update

docker-compose.yaml

@@ -3,7 +3,7 @@ services:
     build:
       context: ./
       dockerfile: ./Dockerfile
-    image: trance0/notenextra:v1.1.11
+    image: trance0/notenextra:v1.1.12
     restart: on-failure:5
     ports:
       - 13000:3000

pages/CSE559A/_meta.js

@@ -22,4 +22,5 @@ export default {
     CSE559A_L17: "Computer Vision (Lecture 17)",
     CSE559A_L18: "Computer Vision (Lecture 18)",
     CSE559A_L19: "Computer Vision (Lecture 19)",
+    CSE559A_L20: "Computer Vision (Lecture 20)",
 }

pages/Math4121/Math4121_L30.md

@@ -0,0 +1,94 @@
+# Math4121 Lecture 30
+
+## Lebesgue Measure
+
+$\mathfrak{M}=\{S\subseteq\mathbb{R}:S\text{ is Lebesgue measurable}\}$ is a $\sigma$-algebra on $\mathbb{R}$ (closed under complementation and countable unions).
+
+### Consequence of Lebesgue Measure
+
+Every open set and closed set is Lebesgue measurable.
+
+#### Inner and Outer Regularity of Lebesgue Measure
+
+Outer regularity:
+$$
+m_e(S)=\inf_{U\text{ open},S\subseteq U}m(U)
+$$
+
+Inner regularity:
+$$
+m_i(S)=\sup_{K\text{ closed},K\subseteq S}m(K)
+$$
+
+Proof:
+
+Inner regularity:
+
+Since $m_i(S)=m(I)-m_e(I\setminus S)$, $S\subseteq I$ for some closed interval $I$. Let $\epsilon>0$ and $U$ be an open set such that $I\setminus S\subseteq U$ and $m(U)<m(I\setminus S)+\epsilon$.
+
+Take $K=I\setminus U$. Then $K\subseteq S$ and $K$ is closed and
+
+$$
+m(K)=m(I)-m(U)>m(I)-m(I\setminus S)-\epsilon
+$$
+
+So $m_i(S)<m(K)+\epsilon$. Since $\epsilon$ is arbitrary, $m_i(S)\leq m_e(S)$.
+
+QED
+
+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.
+
+#### Symmetric difference
+
+The symmetric difference of two sets $S$ and $T$ is defined as
+
+$$
+S\Delta T=(S\setminus T)\cup(T\setminus S)
+$$
+
+_The XOR operation on two sets._
+
+#### Theorem 
+
+If $S\subset I$ is measurable, then for every $\epsilon>0$, $\exists I_1,I_2,\cdots,I_n\subset I$ open intervals such that
+
+$$
+m(S\Delta U)<\epsilon
+$$
+
+where $U=\bigcup_{j =1}^n I_j$.
+
+Proof:
+
+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.
+
+Check:
+
+$$
+m(S\Delta U)=m(S\setminus U)+m(U\setminus S)\leq m(S\setminus K)+m(U\setminus S)<\epsilon
+$$
+
+QED
+
+Recall $\{T_j\}_{j=1}^\infty$ are disjoint measurable sets. Then $T=\bigcup_{j=1}^\infty T_j$ is measurable and
+
+$$
+m(T)=\sum_{j=1}^\infty m(T_j)
+$$
+
+#### Corollary (Better osgood's theorem on Lebesgue measure)
+
+If $S_1\subseteq S_2\subseteq S_3\subseteq\cdots$  are measruable (no need to be closed and bounded) and $S=\bigcup_{j=1}^\infty S_j$, then
+
+$$
+m(S)=\lim_{j\to\infty}m(S_j)
+$$
+
+Proof:
+
+Let $T_1=S_1$ and $T_i=S_i\setminus S_{i-1}$ for $i\geq 2$. Still have $S=\bigcup_{j=1}^\infty T_j$.
+
+Where $T_i$ are disjoint measurable sets. So $m(S)=\sum_{j=1}^\infty m(T_j)$.
+
+So $\lim_{j\to\infty}m(S_j)=\sum_{j=1}^\infty m(T_j)=m(S)$.
+

pages/Math4121/_meta.js

@@ -32,5 +32,9 @@ export default {
     Math4121_L26: "Introduction to Lebesgue Integration (Lecture 26)",
     Math4121_L27: "Introduction to Lebesgue Integration (Lecture 27)",
     Math4121_L28: "Introduction to Lebesgue Integration (Lecture 28)",
-    Math4121_L29: "Introduction to Lebesgue Integration (Lecture 29)"
+    Math4121_L29: "Introduction to Lebesgue Integration (Lecture 29)",
+    Math4121_L30: "Introduction to Lebesgue Integration (Lecture 30)",
+    Math4121_L31: "Introduction to Lebesgue Integration (Lecture 31)",
+    Math4121_L32: "Introduction to Lebesgue Integration (Lecture 32)",
+    Math4121_L33: "Introduction to Lebesgue Integration (Lecture 33)"
 }