update
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@@ -3,7 +3,7 @@ services:
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build:
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context: ./
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dockerfile: ./Dockerfile
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image: trance0/notenextra:v1.1.8
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image: trance0/notenextra:v1.1.9
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restart: on-failure:5
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ports:
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- 13000:3000
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@@ -2,7 +2,7 @@
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## Continue on Measure Theory
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### Borel Mesure
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### Borel Measure
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Finite additivity of Jordan content, i.e. for any $\{S_j\}_{j=1}^N$ pairwise disjoint sets and Jordan measurable, then
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@@ -73,5 +73,3 @@ SVC(3) is Jordan measurable, but $|SVC(3)|=\mathfrak{c}$. so $|\mathscr{P}(SVC(3
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But for any $S\subset \mathscr{P}(SVC(3))$, $c_e(S)\leq c_e(SVC(3))=0$ so $S$ is Jordan measurable.
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However, there are $\mathfrak{c}$ many intervals and $\mathcal{B}$ is generated by countable operations from intervals.
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@@ -1 +1,93 @@
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# Lecture 26
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# Math4121 Lecture 26
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## Lebesgue Measure
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### Lebesgue's Integration
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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.
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$l=l_0<l_1<\cdots<l_n=L$
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Define $S_i=\{x\in[a,b]:l_{i-1}\leq f(x)<l_i\}$
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Defined the characteristic function of set $S$ is $\chi_S(x)=1$ if $x\in S$ and $0$ otherwise.
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Then $f$ lies between the following simple functions:
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$$
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\sum_{i=1}^n l_{i-1}\chi_{S_i}\leq f(x)\leq \sum_{i=1}^n l_i\chi_{S_i}
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$$
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_This representation allows us to measure some weird sets on the $x$-axis by constraining the $y$-axis._
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This is still kind of Riemann sum, but $S_i$ can be very weird (not just intervals).
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If we can "measure" each $S_i$, then we could define the integral of $f$ by
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$$
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\sup_{l_0,\cdots,l_n}\sum_{i=1}^n l_{i-1}m(S_i)\quad \inf_{l_0,\cdots,l_n}\sum_{i=1}^n l_i m(S_i)
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$$
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If we used Jordan content, for $m$ here, this is just a different perspective of Riemann integral.
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If we use Borel measure maybe things would be different (perhaps, better)?
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As we discussed last time, this limits the measurable sets significantly.
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Nonetheless, let's try
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1. Characteristic function of the rational numbers
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$$
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f(x)=\begin{cases}
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1 & x\in\mathbb{Q}\cap[0,1]\\
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0 & \text{otherwise}
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\end{cases}
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$$
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Take partition $0=x_0<x_1<\cdots<x_n=1$
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$S_1=\{x:0\leq f(x)<\epsilon\}=[0,1]\setminus\mathbb{Q}$
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$S_2=\{x:\epsilon\leq f(x)<1\}=\emptyset$
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$S_3=\{x:1\leq f(x)<1+\epsilon\}=\mathbb{Q}\cap[0,1]$
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$$
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\sum_{i=1}^n l_{i-1}m(S_i)=0\cdot 1+\epsilon\cdot 0+1\cdot 0=0
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$$
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$$
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\sum_{i=1}^n l_i m(S_i)=\epsilon\cdot 1+1\cdot 0+(1+\epsilon)\cdot 0=\epsilon
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$$
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So, $0\leq\int_0^1 f(x)dm\leq\epsilon$, here $m$ means the measure we used.
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As $\epsilon$ is arbitrary, we have $\int_0^1 f(x)dm=0$.
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_This shows that $\int \chi_S(x)dm=m(S)$ for any measurable set $S$._
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### Lebesgue's Measure
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#### Definition of Lebesgue measure
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Outer Measure:
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Given $S\subset [a,b]$, let $\mathcal{C}$ be the collection of all countable covers of $S$ by open intervals.
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$$
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m_e(S)=\inf_{C\in\mathcal{C}}m(C)
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$$
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where $m(C)$ is the Borel measure.
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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.
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Properties:
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1. Translation invariant: $m_e(S+a)=m_e(S)$
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2. Countable additivity: $m_e(S\cup T)=m_e(S)+m_e(T)$ if $S\cap T=\emptyset$
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3. $m([0,1])=1$
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**Notice we don't have the difference property here.**
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