11.1 Introduction

Remark 11.1.1 (General Plan).label

1. (1)

   General Measure Theory

   1. (a)

      Caratheodory Construction

   
   
2. (2)

   Integration

   1. (a)

      Measurable Sets and Functions

   2. (b)

      Fatou Lemma

   3. (c)

      Monotone/Dominated Convergence

   
   
3. (3)

   Radon-Nikodym Derivative

Let $E\suf \R^{n}$ be a bounded domain and suppose $\pd\cl{E}$ is piecewise smooth, then $E$ is regular and the volume of $E$ is given by

interpreted as a Riemann Integral. The extension of $\text{vol}(E)=\abs{E}$ to Lebesgue measurable sets is denoted by $m(E)$ called the Lebesgue measure of $E$ with several properties:

1. (1)

   If $E$ is regular then $\abs{E}=m(E)$

2. (2)

   $E\supf E'\implies m(E)\geq m(E')$

3. (3)

   Countable Additivity

The set of measurable functions is closed under taking limits and for $E\suf\R^{n}$ regular, $f\in C^{0}(\cl{E})$ then $\integral{E}{}fdm=\integral{E}{}f(x_{1},...,x_{n})dx_{1}\cds dx_{n}$ Let $P=\curl{x_i}_{0}^{n}$ be partition of $[a,b]$ then $a=x_{0}<x_{1}<\cds<x_{n}=b$. Let $f:[a,b]\to\R$ be bounded then the lower and upper Riemann Sums are given by

and for any refinement $P'=P\cup \curl{y_1,...,y_m}\supf P$ we have

By monotonicity $L(f)=\define \SUP{P}L(f,P),U(f)\define\INF{P}U(f,P)$ exists. Given $f:[a,b]\to \R$ bounded we say that $f$ is Riemann Integrable and write $f\in RI([a,b])$ if $L(f)=U(f)$. If $f\in C^{0}*([a,b])$ then $f\in RI([a,b])$. Indeed for any $\e>0$ there exists a partition such that $\abs{U(f,P_\e)-L(f,P_\e)}<\e$ hence by taking limit $\abs{U(f)-L(f)}<\e$ and since $\e>0$ was arbitrary, $U(f)=L(f)$. ^[1]

Remark 11.1.2 (Deficiencies of the Riemann Integral).label ^[2] Consider the function

then $f\nin RI([0,1])$ since $U(f,[0,1])=\infty$. Additionally for $\curl{r_n}_{1}^{\infty}=\Q\cap [0,1]$ consider the family of functions

then $f_{k}\in RI([0,1])$ for all $k\geq 1$ but $f(x)\define\limit{k\to \infty}f_{k}(x)=\begin{cases}1&x\in \Q\cap [0,1]\\ 0&\text{else}\end{cases} \nin RI([0,1])$ ^[3]

Remark 11.1.3 (Set Theory Background).label Let $X\neq \emp$ be a set we denote the powerset by $\cali{P}(X)=2^{X}=\curl{2:2\suf X}$. Given $E,F\suf X$ denote the difference by $E\del F=E-F=\curl{x\in X:x\in E,x\nin F}$, symmetric difference by $E\triangle F=(E\del F)\cup (F\del E)$, $E^{c}=X\del E$ for $E\suf X$. Given $f:X\to Y$ and $E\suf Y$ then $f\inv(E)=\curl{x\in X:f(x)\in E}$, $f(D)=\curl{f(x):x\in D\suf X}$. We have some properties for $f:X\to Y$ with $E_{\al}\suf X$ for $\al\in A$

1. (1)

   $f\inv(\cups{}{\al\in A }E_{\al} )=\cups{}{\al\in A }f\inv(E_{\al})$

2. (2)

   $f\inv(\caps{}{\al\in A }E_{\al} )=\caps{}{\al\in A }f\inv(E_{\al})$

3. (3)

   $f\inv(E^{c})=\parens{f\inv(E)}^{c}$

An equivalence relation is a relation $R\suf X\ti X$ satisfying

1. (1)

   $\fall x\in X,xRx$

2. (2)

   $\fall x,y\in X, xRy\iff yRx$

3. (3)

   $\fall x,y,z\in X$ such that $xRy$ and $yRz$ then $xRz$

A partial ordering on $X\neq\emp$ is a relation $R\suf X\ti X$ ($R=\leq$) such that

1. (1)

   $x\leq y,y\leq z\implies x\leq z$

2. (2)

   $x\leq y,y\leq x\implies x=y$

3. (3)

   $x\leq x$

The pair $(X,\leq)$ is a poset. A maximal (resp. minimal) element $x\in X$ such that $y\geq x$ (resp. $y\leq x$) implies $y=x$. If $\leq$ is a total ordering if $\fall x,y\in X$ either $x\leq y$ or $y\leq x$ and $(X,\leq)$ is totally (linearly) ordered.

Let $(X,\leq)$ be a poset such that every linearly-ordered subset has a maximal element. Zorn’s Lemma states that $X$ has a maximal element which is equivalent to the Axiom of Choice. Let $A$ be an index set and $\curl{X_\al}_{\al\in A}$ is a nonempty collection of sets. Define the product set $X\define\prods{}{\al\in A}X_{\al}=\curl{f:A\to \cups{}{\al\in A}X_\al:\fall \al\in A,f(\al)\in X_\al}$ which comes with natural projections $\pi_{\al}:X\to X_{\al}$ defined by $\pi_{\al}(f)=f(\al)$. When $A$ is countable $\prods{}{\al\in A}X_{\al}$ is naturally isomorphic to the Cartesian product. AC states that under the above setting $X\neq\emp$.

Furthermore for $X\neq\emp,\curl{E_n}_{1}^{\infty}\suf \cali{P}(X)$ then

^[4]

Definition 11.1.4 (Measures).label Let $E\suf\R^{n}$ we want to defined measure of $E$ denoted $\mu(E)$ that agrees with the Euclidean volume when $E$ is regular enough but has some further properties

1. (1)

   $\mu:\cali{P}(X)\to [0,\infty]$

2. (2)

   $\mu(E)\geq 0$

3. (3)

   $E\suf F\implies \mu(E)\leq \mu(F)$

4. (4)

   Given $\curl{E_n }_{1}^{\infty}$ such that $E_{n}\cap E_{m}=\emp$ for $n\neq m$ then $\mu \parens{\dcups{\infty }{n=1}E_n}=\sums{\infty}{n=1}\mu(E_{n})$

Definition 11.1.5 (Algebra).label Let $X$ be a nonempty set and $\cali{A}\suf 2^{X}$ satisfying:

1. (1)

   $\cal{A}\neq\emp$

2. (2)

   For any $E_{1},...,E_{n}\in \cali{A}$, $\cups{n}{i=1}E_{i}\in \cal{A}$

3. (3)

   For each $E\in \cal{A}$, $E^{c}\in\cal{A}$

Example 11.1.6 (Vitali-ish).label Let $I=[0,1]$ and define an equivalence relation $\sim$ on $I$ by $x\sim y\iff x-y\in\Q$ for all $x,y\in I$. Set $E_{x}=[x]/\cups{}{y\neq x}[y]$ then $I=\cups{}{x\in I}E_{x}$. Let $N\suf [0,1]$ be a set containing precisely one element for each equivalence class which exists by AC. Set $R=Q\cap [0,1]$ and for each $r\in R$ we define $N_{r}=\curl{x+r:x\in N\cap [0,1-r)}\cup \curl{x+r-1:x\in N\cap [1-r,1]}$. By translation invariance and disjoint additivity, $\mu(N_{r})=\mu(N)$ for each $r\in R$. Let $x\in[0,1]$, if $y\in N$ with $y\in [x]$ then $x\in N_{r}$ where $r= \begin{cases}r=x-y&x\geq y\\ r=x-y+1&x<y\end{cases}$. Suppose $x\in N_{r}\cap N_{s}$ for $r\neq s\in R$ then $x-r,x-s$ are disjoint elements of $N$ (one can take $x-r+1$ instead of $x-r$ to land in $I$). This shows each $x\in [0,1]$ belongs to a unique $N_{r}$. Since $N=(N\cap [0,1-r))\sqcup (N\cap [1-r,1])$ we have $\mu(N)=\mu(N_{r})$. Furthermore $[0,1]=\dcups{}{r\in R}N_{r}$ hence $1=\mu([0,1])=\sums{}{r\in R}\mu(N_{r})=\sums{}{r\in R}\mu(N)$

Lemma 11.1.7.label Let $X$ be a nonempty set and $\curl{E_n}_{n\in\N}\suf 2^{X}$ then there exists $\curl{F_n}_{n\in\N}$ pairwise disjoint such that $\cups{}{n\in\N}E_{n}=\cups{}{n\in\N}F_{n}$.

Definition 11.1.8.label Let $\curl{A_\al }_{\al\in J}$ be a family of $\s$-algebras then $\caps{}{\al\in J}A_{\al}$ is a $\s$-algebra so given a family $\cali{E}$ of subsets of $X$ there exists a smallest $\s$-algebra $\cali{M}(\cali{E})$ containing $\cali{E}$ called the generating $\s$-algebra of $\cali{E}$

Lemma 11.1.9.label Let $\cali{E}\suf\cali{M}(\cali{F})$ where $\cali{E},\cali{F}$ are families of subsets of $X$ then $\cali{M}(\cali{E})\suf \cali{M}(\cali{F})$