From 6805462262967083755f4cae846cbca4b2f5099e Mon Sep 17 00:00:00 2001
From: Zheyuan Wu <60459821+Trance-0@users.noreply.github.com>
Date: Mon, 24 Mar 2025 10:54:32 -0500
Subject: [PATCH] update

---
 docker-compose.yaml            |  2 +-
 pages/Math4121/Math4121_L25.md |  4 +-
 pages/Math4121/Math4121_L26.md | 94 +++++++++++++++++++++++++++++++++-
 3 files changed, 95 insertions(+), 5 deletions(-)

diff --git a/docker-compose.yaml b/docker-compose.yaml
index 4d9633c..c517a7c 100644
--- a/docker-compose.yaml
+++ b/docker-compose.yaml
@@ -3,7 +3,7 @@ services:
     build:
       context: ./
       dockerfile: ./Dockerfile
-    image: trance0/notenextra:v1.1.8
+    image: trance0/notenextra:v1.1.9
     restart: on-failure:5
     ports:
       - 13000:3000
diff --git a/pages/Math4121/Math4121_L25.md b/pages/Math4121/Math4121_L25.md
index 9686cd8..28a36a0 100644
--- a/pages/Math4121/Math4121_L25.md
+++ b/pages/Math4121/Math4121_L25.md
@@ -2,7 +2,7 @@
 
 ## Continue on Measure Theory
 
-### Borel Mesure
+### Borel Measure
 
 Finite additivity of Jordan content, i.e. for any $\{S_j\}_{j=1}^N$ pairwise disjoint sets and Jordan measurable, then
 
@@ -73,5 +73,3 @@ SVC(3) is Jordan measurable, but $|SVC(3)|=\mathfrak{c}$. so $|\mathscr{P}(SVC(3
 But for any $S\subset \mathscr{P}(SVC(3))$, $c_e(S)\leq c_e(SVC(3))=0$ so $S$ is Jordan measurable.
 
 However, there are $\mathfrak{c}$ many intervals and $\mathcal{B}$ is generated by countable operations from intervals.
-
-
diff --git a/pages/Math4121/Math4121_L26.md b/pages/Math4121/Math4121_L26.md
index 9ba5345..7753ac1 100644
--- a/pages/Math4121/Math4121_L26.md
+++ b/pages/Math4121/Math4121_L26.md
@@ -1 +1,93 @@
-# Lecture 26
\ No newline at end of file
+# Math4121 Lecture 26
+
+## Lebesgue Measure
+
+### Lebesgue's Integration
+
+Partition on the y-axis, let $l$ be the minimum of $f(x)$ on the $y$-axis, $L$ be the maximum of $f(x)$ on the $y$-axis.
+
+$l=l_0<l_1<\cdots<l_n=L$
+
+Define $S_i=\{x\in[a,b]:l_{i-1}\leq f(x)<l_i\}$
+
+Defined the characteristic function of set $S$ is $\chi_S(x)=1$ if $x\in S$ and $0$ otherwise.
+
+Then $f$ lies between the following simple functions:
+
+$$
+\sum_{i=1}^n l_{i-1}\chi_{S_i}\leq f(x)\leq \sum_{i=1}^n l_i\chi_{S_i}
+$$
+
+_This representation allows us to measure some weird sets on the $x$-axis by constraining the $y$-axis._
+
+This is still kind of Riemann sum, but $S_i$ can be very weird (not just intervals).
+
+If we can "measure" each $S_i$, then we could define the integral of $f$ by
+
+$$
+\sup_{l_0,\cdots,l_n}\sum_{i=1}^n l_{i-1}m(S_i)\quad \inf_{l_0,\cdots,l_n}\sum_{i=1}^n l_i m(S_i)
+$$
+
+If we used Jordan content, for $m$ here, this is just a different perspective of Riemann integral.
+
+If we use Borel measure maybe things would be different (perhaps, better)?
+
+As we discussed last time, this limits the measurable sets significantly.
+
+Nonetheless, let's try
+
+1. Characteristic function of the rational numbers
+
+$$
+f(x)=\begin{cases}
+1 & x\in\mathbb{Q}\cap[0,1]\\
+0 & \text{otherwise}
+\end{cases}
+$$
+
+Take partition $0=x_0<x_1<\cdots<x_n=1$
+
+$S_1=\{x:0\leq f(x)<\epsilon\}=[0,1]\setminus\mathbb{Q}$
+
+$S_2=\{x:\epsilon\leq f(x)<1\}=\emptyset$
+
+$S_3=\{x:1\leq f(x)<1+\epsilon\}=\mathbb{Q}\cap[0,1]$
+
+$$
+\sum_{i=1}^n l_{i-1}m(S_i)=0\cdot 1+\epsilon\cdot 0+1\cdot 0=0
+$$
+
+$$
+\sum_{i=1}^n l_i m(S_i)=\epsilon\cdot 1+1\cdot 0+(1+\epsilon)\cdot 0=\epsilon
+$$
+
+So, $0\leq\int_0^1 f(x)dm\leq\epsilon$, here $m$ means the measure we used.
+
+As $\epsilon$ is arbitrary, we have $\int_0^1 f(x)dm=0$.
+
+_This shows that $\int \chi_S(x)dm=m(S)$ for any measurable set $S$._
+
+### Lebesgue's Measure
+
+#### Definition of Lebesgue measure
+
+Outer Measure:
+
+Given $S\subset [a,b]$, let $\mathcal{C}$ be the collection of all countable covers of $S$ by open intervals.
+
+$$
+m_e(S)=\inf_{C\in\mathcal{C}}m(C)
+$$
+
+where $m(C)$ is the Borel measure.
+
+Recall such $C$ are Boreal measurable because open interval are in $\mathcal{B}$ and $\mathcal{B}$ (being a sigma algebra) is closed under countable unions.
+
+Properties:
+
+1. Translation invariant: $m_e(S+a)=m_e(S)$
+2. Countable additivity: $m_e(S\cup T)=m_e(S)+m_e(T)$ if $S\cap T=\emptyset$
+3. $m([0,1])=1$
+
+**Notice we don't have the difference property here.**
+