update

pages/Math4121/Math4121_L19.md

@@ -1 +1,44 @@
-# Lecture 19
+# Math 4121 Lecture 19
+
+## Continue on the "small set"
+
+### Cantor set
+
+#### Theorem: Cantor set is perfect, nowhere dense
+
+Proved last lecture.
+
+_Other construction of the set by removing the middle non-zero interval $(\frac{1}{n},n>0)$ and take the intersection of all such steps is called $SVC(n)$_
+
+Back to $\frac{1}{3}$ Cantor set.
+
+Every step we delete $\frac{2^{n-1}}{3^n}$ of the total "content".
+
+Thus, the total length removed after infinitely many steps is:
+
+$$
+\sum_{n=1}^{\infty} \frac{2^{n-1}}{3^n} = \frac{1}{3}\sum_{n=0}^{\infty} \left(\frac{2}{3}\right)^n=1
+$$
+
+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._
+
+#### Monotonicity of outer content
+
+If $S\subseteq T$, then $c_e(S)\leq c_e(T)$.
+
+Proof: 
+
+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
+
+#### Theorem Osgorod's Lemma
+
+If $S$ is closed and bounded, then
+
+$$
+\lim_{k\to \infty} c_e(S_k)=c_e(S)
+$$
+

pages/Math4121/_meta.js

@@ -21,9 +21,7 @@ export default {
     Math4121_L16: "Introduction to Lebesgue Integration (Lecture 16)",
     Math4121_L17: "Introduction to Lebesgue Integration (Lecture 17)",
     Math4121_L18: "Introduction to Lebesgue Integration (Lecture 18)",
-    Math4121_L19: {
-        display: 'hidden'
-    },
+    Math4121_L19: "Introduction to Lebesgue Integration (Lecture 19)",
     Math4121_L20: {
         display: 'hidden'
     },