show that
 
@@ -61,13 +62,14 @@ $$
 
 by measurability of $S_1$ again.
 
-QED
+</details>
 
 #### Theorem 5.10 (Countable union/intersection of Lebesgue measurable sets is Lebesgue measurable)
 
 Any countable union/intersection of Lebesgue measurable sets is Lebesgue measurable.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Let $\{S_j\}_{j=1}^{\infty}\subset\mathfrak{M}$. Definte $T_j=\bigcup_{k=1}^{j}S_k$ such that $T_{j-1}\subset T_j$ for all $j$.
 
@@ -109,7 +111,7 @@ Therefore, $m_e(X\cap S)=m_e(X)$.
 
 Therefore, $S$ is measurable.
 
-QED
+</details>
 
 #### Corollary from the proof
 

content/Math4121/Math4121_L30.md

@@ -20,7 +20,8 @@ $$
 m_i(S)=\sup_{K\text{ closed},K\subseteq S}m(K)
 $$
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Inner regularity:
 
@@ -34,7 +35,7 @@ $$
 
 So $m_i(S)<m(K)+\epsilon$. Since $\epsilon$ is arbitrary, $m_i(S)\leq m_e(S)$.
 
-QED
+</details>
 
 We can approximate $m(S)$ from outside by open sets. If we are just concerned with "approximating" $m(S)$, we can use finite union of intervals.
 
@@ -58,7 +59,8 @@ $$
 
 where $U=\bigcup_{j =1}^n I_j$.
 
-Proof:
+<details>
+<summary>Proof</summary>
 
 Let $\epsilon>0$ and $m(V)<m(S)+\frac{\epsilon}{2}$. Let $K\subseteq S$ be closed set such that $m(S)-\frac{\epsilon}{2}<m(K)$. $V$ is an open cover of closed and bounded set $K$. By Heine-Borel theorem, $K$ has a finite subcover. Let $I_1,I_2,\cdots,I_n$ be the open intervals in the subcover.
 
@@ -68,7 +70,7 @@ $$
 m(S\Delta U)=m(S\setminus U)+m(U\setminus S)\leq m(S\setminus K)+m(U\setminus S)<\epsilon
 $$
 
-QED
+</details>
 
 Recall $\{T_j\}_{j=1}^\infty$ are disjoint measurable sets. Then $T=\bigcup_{j=1}^\infty T_j$ is measurable and