From 74bda1a276de5dc114b0b8c9644cea904f4cf4e5 Mon Sep 17 00:00:00 2001 From: skityl Date: Tue, 31 Mar 2015 21:18:19 +1100 Subject: [PATCH] Added measure theory notes --- Measure Theory/tex/measure.tex | 1324 ++++++++++++++++++++++++++++ Measure Theory/tex/owmaths.cls | 72 ++ Measure Theory/tex/owshortcuts.sty | 39 + 3 files changed, 1435 insertions(+) create mode 100644 Measure Theory/tex/measure.tex create mode 100755 Measure Theory/tex/owmaths.cls create mode 100644 Measure Theory/tex/owshortcuts.sty diff --git a/Measure Theory/tex/measure.tex b/Measure Theory/tex/measure.tex new file mode 100644 index 0000000..2f80936 --- /dev/null +++ b/Measure Theory/tex/measure.tex @@ -0,0 +1,1324 @@ +\documentclass{owmaths} + +\usepackage{owshortcuts} +\usepackage[all]{xy} +\usepackage{csquotes} + +\begin{document} + +\subject{Measure Theory} +\author{Edward McDonald} +\title{Elementary measure theory from an abstract point of view} +\studentno{616} + + +\setlength\parindent{0pt} + +\section{Introduction} +The purpose of these notes is to go over the elementary concepts of +abstract measure theory. There will be few original ideas of proofs. Instead +of being original, the purpose of these notes is to: +\begin{enumerate} + \item{} Cover the basic ideas of measure theory without getting bogged + down in specific examples, like the Carath\'erodory construction, or Lebesgue + measure + \item{} To include all proofs, so the measure theory is presented + as a coherent whole + \item{} To present ideas in as abstract a manner as possible +\end{enumerate} + +\section{Concepts from order theory} +As a reminder, we provide the definition of a partially ordered set. +\begin{definition} + A \emph{partially ordered set} (a.k.a. a poset) is a pair $(P,\leq)$ + where $\leq$ is a relation, such that: + \begin{enumerate} + \item{} $\leq$ is reflexive: for all $a \in P$, $a \leq a$. + \item{} $\leq$ is antisymmetric: if $a,b \in P$ with $a\leq b$ and $b\leq a$, then $a = b$. + \item{} $\leq$ is transitive: if $a,b,c \in P$, with $a \leq b$ and $b\leq c$, then $a\leq c$. + \end{enumerate} +\end{definition} +We shall make extensive use of the concepts of least upper bound +and greatest lower bound. Thus we introduce a specific notation for them. +\begin{definition} + Let $\{a_i\}_{i \in I}$ be an indexed subset of a partially ordered set $(P,\leq)$. Then + an element $b \in P$ is set to be the least upper bound of $\{a_i\}_{i \in I}$ if, + for any $c \in P$ such that $a_i \leq c$ for all $i \in I$, we have $b \leq c$. + We write symbollically, + \begin{equation*} + b = \bigvee_{i \in I} a_i. + \end{equation*} + If $x,y \in P$, we write the least upper bound of $\{x,y\}$ as $x \vee y$. + + Similarly, the greatest lower bound of a set $\{a_i\}_{i \in I}$ is denoted + \begin{equation*} + \bigwedge_{i \in I} a_i. + \end{equation*} + + Theese operations are called \emph{meet} and \emph{join} respectively. +\end{definition} +In a general poset, there is no reason why the meet of +join of an arbitrary subset should exist. A lattice +is a poset where we assume that greatest lower bounds +and least upper bounds of pairs of elements always exist. That is, +\begin{definition} + A \emph{lattice} is a poset $(P,\leq)$ where for any $a,b \in P$, + we have $a\vee b,a\wedge b \in P$. + + A lattice is distributive if for any $a,b,c \in P$, we have + \begin{equation*} + a \vee (b \wedge c) = (a\vee b)\wedge (a\vee c) + \end{equation*} + or + \begin{equation*} + a \wedge (b \vee c) = (a \wedge b) \vee (a \wedge c). + \end{equation*} +\end{definition} +\begin{proposition} + One notion of distributivity implies the other. +\end{proposition} +\begin{proof} + Suppose that for all $a,b,c \in P$, we have + \begin{equation*} + a \vee (b \wedge c) = (a \vee b)\wedge (a \vee c). + \end{equation*} + + Then compute, + \begin{align*} + (a \wedge b) \vee (a \wedge c) &= (a\vee(a\wedge c)) \wedge (b\vee (a\wedge c))\\ + &= a \wedge (b \vee a) \wedge (b\vee c)\\ + &= a \wedge (b \vee c). + \end{align*} + Here we have used the associative laws and absorption law, that $a \wedge (b\vee a) = a$. + These are easy to prove. +\end{proof} + + + +All posets that we deal with in these notes are bounded. So we define a bounded +poset. +\begin{definition} + A poset $(P,\leq)$ is bounded if there exist elements $0,1 \in P$ + such that $0\leq a \leq 1$ for all $a \in P$. +\end{definition} + +So far we have defined posets as abstract sets with relations. However, we shall +mostly be interested in \emph{concrete posets}. That is, given a set +$X$, any subset $A \subseteq \mathcal{P}(X)$ is a poset, +with the partial order being given by inclusion. + +In order to introduce Boolean algebras, we need to introduce complements. +This is a concept that makes sense for bounded posets. +\begin{definition} + Suppose that $(P,\leq,0,1)$ is a bounded poset. Then given $a \in P$, + an element $b \in P$ is called a complement of $b$ if + \begin{align*} + a \wedge b &= 0\\ + a \vee b &= 1 + \end{align*} +\end{definition} +\begin{proposition} + Complements in bounded distributive lattices are unique, if they exist. +\end{proposition} +\begin{proof} + Let $(P,\leq,0,1)$ be a bounded distributive lattice. Suppose that $a \in P$, + and $b,c \in P$ are complements of $a$. Then, + \begin{equation*} + (a \vee b) \wedge c = c + \end{equation*} + Hence, + \begin{equation*} + (a \wedge c) \vee (b \wedge c) = c + \end{equation*} + Thus $b \wedge c = c$, so $b \leq c$. By symmetry, $c \leq b$. So $c = b$. +\end{proof} + +Now we introduce the important concept of a Boolean algebra. +\begin{definition} + A \emph{Boolean algebra} is a bounded distributive lattice + such that every element has a complement. The complement of an element + $a$ is denoted $a'$. +\end{definition} +\begin{remark} + The unary $'$ operation is well defined, since complements are unique + in a distributive lattice. It is easy to see that $'$ is involutive, that is + $a'' = a$ for any $a$. Also, $0' = 1$. +\end{remark} +\begin{proposition} + Suppose that $B$ is a Boolean algebra, and $a,b \in B$. Then $(a\vee b)' = a'\wedge b'$ + and $(a \wedge b)' = a' \vee b'$. +\end{proposition} +\begin{proof} + This is a simple verification. By uniquenss, we need to check that + \begin{equation*} + (a \wedge b) \wedge (a' \vee b') = 1 + \end{equation*} + and that + \begin{equation*} + (a \wedge b) \vee (a' \vee b') = 0. + \end{equation*} + This follows from distributivity. +\end{proof} + +\begin{definition} + Given a set $X$, a Boolean algebra on $X$ is a subset of $\mathcal{P}(X)$ + ordered by inclusion, with meet and join given by intersection and union, + bounds $\emptyset$ and $X$ + and complement given by set complementation. +\end{definition} + + +For measure theory, Boolean algebras are not enough. We must instead +introduce the concept of a Boolean $\sigma$-algebra. +\begin{definition} + A Boolean $\sigma$-algebra is a Boolean algebra that is closed + under countable meets and countable joins. That is, + a Boolean algebra + $A$ is a Boolean $\sigma$-algebra if for any $\{a_i\}_{i=1}^\infty \subseteq A$, we have + \begin{equation*} + \bigwedge_{i=1}^\infty a_i,\;\bigvee_{i=1}^\infty a_i \in A. + \end{equation*} + + Furthermore we ask that countable least upper bounds and greatest lower bounds + satisfy \emph{infinite distributive laws}: for any countable subsets $\{a_i\}_{i=1}^\infty$ + and $\{a_j\}_{j=1}^\infty$, we have + \begin{align*} + \left(\bigvee_{i=1}^\infty a_i\right)\wedge\left(\bigvee_{j=1}^\infty b_j \right) &= \bigvee_{i,j=1}^\infty (a_i \wedge b_j)\\ + \left(\bigwedge_{i=1}^\infty a_i\right)\vee\left(\bigwedge_{j=1}^\infty b_j\right) &= \bigwedge_{i,j=1}^\infty (a_i \vee b_j) + \end{align*} +\end{definition} +\begin{proposition} + In a Boolean $\sigma$-algebra, De Morgan's laws generalise to countable meets and joins. +\end{proposition} +\begin{proof} + Again, by uniqueness of complements, we only need to check that + \begin{equation*} + \left(\bigvee_{i=1}^\infty a_i\right)' = \bigwedge_{i=1}^\infty a_i'. + \end{equation*} + and similarly with joins and meets interchanged. This is a simple verification. +\end{proof} + +\begin{definition} + A $\sigma$-algebra on a set $X$ is a subset of $\mathcal{P}(X)$ + with meet and join given by intersection and union, with bounds $\emptyset$ and $X$ + and complement given by set complementation. +\end{definition} + +$sigma$-algebras on sets are usually specified by a generating set. This is justified +by the following proposition. +\begin{proposition} + Let $X$ be a set, and let $\mathcal{C} \subseteq \mathcal{P}(X)$. There + is a unique smallest $\sigma$-algebra containing $\mathcal{C}$, which we + denote $\sigma(\mathcal{C})$. +\end{proposition} +\begin{proof} + The key here is that $\mathcal{P}(X)$ is a $\sigma$-algebra + containing $\mathcal{C}$, and an intersection of an arbitrary family + of $\sigma$-algebras is again a $\sigma$-algebra. + + Hence, take the intersection of all the $\sigma$-algebras containing $\mathcal{C}$. +\end{proof} + + +\section{Monotone class theorem} +$\sigma$-algebras on sets are the posets which appear most frequently in measure theory. +However, also of great importance are $\pi$-classes and $d$-classes, which +are connected to $\sigma$-algebras by the monotone class theorem. + +\begin{definition} + Suppose that $X$ is a set. A $\pi$-class on $X$ is a non-empty collection of subsets + of $X$ closed under intersection. +\end{definition} + +\begin{definition} + Suppose that $X$ is a set. A $d$-class on $X$ (also called a Dynkin class, or a monotone class, but + we will exclusively call it a $d$-class) is a collection of subsets + of $X$ closed under increasing countable union, that is, for any countable collection + of subsets, $\{A_i\}_{i=1}^\infty \subseteq \mathcal{P}(X)$ with $A_i \subseteq A_j$ + for $i \leq j$, we have + \begin{equation*} + \bigcup_{i=1}^\infty A_i + \end{equation*} + in the $d$-class. + + A $d$ class must contain $X$. + + A $d$-class also must be closed under relative complements. That is, if $A,B$ + are in the $d$-class, with $B \subseteq A$ then $A \setminus B$ is in the $d$-class. + +\end{definition} + +An important feature of $d$-classes is that they can be generated by sets, similar +to $\sigma$-algebras. +\begin{proposition} + Let $X$ be a set, and let $\mathcal{C} \subseteq \mathcal{P}(X)$ + be a collection of subsets of $X$. There is a unique smallest $d$-class + containing $\mathcal{C}$, which we denote $d(\mathcal{C})$. +\end{proposition} +\begin{proof} + This is identitical to the case with $\sigma$-algebras. Since $\mathcal{P}(X)$ + is a $d$-class containing $\mathcal{C}$, and any intersection of a family of $d$-classes + is again a $d$-class, simply take the intersection of all $d$-classes containing $\mathcal{C}$. +\end{proof} + +Now we state and prove the important monotone class theorem: +\begin{theorem} + Let $X$ be a set, and suppose that $\mathcal{C}$ is a $\pi$-class + on $X$. Then $\sigma(\mathcal{C}) = d(\mathcal{C})$. +\end{theorem} +\begin{proof} + We clearly have that $d(\mathcal{C}) \subseteq \sigma(\mathcal{C})$, + since any $\sigma$-algebra is a $d$-class. So we must prove + the reverse inclusion. We can accomplish this by proving that $d(\mathcal{C})$ + is a $\sigma$-algebra. + + Since $X \in \mathcal{C}$, we have that $X \setminus A \in d(\mathcal{C})$ + for any $A \in d(\mathcal{C})$ since $d(\mathcal{C})$ is closed under + relative complements. Hence $d(\mathcal{C})$ is closed under complementation. + + Let us show that $d(\mathcal{C})$ is closed under finite intersections. Let + \begin{equation*} + \mathcal{D}_1 = \{A \in d(\mathcal{C}) \;:\;A\cap C \in d(\mathcal{C}),\text{ for all }C \in \mathcal{C}\}. + \end{equation*} + It is easily verified that $\mathcal{D}$ is a $d$-class, and since $\mathcal{C}$ is a $\pi$-class, we + have $\mathcal{C} \subseteq \mathcal{D}_1$, so $d(\mathcal{C}) \subseteq \mathcal{D}_1$. + Hence, $\mathcal{D}_1 = d(\mathcal{C})$. + + Now define, + \begin{equation*} + \mathcal{D}_2 = \{A \in d(\mathcal{C})\;:\;A\cap B \in \mathcal{D}_1,\text{ for all }B \in d(\mathcal{C})\}. + \end{equation*} + We have $\mathcal{C} \subseteq \mathcal{D}_2$, and again it is easily + verified that $\mathcal{D}_2$ is a $d$-class. Hence $\mathcal{D}_2 = d(\mathcal{C})$. + + Thus $d(\mathcal{C})$ is closed under finite intersection. Let $A,B \in d(\mathcal{C})$. + Then $X \setminus A,X\setminus B \in \mathcal{C}$. Hence $A \cup B = X\setminus ((X\setminus A)\cap (X\setminus B))$. + + So $d(\mathcal{C})$ is closed under finite unions. + + Given any countable subset, $\{A_i\}_{i=1}^\infty$, we can write $\bigcup_{i=1}^\infty A_i$ + as an increasing union by defining $B_i = \bigcup_{j=1}^i A_i$, and then + $\bigcup_{i=1}^\infty A_i = \bigcup_{i=1}^\infty B_i$. Hence + since $d(\mathcal{C})$ is closed under finite union, and countable increasing union, + it is closed under countable union. By taking complements, it is closed + under countable intersection. Hence $d(\mathcal{C})$ is a $\sigma$-algebra, + so finally we have $\sigma(\mathcal{C}) = d(\mathcal{C})$. +\end{proof} + +\section{Measurable functions} +First we define measurable functions on a concrete $\sigma$-algebra, and then talk +about how the concept is generalised to abstract $\sigma$-algebras. +Despite the name, one does not need a measure to define measurable functions, +and the concept is one that is purely to do with $\sigma$-algebras, and is therefore +within the topic of order theory. +\begin{definition} + Let $(X,\mathcal{A})$ and $(Y,\mathcal{B})$ be two sets equipped + with $\sigma$-algebras. A function $f:X\rightarrow Y$ + is measurable if, for each $B \in \mathcal{B}$, $f^{-1}(B) \in \mathcal{A}$. +\end{definition} + +There is a very important lemma, which we shall use frequently: +\begin{lemma} + Suppose that $(X,\mathcal{A})$, $(Y,\mathcal{B})$ are $\sigma$-algebras. + Further suppose that $\mathcal{B} = \sigma(\mathcal{C})$ for some + collection $\mathcal{C}$ of subsets of $Y$. Then $f:X\rightarrow Y$ + is measurable if and only if $f^{-1}(C) \in \mathcal{A}$ for each $C \in \mathcal{C}$. +\end{lemma} +\begin{proof} + One direction of implication is clear, so suppose that $f^{-1}(C) \in \mathcal{A}$ + for each $C \in \mathcal{C}$. Let $\mathcal{F}$ be the set of all $A \subset Y$ + such that $f^{-1}(A) \in \mathcal{A}$. We have $\mathcal{C} \subseteq \mathcal{F}$ + by construction. However, it is easy to see that $\mathcal{F}$ is a $\sigma$-algebra. + Hence, $\sigma(\mathcal{C}) \subseteq F$. And we are done. +\end{proof} + + +\begin{proposition} + The compostion of measurable functions is measurable. +\end{proposition} +\begin{proof} + This is obvious. +\end{proof} + +The correct way to think about measurable functions is, for a measurable +function $f:X\rightarrow Y$, we have a map $f^{-1}:\mathcal{B}\rightarrow\mathcal{A}$, +and because preimages preserve unions, intersections and complements, this is an +\emph{algebra homomorphism}. We define this now. +\begin{definition} + Let $A$ and $B$ be abstract Boolean $\sigma$-algebras. A function $f:A\rightarrow B$ + is an algebra homomorphism if, for any collection $\{a_i\}_{i=1}^\infty \subseteq A$, we have + \begin{align*} + f\left(\bigvee_{i=1}^\infty a_i \right) &= \bigvee_{i=1}^\infty f(a_i)\\ + f\left(\bigwedge_{i=1}^\infty a_i\right) &= \bigwedge_{i=1}^\infty f(a_i). + \end{align*} + And furthermore, for any $a \in A$ we have $f(a') = f(a)'$, and $f(1) = 1$ + and $f(0) = 0$. +\end{definition} + + +The concrete case of a measurable function motivates the abstract case: +\begin{definition} + Given two Boolean $\sigma$-algebras, $A$ and $B$, a measurable function + $f:A\rightarrow B$ is an algebra homomorphism, from $B$ to $A$. +\end{definition} + + +\section{Products of $\sigma$-algebras} +Given two concrete $\sigma$-algebras, $(X,\mathcal{A})$ and $(Y,\mathcal{B})$, +it is natural to want to find a compatibile $\sigma$-algebra on the +cartesian product $X \times Y$. This is described in the product sigma +algebra construction, which actually works for arbitrarily many factors. +The ``compatibility" is provided by the universal property. +\begin{definition} + Let $\{(X_i,\mathcal{A}_i)\}_{i\in I}$ be an indexed collection of concrete + $\sigma$-algebras. Define, + \begin{equation*} + \prod_{i \in I} \mathcal{A}_i := \{ \prod_{i \in I} A_i\;:\;A_i \in \mathcal{A}_i,\text{ and }A_i = X_i\text{ for all but finitely many }i \in I\} + \end{equation*} + and then, + \begin{equation*} + \bigotimes_{i \in I} \mathcal{A}_i := \sigma\left(\prod_{i \in I} \mathcal{A}_i\right). + \end{equation*} + This is a $\sigma$-algebra on $\prod_{i\in I} X_i$. + + The pair $\left(\prod_{i \in I} X_i,\bigotimes_{i \in I} \mathcal{A}_i\right)$ is + called the product $\sigma$-algebra. + + Let the projection onto the $X_j$ factor be denoted $\pi_j:\prod_{j \in I} X_i \rightarrow X_i$ +\end{definition} + +The crucial theorem characterising the product is the \emph{universal property}, defined +and proved below. +\begin{proposition} + Let $\{(X_i,\mathcal{A}_i)\}_{i \in I}$ be an indexed collection of $\sigma$-algebras. + Let $(Y,\mathcal{B})$ be a $\sigma$-algebra, and suppose that for each $i \in I$, + there is a measurable function $f_i:Y\rightarrow X_i$. Then there exists a unique + measurable function $f:Y\rightarrow \prod_{i \in I} X_i$, such that for each $i$ + the following diagram commutes. + \begin{displaymath} + \xymatrix{ + & + (Y,\mathcal{B}) \ar@{.>}[d]^f \ar[ld]^{f_i}& \\ + (X_i,\mathcal{A}_i) & + \left(\prod_{i\in I} X_i,\bigotimes_{i \in I} \mathcal{A}_i\right)\ar[l]^{\pi_i}& + } + \end{displaymath} +\end{proposition} +\begin{proof} + Define the function $f = (f_i)_{i \in I}$. We need to show that $f^{-1}(C) \in \mathcal{B}$ + for each $\mathcal{C} \in \prod_{i \in I} \mathcal{A}_i$. So we can say that $C = \prod_{i \in I} A_i$ + for some selection $A_i \in \mathcal{A}_i$, where $A_i \neq X_i$ for only finitely + many $i$. Then + \begin{equation*} + f^{-1}(C) = \bigcap_{\{i \in I\;:\;A_i \neq X_i\}}f_{i}^{-1}(A_i) \in \mathcal{B}. + \end{equation*} +\end{proof} + +The product $\sigma$-algebra satisfies numerous nice properties. We document a +few now. +\begin{proposition} + Suppose that $(X,\mathcal{A})$ and $(Y,\mathcal{B})$ are concrete $\sigma$-algebras. + Let $E \subseteq X\times Y$. Then for each $x \in X$ and $y \in Y$, + define + \begin{align*} + E^x &:= \{y \in Y\;:\;(x,y) \in E\}\\ + E^y &:= \{x \in X\;:\;(x,y) \in E\}. + \end{align*} + $E^x$ is called the $x$-slice of $E$, and $E^y$ is the $y$-slice of $E$. + If $E \in \mathcal{A} \otimes \mathcal{B}$, then $E^x \in \mathcal{B}$ + and $E^y \in \mathcal{A}$. +\end{proposition} +\begin{proof} + It is sufficient to prove the following. Denote by $\mathcal{F}$ + the following set, + \begin{equation*} + \mathcal{F} = \{E \subseteq X\times Y\;:\;\forall (x,y) \in X\times Y, E^x \in \mathcal{B},E^y \in \mathcal{A}\}. + \end{equation*} + We wish to prove that $\mathcal{F} = \mathcal{A}\otimes \mathcal{B}$. + + We have $\mathcal{A} \times \mathcal{B} \subseteq \mathcal{F}$. + + If $A,B \in \mathcal{F}$, with $B \subseteq A$, then $A \setminus B \in \mathcal{F}$ + since $(A \setminus B)^x = A^x \setminus B^x$. + + If $\{A_n\}_{n=1}^\infty \subseteq \mathcal{F}$ is an increasing sequence, + then $\left(\bigcup_{n=1}^\infty A_n\right)^x = \bigcup_{n=1}^\infty A_n^x \in \mathcal{F}$. + + Hence $\mathcal{F}$ is a $d$-class, so $d(\mathcal{A}\times\mathcal{B}) \subseteq \mathcal{F}$. + + However, $\mathcal{A}\times\mathcal{B}$ is a $\pi$-class, so + by the monotone class theorem, we therefore have that $\mathcal{A}\otimes\mathcal{B} = \sigma(\mathcal{A}\times\mathcal{B}) \subseteq \mathcal{F}$. + +\end{proof} + + +\begin{proposition} + Suppose that $(X,\mathcal{A})$, $(Y,\mathcal{B})$ and $(E,\mathcal{E})$ + are $\sigma$-algebras. + + The a function $f:X\times Y\rightarrow E$ is $\mathcal{A}\otimes \mathcal{B}$-measurable + if and only if, for every $x \in X$ and $y \in Y$, we have that + the function $f^x:y\mapsto f(x,y)$ is $\mathcal{B}$-measurable + and the function $f^y:x\mapsto f(x,y)$ is $\mathcal{A}$-measurable. +\end{proposition} +\begin{proof} + First suppose that $f:X\times Y\rightarrow E$ + is $\mathcal{A}\otimes \mathcal{B}$-measurable. Then fix $x \in X$. Let $F \in \mathcal{E}$. + + Hence, + \begin{equation*} + (f^x)^{-1}(F) = \{y\;:\;f(x,y) \in F\}. + \end{equation*} + But the right hand side is simply, + \begin{equation*} + \{(x,y)\;:\;f(x,y) \in F\}^x = f^{-1}(F)^x. + \end{equation*} + However by assumption $f^{-1}(F) \in \mathcal{A}\otimes\mathcal{B}$. + Hence, $f^{-1}(F) \in \mathcal{B}$. Hence $f^x$ is $\mathcal{B}$-measurable. + + By symmetry, for each $y \in Y$ we have $f^y$ is $\mathcal{A}$-measurable + + + +\end{proof} +\section{Real measurable functions} +We will now focus on measurable real valued functions on a concrete $sigma$-algebra. +This will motivate our definitions for measurable real valued functions +on an abstract $\sigma$-algebra. We shall use the machinery of product measures +here. + +The real numbers, $\Rl$, are given a $\sigma$-algebra, generated +by all the half-open intervals of the form $(a,b]$, for $a < b \in \Rl$. It is +easy to see that this is equivalently generated by rays, $(a,\infty)$, $[a,\infty)$ +or open intervals $(a,b)$ or closed intervals $[a,b]$, or open sets. + +Given a concrete $\sigma$-algebra $(X,\mathcal{A})$, let $\mathcal{M}(X,\mathcal{A})$ +denote the set of measurable functions from $X$ to $(\Rl,\mathcal{B}(\Rl))$. + +\begin{lemma} + Let $X = \prod_{i=1}^\infty[0,\infty]$, where each factor has the Borel + $\sigma$-algebra, and $X$ has the product algebra. Then the following functions + are measurable: + \begin{enumerate} + \item{} $(x_n)_{n=1}^\infty \mapsto \sup_n x_n$ + \item{} $(x_n)_{n=1}^\infty \mapsto \inf_n x_n$ + \item{} $(x_n)_{n=1}^\infty \mapsto \limsup_n x_n$ + \item{} $(x_n)_{n=1}^\infty \mapsto \liminf_n x_n$. + \end{enumerate} +\end{lemma} +\begin{proof} + We only need to prove $1$. Let + \begin{equation*} + f((x_n)_{n=1}^\infty) = \sup_n x_n. + \end{equation*} + We compute $f^{-1}((a,\infty])$ for some $a > 0$. This is simply, + \begin{equation*} + f^{-1}((a,\infty]) = \bigcup_{n=1}^\infty \prod_{i=1}^{n-1} [0,\infty]\times (a,\infty])\times\prod_{i=n+1}^\infty [0,\infty) + \end{equation*} + Which is in the product $\sigma$-algebra. +\end{proof} + +\begin{lemma} + Let $X = \Rl^2$, given the $\sigma$-algebra $\mathbb{B}(\Rl)\otimes \mathbb{B}(\Rl)$. + Then the function $(x,y)\mapsto x+y$ is measurable. +\end{lemma} +\begin{proof} + This is true + because addition is continuous. +\end{proof} + + +\begin{proposition} + $\mathcal{M}(X,\mathcal{A})$ is a real vector space, when provided + with pointwise function addition and scalar multiplication. +\end{proposition} +\begin{proof} + Let $f \in \mathcal{M}(X,\mathcal{A})$. Given $\alpha \in \Rl$, if $\alpha = 0$, + we clearly have $\alpha f \in \mathcal{M}(X,\mathcal{A})$, so assume $\alpha \neq 0$. + + First assume that $\alpha > 0$. We can easily compute, $(\alpha f)^{-1}((a,b]) = f^{-1}(a/\alpha,b/\alpha]) \in \mathcal{A}$. + + Now if $\alpha < 0$, we have $(\alpha f)^{-1}((a,b]) = f^{-1}([b/\alpha,a/\alpha)) \in \mathcal{A}$. + + Now if $f,g \in \mathcal{M}(X,\mathcal{A})$, we can compute $f+g$ as a composition, + $x \mapsto (f(x),g(x))\mapsto f(x) + g(x)$. This is a composition of measurable maps. +\end{proof} + +\begin{proposition} + Given a sequence $\{f_n\}_{n=1}^\infty \subseteq \mathcal{M}(X,\mathcal{A})$. + Then $\inf_{n} f_n ,\sup_n f_n \in \mathcal{M}(X,\mathcal{A})$. +\end{proposition} +\begin{proof} + Since each $f_n$ is measurable, the product map $x\mapsto (f_n(x))_{n\in\Ntrl}$ + to $\Rl^\Ntrl$ is measurable. +\end{proof} + +Now, we use these proofs to motivate our construction of $\mathcal{M}(A)$, +where $A$ is a Boolean $\sigma$-algebra. + +\section{Measure Algebras} +Now that we have all the order theory out of the way, we can finally +get to the theory of measures. The elementary concept +that we introduce here is a \emph{measure algebra}. +\begin{definition} + A measure algebra is a pair $(A,\mu)$, where $A$ is a Boolean $\sigma$-algebra and + $\mu:A\rightarrow[0,\infty]$ is such that: + \begin{enumerate} + \item{} $\mu(a) < \infty$ for some $a \in A$. + \item{} If $\{a_i\}_{i=1}^\infty \subseteq A$ is a sequence that is + \emph{pairwise disjoint} (meaning $a_i \wedge a_j = 0$ for $i\neq j$) + we have + \begin{equation*} + \mu\left(\bigvee_{i=1}^\infty a_i\right) = \sum_{i=1}^\infty \mu(a_i). + \end{equation*} + \end{enumerate} +\end{definition} +\begin{remark} + Given a $sigma$-algebra $\mathcal{A}$ of subsets of a set $X$, and a measure + algebra $(\mathcal{A},\mu)$, the triple $(X,\mathcal{A},\mu)$ is called a measure + space. Measure spaces are special cases of measure algebras, so we prove + as many things as possible for measure algebras before later on specialising + to measure spaces. +\end{remark} +\begin{proposition} + Let $(A,\mu)$ be a measure algebra. Then $\mu(0) = 0$. +\end{proposition} +\begin{proof} + Let $a \in A$ with $\mu(a) < \infty$. By countable disjoint additivity, we have + \begin{equation*} + \mu(a\vee 0 \vee 0 \vee 0 \vee \cdots) = \mu(a) + \mu(0) + \mu(0) + \cdots. + \end{equation*} + Hence, + \begin{equation*} + \mu(0) + \mu(0) + \mu(0) + \cdots = 0. + \end{equation*} + Thus $\mu(0) = 0$. +\end{proof} +\begin{proposition} + Measure are finitely disjointly additive. That is, if $(A,\mu)$ is a measure + algebra, and $a,b \in A$ with $a \wedge b = 0$, then $\mu(a \vee b) = \mu(a) + \mu(b)$. +\end{proposition} +\begin{proof} + Write $a \vee b = a \vee b \vee 0 \vee 0 \vee 0 \vee \cdots$, then use + countable disjoint additivity and $\mu(0) = 0$. +\end{proof} +\begin{proposition} + Let $(A,\mu)$ be a measure algebra, and $a,b \in A$. Then, + \begin{equation*} + \mu(a \vee b) + \mu(a \wedge b) = \mu(a) + \mu(b). + \end{equation*} + If $\mu(a \wedge b) < \infty$, we have the familiar inclusion-exclusion principle, + $\mu(a\vee b) = \mu(a) + \mu(b) - \mu(a \wedge b)$. +\end{proposition} +\begin{proof} + We can write, $a \vee b = a \vee ( b \wedge a')$, and $a \vee b = (a \wedge b) \vee (a \wedge b') \vee (b \wedge a')$. + Then by disjoint additivity, we have + \begin{equation*} + \mu(a \vee b) = \mu(a\wedge b) + \mu(a \wedge b') + \mu(a' \wedge b). + \end{equation*} + Hence, + \begin{equation*} + \mu(a \vee b)+\mu(a\wedge b) = (\mu(a\wedge b) + \mu(a \wedge b')) + (\mu(a \wedge b) + \mu(a'\wedge b)) + \end{equation*} + But by disjoint additvity, the right hand side is simply $\mu(a) + \mu(b)$. + +\end{proof} +\begin{proposition} + Suppose that $(A,\mu)$ is a measure algebra, and $\{a_i\}_{i=1}^\infty$ + is an increasing sequence of elements of $A$, that is, $a_i \leq a_j$ + for $i \leq j$. Then + \begin{equation*} + \mu\left(\bigvee_{i=1}^\infty a_i\right) = \lim_{i\rightarrow\infty} \mu(a_i). + \end{equation*} +\end{proposition} +\begin{proof} + We ``disjointify" the sequence $\{a_i\}_{i=1}^\infty$. Let $b_i = a_i \wedge a_{i-1}'$ + for $i>1$. Then $\bigvee_{i=1}^\infty b_i = \bigvee_{i=1}^\infty a_i$, but + the sequence $\{b_i\}_{i=1}^\infty$ is pairwise disjoint. Hence, + \begin{align*} + \mu\left(\bigvee_{i=1}^\infty a_i\right) &= \mu\left(\bigvee_{i=1}^\infty b_i\right)\\ + &= \sum_{i=1}^\infty \mu(b_i)\\ + &= \lim_{n\rightarrow\infty} \sum_{i=1}^n \mu(b_i)\\ + &= \lim_{n\rightarrow \infty} \mu\left(\bigvee_{i=1}^n b_i\right)\\ + &= \lim_{n\rightarrow\infty} \mu(a_n). + \end{align*} +\end{proof} +\begin{proposition} + Let $(A,\mu)$ be a measure algebra. Let $\{a_i\}_{i=1}^\infty$ be a decreasing + sequence of elements of $a$, that is $a_i \leq a_j$ for $i \geq j$, furthermore + assume that $\mu(a_1) < \infty$. Then + \begin{equation*} + \mu\left(\bigwedge_{i=1}^\infty a_i\right) = \lim_{n\rightarrow\infty} \mu(a_n). + \end{equation*} +\end{proposition} +\begin{proof} + We have, + \begin{align*} + \mu\left(a_1\wedge \left(\bigwedge_{i=1}^\infty a_i\right)'\right) &= \mu\left(\bigvee_{i=1}^\infty a_1\wedge a_i' \right)\\ + &= \lim_{n\rightarrow\infty} \mu(a_1\wedge a_n')\\ + &= \lim_{n\rightarrow\infty} \mu(a_1)-\mu(a_n). + \end{align*} + Hence, subtract $\mu(a_1)$ from both sides and the result follows. +\end{proof} + +There is an important property of measure spaces, called $\sigma$-finiteness. +\begin{definition} + A measure space $(X,\mathcal{A},\mu)$ is $\sigma$-finite if there is + a sequence $\{A_n\}_{n=1}^\infty \subseteq \mathcal{A}$ such + that $\mu(A_n) < \infty$ for each $n$, but $\bigcup_{n=1}^\infty A_n = X$. +\end{definition} + +\begin{proposition} + Suppose that $(X,\mathcal{A})$ is a $\sigma$-algebra, where $\mathcal{A} = \sigma(\mathcal{C})$ + for some $\pi$-class $\mathcal{C}$. If two $\sigma$-finite measures $\mu$ and $\nu$ on $\mathcal{A}$ + agree on $\mathcal{C}$, then they are equal. +\end{proposition} +\begin{proof} + First consider the case where $\mu$ and $\nu$ are finite. Then, let $\mathcal{F}$ + be the collection of all $F \in \mathcal{A}$ such that $\mu(A) = \nu(A)$. We see + that if $A,B \in \mathcal{F}$, with $B \subseteq A$, then + \begin{equation*} + \mu(A\setminus B) = \mu(A)-\mu(B) = \nu(A)-\nu(B) = \nu(A\setminus B). + \end{equation*} + So $A\setminus B \in \mathcal{F}$. + + Now if $\{A_n\}_{n=1}^\infty$ is an increasing sequence in $\mathcal{F}$, we + have + \begin{equation*} + \mu\left(\bigcup_{n=1}^\infty A_n\right) = \lim_{n\rightarrow\infty} \mu\left(A_n\right) = \lim_{n\rightarrow\infty} \nu(A_n) = \nu\left(\bigcup_{n=1}^\infty\right) + \end{equation*} + + Hence $\mathcal{F}$ is a $d$-class, so $\mathcal{A} = \sigma(\mathcal{C}) = d(\mathcal{C}) \subseteq \mathcal{F}$. + + Now assume that $\mu$ and $\nu$ are $\sigma$-finite. We can find an + increasing sequence $\{A_n\}_{n=1}^\infty$ such that $\mu(A_n),\nu(A_n) <\infty$ + and $X = \bigcup_{n=1}^\infty A_n$. Then, if $A \in \mathcal{F}$ since the + measures $\mu_n(A) = \mu(A_n\cap A$ and $\nu_n(A) = \nu(A\cap A_n)$, we have + \begin{equation*} + \mu(A) = \lim_{n\rightarrow\infty} \mu(A\cap A_n) = \lim_{n\rightarrow\infty} \nu(A\cap A_n) = \nu(A). + \end{equation*} +\end{proof} + +\section{Integration of non-negative functions} +Let $X$ be any set, and let $f:X\rightarrow [0,\infty)$. We can find a sequence +of functions of finite range that converges to $f$ pointwisely and monotonically +from below. + +Let +\begin{equation*} + s_n = \sum_{k=1}^{2^{2n}} \frac{k}{2^n}\chi_{f^{-1}([k/2^n,(k+1)/2^n))} +\end{equation*} + +Now let $(X,\mathcal{A},\mu)$ be a measure space. A function $s:X\rightarrow[0,\infty]$ +is called simple if it measurable and has finite image. If +\begin{equation*} + s = \sum_{k=1}^n a_k \chi_{A_k} +\end{equation*} +for some disjoint selection $A_k \in \mathcal{A}$, we define the integral of $s$ +to be +\begin{equation*} + \int_X s\;d\mu := \sum_{k=1}^n a_k\mu(A_k) +\end{equation*} +where we assert only for the purposes of this definition that $0\cdot\infty = 0$. + +Now if $f:X\rightarrow [0,\infty]$ is any measurable function, we define +\begin{equation*} + \int_X f\;d\mu = \sup\{\int_X s\;d\mu\;:\;s\text{ is a simple function },s \leq f\} +\end{equation*} + +Our first proposition is obvious, +\begin{proposition} + Let $f,g:X\rightarrow[0,\infty]$ be measurable, and $f \leq g$, then + \begin{equation*} + \int_X f\;d\mu \leq \int_X g\;d\mu. + \end{equation*} +\end{proposition} +\begin{proof} + This follows from the fact + \begin{equation*} + \{s\;:\;\text{ is a simple function },s \leq f\} + \end{equation*} + is a subset of + \begin{equation*} + \{s\;:\;\text{ is a simple function },s \leq g\} + \end{equation*} +\end{proof} + +\begin{proposition} + Suppose that $\{f_n\}_{n=1}^\infty$ is a monotonically increasing + sequence of measurable functions from $X$ to $[0,\infty)$. Then + \begin{equation*} + \lim_{n\rightarrow\infty} \int_Xf_n\;d\mu = \int_X \lim_{n\rightarrow\infty} f_n\;d\mu. + \end{equation*} +\end{proposition} +\begin{proof} + Let $f = \lim_{n\rightarrow \infty} f_n$. For any $n$, we have $f_n\leq f$. Hence, + \begin{equation*} + \int_{X} f_n\;d\mu \leq \int_X f\;d\mu. + \end{equation*} + Thus, + \begin{equation*} + \lim_{n\rightarrow\infty} \int_X f_n\;d\mu \leq\int_X f\;d\mu. + \end{equation*} + + So we must prove the reverse inequality. + + Let $s$ be a simple function with $s \leq f$. + Let $s = \sum_{k=1}^N s_k \chi_{B_k}$, where the $B_k$ + are pairwise disjoint. + We would like to show that + $\lim_{n\rightarrow\infty} f_nd\mu \geq \int_X s\;d\mu$. To this end, + let $c \in (0,1)$. Define + \begin{equation*} + A_n = \{x\;:\;f_n \geq cs\}. + \end{equation*} + We have, + \begin{equation*} + f_n \geq f_n \chi_{A_n} \geq cs\chi_{A_n}. + \end{equation*} + Hence, + \begin{equation*} + \int_X f_n\;d\mu \geq \sum_{k=1}^N cs_k \mu(B_k \cap A_n). + \end{equation*} + Now we take the limit. Note that since $f_n$ converges pointwisely, + $\bigcup_n A_n = X$. Thus, $\lim_{n\rightarrow\infty} \mu(B_k \cap A_n) = \mu(B_k)$. + + Thus, + \begin{equation*} + \lim_{n\rightarrow\infty} \int_X f_n\;d\mu \geq c\int_X s\;d\mu. + \end{equation*} + But $c$ is arbitrary, so we have our result. +\end{proof} + + +The next result is called Fatou's lemma. +\begin{lemma} + Let $\{f_n\}_{n=1}^\infty$ be a sequence of measurable functions from $X$ to $[0,\infty)$. We have + \begin{equation*} + \int_X \liminf_{n} f_n\;d\mu \leq \liminf_{n} \int_X f_n\;d\mu. + \end{equation*} +\end{lemma} +\begin{proof} + For any $k \geq 0$, and $j\geq n$, we have + \begin{equation*} + \inf_{n\geq k} f_n \leq f_j. + \end{equation*} + Hence, we integrate this, + \begin{equation*} + \int_X \inf_{n\geq k} f_n \leq \int_{X} f_j\;d\mu. + \end{equation*} + But since $j \geq k$ is arbitrary, this implies + \begin{equation*} + \int_X \inf_{n\geq k} f_n\;d\mu \leq \inf_{n\geq k} \int_X f_n \;d\mu. + \end{equation*} + Hence, + \begin{equation*} + \lim_{k\rightarrow\infty} \int_X \inf_{n\geq k} f_n\;d\mu \leq \lim_{k\rightarrow\infty} \inf_{n\geq k} \int_X f_n\;d\mu. + \end{equation*} + So by the monotone convergence theorem, + \begin{equation*} + \int_X \liminf_n f_n\;d\mu \leq \liminf_n \int_X f_n\;d\mu. + \end{equation*} +\end{proof} + +The next theorem generalises the monotone convergence theorem, and is called +the convergence from beneath theorem: +\begin{proposition} + Suppose that $\{f_n\}_{n=1}^\infty$ is sequence of measurable + functions from $X$ to $[0,\infty]$. Suppose that $\lim_{n\rightarrow\infty} = f$, + and for each $n$, we have $f_n \leq f$. Then + \begin{equation*} + \lim_{n\rightarrow\infty} \int_X f_n\;d\mu = \int_X f\;d\mu. + \end{equation*} +\end{proposition} +\begin{proof} + For each $n$, we have $f_n \leq f$, hence + \begin{equation*} + \limsup_n \int_X f_n \;d\mu \leq \int_X f\;d\mu. + \end{equation*} + But also $\liminf_n f_n = f$, so by Fatou's lemma, + \begin{equation*} + \int_X f\;d\mu = \int_X \liminf_n f_n\;d\mu \leq \liminf_n \int_X f_n\;d\mu. + \end{equation*} + So we are done. +\end{proof} + + +The monotone convergence theorem justifies our definition of the integral +in the case of abstract measure algebras. Suppose that $(A,\mu)$ is a measure algebra, +and $f:A\rightarrow \mathcal{B}([0,\infty])$ is a measurable function (that is, +$f$ is a $\sigma$-algebra homomorphism from $\mathcal{B}([0,\infty])$ to $A$. Then define +\begin{equation*} + \int_X f\;d\mu := \lim_{n\rightarrow\infty} \sum_{k=1}^{2^{2n}} \frac{k}{2^n}\mu(f(k/2^n,(k+1)/2^n]) +\end{equation*} + +Now using this definition, we prove some further properties. +\begin{proposition} + Let $(A,\mu)$ be a $\sigma$-algebra. Let $f,g:A\rightarrow\mathcal{B}([0,\infty])$ +\end{proposition} + +\section{Product measures and Tonelli's theorem} +Let $(X,\mathcal{A},\mu)$ and $(Y,\mathcal{B},\nu)$ be $\sigma$-finite measure +spaces. We would like to define a compatible measure $\mu\times\nu$ on $(X\times Y,\mathcal{A}\otimes \mathcal{B})$. + +We can specify this uniquely by saying that for each $A\times B \in \mathcal{A}\times \mathcal{B}$, +we have $(\mu\times \nu)(A\times B) = \mu(A)\nu(B)$. This determines +the measure uniquely, since it is obvious that any measure satisfying this equality +is $\sigma$-finite, and $\mathcal{A}\times\mathcal{B}$ is a $\pi$-class. + +So we only need to prove that $\mu\times\nu$ exists. First we need to +prove the following, +\begin{lemma} + Let $E \in \mathcal{A}\otimes\mathcal{B}$. Then the function $X\rightarrow [0,\infty]$ + given by $x\mapsto \nu(E^x)$ is $\mathcal{A}$ measurable. +\end{lemma} +\begin{proof} + + Let $\mathcal{F}$ be the set of all $E \subseteq X\times Y$ such that $x \mapsto \nu(E^x)$ + is $\mathcal{A}$-measurable. + + We clearly have $\mathcal{A}\times\mathcal{B} \subseteq \mathcal{F}$. + + Suppose that $\{A_n\}_{n=1}^\infty$ is an increasing sequence in $\mathcal{F}$. Then $\nu\left(\bigcup_{n=1}^\infty A_n\right)^x = \nu\left(\bigcup_{n=1}^\infty A_n^x\right)$. + But by countable additivity, this is $\sum_{n=1}^\infty \nu(A_n^x)$. + This is a supremum of measurable functions, hence is measurable. + + Now assume that $\nu$ is a finite measure. + + If $A,B \in \mathcal{F}$, then we have $\nu(A\setminus B)^x = \nu(A^x\setminus B^x) = \nu(A^x) - \nu(B^x)$, + hence $A \setminus B \in \mathcal{F}$. + + Now if $\nu$ is $\sigma$-finite, we have for each $B \in \mathcal{B}$, a sequence + of sets of finite measure $A_n$ such that $\nu(B) = \lim_{n\rightarrow\infty} \nu(A_n\cap B)$. + + Hence, if $A,B \in \mathcal{F}$, then + \begin{align*} + \nu(A\setminus B)^x &= \lim_{n\rightarrow\infty}\\ + &= \nu((A\cap A_n)^x\setminus (B\cap A_n)^x)\\ + &= \lim_{n\rightarrow\infty} \nu(A\cap A_n)^x-\nu(B\cap A_n)^x\\ + &= \lim_{n\rightarrow\infty} \nu(A_n\cap(A\setminus B))^x\\ + \end{align*} + Hence, $A\setminus B \in \mathcal{F}$. + +\end{proof} + +\begin{proposition} + Let $(X,\mathcal{A},\mu)$ and $(Y,\mathcal{B},\nu)$ be $\sigma$-finite + measure spaces. Then there exists a product measure on $(X\times Y,\mathcal{A}\otimes \mathcal{B})$. +\end{proposition} +\begin{proof} + Define, for each $E \in \mathcal{A}\otimes \mathcal{B}$, + \begin{equation*} + (\mu\times\nu)(E) := \int_X \nu(E^x)\;d\mu(x). + \end{equation*} + This is a measure by the monotone convergence theorem. That is, if $\{A_n\}_{n=1}^\infty$ + is a pairwise disjoint subset of $\mathcal{A}\otimes\mathcal{B}$, then + \begin{equation*} + (\mu\times\nu)\left(\bigcup_{n=1}^\infty A_n\right) = \sum_{n=1}^\infty \int_{X} \nu(A_n^x)\;d\mu(x). + \end{equation*} + and we have $(\mu\times\nu)(\emptyset) = 0$ +\end{proof} + +Now we prove the famous and important Tonelli's theorem: +\begin{proposition} + Let $(X,\mathcal{A},\nu)$ and $(Y,\mathcal{B},\mu)$ be measure spaces, + and let $f:X\times Y\rightarrow[0,\infty]$ be $\mathcal{A}\otimes\mathcal{B}$ + measurable. Then + \begin{equation*} + \int_{X\times Y} f\;d(\mu\times\nu) = \int_Y \int_X f(x,y)\;d\mu(x)d\nu(y) = \int_X \int_Y f(x,y)\; d\nu(y)d\mu(x) + \end{equation*} +\end{proposition} +\begin{proof} + This is true by definition when $f$ is an indicator function. Hence by linearity + it is true when $f$ is simple. Then by the monotone convergence theorem, + it is true for any non-negative measurable function $f$. +\end{proof} + +\section{$L^p$ spaces} +So far we have dealt only with non-negative measurable functions. We define +spaces of $\Cplx$ valued functions by noting that for any $p > 0$, the function +$x\mapsto |x|^p$ is measurable on $\Cplx$ with the Borel $\sigma$-algebra. + +Hence define, +\begin{definition} + Given $p > 0$ and a measure space $(X,\mathcal{A},\mu)$, define the set + \begin{equation*} + \mathcal{L}^p(X,\mu) := \{f:X\rightarrow\Cplx\;:\;f\text{ is }\mathcal{B}(\Cplx)\text{ measurable and},\int_X |f|^p\;d\mu < \infty\} + \end{equation*} +\end{definition} + +Note that for any non-negative real numbers $a$ and $b$, we have the inequaluty $(a+b)^p \leq a^p + b^p$. +Hence, $\mathcal{L}^p(X,\mu)$ is a complex vector space. Define the quantity, +\begin{equation*} + \|f\|_p = \left(\int_X |f|^p\;d\mu\right)^{1/p} +\end{equation*} +Hence we have $\|f+g\|_p^p \leq \|f\|_p^p + \|g\|_p^p$. + +Given $f \in \mathcal{L}^1(X,\mu)$, we can write $f$ +as $f = (\max\{\Re(f),0\}-\max\{0,-\Re{f}\}) + i(\max\{\Im(f),0\}-\max\{0,-\Im(f)\})$, +then define +\begin{equation*} + \int_X f\;d\mu = \int_X \max\{\Re(f),0\}\;d\mu - \int_X \max\{0,-\Re(f)\}\;d\mu + i\left(\int_X \max\{\Im(f),0\}\;d\mu - \int_X \max\{0,-\Im(f)\}\;d\mu\right) +\end{equation*} + +Now we can state the important dominated convergence theorem, +\begin{proposition} + Let $\{f_n\}_{n=1}^\infty \subset \mathcal{L}^1(X,\mu)$ be such that + there exists a function $g \in \mathcal{L}^1(X,\mu)$ with $|f_n| \leq g$ + for all $n$. Suppose that for every $x$, $f_n(x)\rightarrow f(x)$. Then + $\|f_n-f\|_1\rightarrow 0$, and hence + \begin{equation*} + \lim_{n\rightarrow\infty} \int_X f_n \;d\mu = \int_X f\;d\mu. + \end{equation*} +\end{proposition} +\begin{proof} + We have $|f-f_n| \leq 2g$, so by Fatou's lemma, + \begin{equation*} + \int_X \liminf_{n} 2g-|f-f_n| \;d\mu \leq \liminf_n\int_X 2g-|f-f_n|\;d\mu. + \end{equation*} + Hence, + \begin{equation*} + \limsup_n \int_X |f-f_n|\;d\mu \leq 0. + \end{equation*} +\end{proof} + +Crucial to defining $L^p$ spaces is the following inequality, +called the H\"older inequality. + +First we prove a lemma, +\begin{lemma} + Let $a,b \geq 0$, and $p > 1$ with $1/p+1/q = 1$. Then + \begin{equation*} + ab \leq \frac{a^p}{p}+\frac{b^q}{q} + \end{equation*} +\end{lemma} +\begin{proof} + Consider the rectanble $[0,a]\times [0,b]$. The area graph of the function $x\mapsto x^{p-1}$ + on the interval $[0,a]$ covers some fraction of $[0,a]\times [0,b]$. The remaining + part of the rectangle is covered by the graph if the inverse function, $x\mapsto x^{\frac{1}{p-1}}$. + Hence, + \begin{equation*} + ab\leq \int_0^a x^{p-1}\;dx + \int_0^{b} x^{\frac{1}{p-1}}\;dx = \frac{a^p}{p}+\frac{b^q}{q}. + \end{equation*} +\end{proof} +\begin{proposition} + Suppose $p \geq 1$, and $1/p + 1/q = 1$. Then, + \begin{equation*} + \|fg\|_1 \leq \|f\|_p\|g\|_q. + \end{equation*} +\end{proposition} +\begin{proof} + If $\|f\|_p = 0$ or $\|g\|_p = 0$, the result is easy. So assume + that $\|f\|_p = \|g\|_p = 1$. Then we have, for each $x \in X$, + \begin{equation*} + |f(x)g(x)| \leq \frac{|f(x)|^p}{p}+\frac{|g(x)|^q}{q} + \end{equation*} + Integrate over $p$ to obtain the result. +\end{proof} + +The next result is called Minkowski's inequality, +\begin{proposition} + Let $p\geq 1$. Then $\|f+g\|_p \leq \|f\|_p + \|g\|_p$. +\end{proposition} +\begin{proof} + Assume $\|f+g\|_p \neq 0$, because then it is obvious. + + We compute, + \begin{equation*} + \|f+g\|_p^p \leq \int_{X} |f||f+g|^{p-1}\;d\mu + \int_X |g||f+g|^{p-1}\;d\mu + \end{equation*} + + Now we use H\"older's inequality, + \begin{equation*} + \|f+g\|_p^p \leq (\|f\|_p+\|g\|_p)\frac{\|f+g\|_p^p}{\|f+g\|_p}. + \end{equation*} + And the result follows. +\end{proof} + +Now we introduce the $L^p$ spaces. Define an equivalence +relation on $\mathcal{L}^p(X,\mu)$ such that two +functions are equivalent if they differ on a set of measure $0$. +This space is called $L^p(X,\mu)$. The algebraic operations and norms +on $\mathcal{L}^p(X,\mu)$ are well defined for $L^p(X,\mu)$ + +\begin{proposition} + Let $p > 0$. Then when equipped with the metric, + $d_p(f,g) = \|f-g\|_p^p$, $L^p(X,\mu)$ is a complete metric space. +\end{proposition} +\begin{proof} + If we suppose that $L^p(X,\mu)$ is complete, then given every series + $\sum_n u_n$ such that $\sum_n \|u_n\|_p^p$ converges, then $\sum_{n} u_n$ + converges. This follows from the bound, + \begin{equation*} + \|u_n + u_{n+1} + u_{n+2} + \cdots + u_{n+m}\|^p_p \leq \|u_n\|_p^p + \|u_{n+1}\|_p^p + \cdots + \|u_{n+m}\|_p^p. + \end{equation*} + Conversely, if $L^p(X,\mu)$ has the property that every series of the form + $\sum_{n} u_n$ where $\sum_{n} \|u_n\|_p^p$ converges, then $\sum_{n} u_n$ + converges in the $L^p$ space, then we prove that $L^p(X,\mu)$ is complete. + + Let $\{u_n\}_{n\in \Ntrl}$ be a Cauchy sequence in the $d_p$ metric. Choose + a subsequence $\{u_{\varphi(n)}\}_{n\in\Ntrl}$ where $\varphi:\Ntrl\rightarrow\Ntrl$ + is increasingly rapidly enough such that $d_p(u_{\varphi(n)},u_{\varphi(n+1)}) < 2^{-n}$. + Then the series $\sum_{n} u_{\varphi(n+1)}-u_{\varphi(n)}$ converges. Hence $\{u_n\}_n$ + converges, so $L^p(X,\mu)$ is complete. + + + Hence to prove that $L^p(X,\mu)$ is complete, it is sufficient to prove that every + series of the form $\sum_n u_n$ where $\sum_n \|u_n\|_p^p$ converges + is convergent. By the monotone convergence theorem, we have + + \begin{equation*} + \lim_{n\rightarrow\infty} \int_X \left(\sum_{k=1}^n |u_k|\right)^p\;d\mu = \int_X \left(\sum_{k=1}^\infty|u_k|\right)^p\;d\mu \leq \sum_{k=1}^\infty \|u_k\|_p^p. + \end{equation*} + + Hence, we have an almost everywhere pointwise converging sum $\sum_{n} |u_n|$. Hence + almost everywhere, the function $f = \sum_{n} u_n$ is defined. The functions, + $f_N = \left(\sum_{n=1}^N u_n\right)^p$ are pointwise dominated by $G = \left(\sum_{n} |u_n|\right)^p$, + which is $L^1$. Hence $f$ is in $L^1$, and the functions $g_n = \sum_{k=n}^\infty u_n$ + are dominated $G$ also, so $f_n\rightarrow f$ in the $L^p$ sense. +\end{proof} + +\section{Signed Measures and Radon-Nykodym} +A simple generalisation of the measure concept is provided +by the idea of a signed measure. +\begin{definition} + Let $A$ be a Boolean $\sigma$-algebra. A function $\nu:A\rightarrow [-\infty,\infty]$ + is called a signed measure if, + \begin{enumerate} + \item{} $\nu(0) = 0$. + \item{} The values $-\infty$ and $+\infty$ are both not simultaneously obtained. + \item{} For any sequence $\{a_k\}_{k=1}^\infty \subseteq A$ with $a_k \wedge a_j = 0$ + for all $k\neq j$, we have + \begin{equation*} + \nu\left(\bigvee_{i=1}^\infty a_i\right) = \sum_{i=1}^\infty \nu(a_i). + \end{equation*} + \end{enumerate} +\end{definition} + +\begin{definition} + An element $p$ of a signed measure algebra $(A,\nu)$ is called \emph{hereditarily positive} + if for any $e \leq p$, we have $\nu(e) \geq 0$. Similarly an element $n$ + is called \emph{hereditarily negative} if for any $f \leq n$ we have $\nu(f) \leq 0$. +\end{definition} +Note that the elements which are both hereditarily negative and hereditarily +positive are exactly null sets. + +\begin{lemma} +\label{positiveImpliesContainsPositive} +Let $(A,\nu)$ be a signed measure algebra, suppose that $a \in A$ has $\nu(A) > 0$. +Then there exists a hereditarily positive element $p \leq a$ with $\nu(p) > 0$. +\end{lemma} +\begin{proof} + If $a$ is hereditarily positive, then we are done. + + Otherwise, there is some $a_1 < a$ with $\nu(a_1) < 0$. + Choose $n_1$ as the smallest positive integer such that $\nu(a_1) < -1/n_1$. + + Now consider $a \wedge a_1'$. We must have $\nu(a\wedge a_1') > 0$. + If $a \wedge a_1'$ is hereditarily positive, we are done. + + Otherwise, there is some $a_2 < a \wedge a_1'$ with $\nu(a_2) < 0$. + + Choose $n_2$ as the smallest positive integer such that $\nu(a_2) < -1/n_2$. + + Continue inductively, building elements $a_k \leq a\wedge \bigwedge_{i=1}^{k-1} a_i'$, + with $n_k$ the smallest positive integer such that $\nu(a_k) \leq -1/n_k$. + + If no such $n_k$ exists, then we are done, and we can choose $p = a \wedge \bigwedge_{i=1}^{k-1} a_i'$. + + Otherwise, the process continues indefinitely. Then define, + \begin{equation*} + p := a \wedge \bigwedge_{i=1}^\infty a_i' = a\wedge\left(\bigvee_{i=1}^\infty a_i\right)'. + \end{equation*} + + Note that we have, + \begin{equation*} + \nu(p) = \nu(a) - \sum_{i=1}^\infty \nu(a_i). + \end{equation*} + Hence we cannot have $\sum_i \nu(a_i) = -\infty$. Thus + the sum $\sum_i 1/n_i$ converges. + + Therefore the sequence $n_i \rightarrow \infty$ as $i\rightarrow\infty$. + + Now let $e \leq p$. If $\nu(e) < 0$, there is some $n_i$ + with $\nu(e) < -1/n_i$. Choose $k > i$ such that $n_k > n_i$. But + then $e_k$ was chosen so that $n_k$ is the least integer such that $\nu(e_k) < -1/n_k$. + + However $\nu(e_k \vee e) < -1/n_i$, so $n_k$ cannot be minimal. + + This is a contradiction, so we must have $\nu(e) \geq 0$. Hence $p$ + is hereditarily positive. +\end{proof} + +Now we prove the Hahn decomposition for measure algebras: +\begin{proposition} + Suppose that $(A,\nu)$ is a signed measure algebra. Then there + exists $p,n \in A$ with $p \vee n = 1$, $\nu(p \wedge n) = 0$ + and $p$ hereditiarily positive and $n$ hereditarily negative. +\end{proposition} +\begin{proof} + Assume without loss of generality that $\nu$ does not take the value $+\infty$. + + Define + \begin{equation*} + \beta = \sup\{\nu(a)\;:\;a\text{ is hereditarily positive.}\}. + \end{equation*} + We see that $\beta$ exists because $0$ is hereditarily positive. Suppose + that $\{a_k\}_{k=1}^\infty$ is a sequence with $\nu(a_k)\rightarrow\beta$. + + Since the join of any two hereditarily positive elements is + hereditarily positive, we may take $a_k$ to be an increasing sequence. + Let + \begin{equation*} + p := \bigvee_{k=1}^\infty a_k. + \end{equation*} + See that if $e \in A$, we have + \begin{equation*} + \nu(p\wedge e) = \lim_{k\rightarrow\infty} \nu(a_k\wedge e) \geq 0. + \end{equation*} + So $p$ is hereditarily positive, and $\nu(p) \leq \beta$. But since + \begin{equation*} + \nu(p) = \lim_{k\rightarrow\infty} \nu(a_k) + \end{equation*} + we must have $\nu(p) = \beta$. + + Now let $n = p'$. Let $f \leq n$. Then if $\nu(f) > 0$ by lemma \ref{positiveImpliesContainsPositive}, we must have + \begin{equation*} + \nu(p\vee f) > \nu(p) > \beta + \end{equation*} + and $p \vee f$ is hereditarily positive, so this is impossible. + + Hence $\nu(f) \leq 0$, and so $n$ is hereditarily negative. + + Since $p \wedge n \leq n$ and $p \wedge n \leq p$, $p\wedge n$ + is both hereditarily positive and negative, hence null. +\end{proof} + +Two positive measures $\mu_1$ +and $\mu_2$ are said to be mutually singular if $\mu_1(a) > 0$ +implies that $\mu_2(a) = 0$, and vice versa. We write $\mu_1 \bot \mu_2$. + +This allows us to state and prove the Jordan decomposition: +\begin{proposition} + Let $(A,\nu)$ be a signed measure algebra. There exist unique positive + measures $\nu_+$ and $\nu_-$ with $\nu_+\bot \nu_-$ + and $\nu = \nu_+-\nu_-$. +\end{proposition} +\begin{proof} + Let $p,n$ be the Hahn decomposition of $A$. Define $\nu_+(a) = \nu(a\wedge p)$ + and $\nu_-(a) = -\nu(a\wedge n)$. + +\end{proof} + +\begin{lemma} + Suppose that $A$ is a Boolean $\sigma$-algebra, and $\lambda$ and $\mu$ + are two finite measures where $\lambda$ is positive, and which are not mutually singular. Then there + exists an $\varepsilon > 0$ and an element $e \in A$ such that $e$ + is hereditarily positive for $\lambda-\varepsilon\mu$, and $\mu(e) > 0$. +\end{lemma} +\begin{proof} + Define the measure $\nu_k = \lambda-(1/k)\mu$. Let $1 = p_k\vee n_k$ + be the associated Hahn decomposition. Then let $p = \bigvee_k p_k$ + and $n = p' = \bigwedge_k n_k$. + + Then $n$ is negative for every $\nu_k$, thus $\lambda(n) \leq 0$. Hence $\lambda(n) = 0$. + + If $\mu(p) = 0$, then $\mu$ and $\lambda$ are mututally singular. + Hence $\mu(p_k) > 0$ for some $k$. Thus we have a solution + for $\varepsilon = 1/k$ and $e = p_k$. +\end{proof} + +At last we can prove the important Radon-Nikodym theorem, also +called the Lebesque decomposition. +\begin{proposition} + Let $(X,\mathcal{A},\nu)$ be a $\sigma$-finite signed measure space. Let + $\mu$ be a positive measure on $(X,\mathcal{A})$. Then there exists a + unique decomposition, + \begin{equation*} + \nu(A) = \lambda(A) + \int_A f\;d\mu. + \end{equation*} + where $f \in \mathcal{L}^1(X,\mu)$ and $\lambda$ is a positive measure with $\lambda \bot \nu$. +\end{proposition} +\begin{proof} + Assume first that $\nu(X) < \infty$ and $\nu$ is a positive measure. + Let + \begin{equation*} + \mathcal{F} = \left\{f \;:\; \int_A f\;d\mu \leq \nu(A)\right\}. + \end{equation*} + Note that if $f,g \in \mathcal{F}$, then $\max\{f,g\} \in \mathcal{F}$. + + Let + \begin{equation*} + a = \sup\left\{\int_X f\;d\mu \;:\; f \in \mathcal{F}\right\}. + \end{equation*} + We have $a \leq \nu(X)$. Suppose that $\{g_n\}_{n=1}^\infty$ + is a sequence of measurable functions which approximate the supremum. That is, + \begin{equation*} + a = \lim_{n\rightarrow\infty} \int_X g_n\;d\mu. + \end{equation*} + + We can choose $g_n$ to be an increasing sequence, since $\mathcal{F}$ + is closed under taking finite maximums. Then let $f = \lim_{n\rightarrow\infty} g_n$. + + Hence by the monotone convergence theorem we have + \begin{equation*} + \int_A f\;d\mu = \lim_{n\rightarrow\infty} \int_A g_n\;d\mu \leq \nu(A). + \end{equation*} + Hence $f \in \mathcal{F}$, and + \begin{equation*} + \int_X f\;d\mu = a. + \end{equation*} + + + Now define the measure $\lambda$, + \begin{equation*} + \lambda(A) = \nu(A) - \int_Af\;d\mu + \end{equation*} + $\lambda$ is a positive measure since $f \in \mathcal{F}$. + + If $\lambda$ is not mutually singular to $\mu$, then there exists $\varepsilon > 0$ + and an element $E \in \mathcal{A}$ such that $E$ is hereditarily positive for + $\lambda - \varepsilon \mu$ and $\mu(E) > 0$. + + Let $a \in A$. Then, + \begin{equation*} + \varepsilon\mu(E\cap A) \leq \lambda(A) = \nu(A) - \int_A f\;d\mu. + \end{equation*} + Then consider + $f_\varepsilon := f+\varepsilon\chi_E$. We have $\int_X f_\varepsilon \;d\mu > \int_X f\;d\mu$. + But this is impossible. + + Hence $\lambda \bot \mu$. + + Now we generalise to the case where $\nu$ is a $\sigma$-finite positive measure + by finding an increasing sequence $\{A_k\}_{k=1}^\infty \subseteq \mathcal{A}$ + and finding a decomposition $d\lambda_k+f_kd\mu$ for the measure + obtained by restricting $\nu$ to $A_k$. By taking limits, this + gives us a decomposition for all of $X$. + + Now use the Jordan decomposition to increase to the case where $\nu$ is signed. +\end{proof} + +\section{Young's inequality} +Suppose that $(G,+)$ is a locally compact abelian group. The general +theory of locally compact abelian groups tells us that $G$ possesses +a unique (up to scaling) translation invariant measure +which is \emph{inner regular} and \emph{outer regular}. Denote +this measure by $\mu$. Given $f,g \in C_c(G)$, define +\begin{equation*} + (f*g)(x) := \int_G f(x-y)g(y)\;d\mu(y). +\end{equation*} + +\begin{proposition} + Let $1 \leq p,q,r \leq \infty$ with $1+1/r = 1/p + 1/q$. Then if + $f \in \mathcal{L}^p(G,\mu)$ and $g \in \mathcal{L}^q(G,\mu)$, + then $f*g \in \mathcal{L}^r(G,\mu)$, with + \begin{equation*} + \|f*g\|_r \leq \|f\|_p\|g\|_q. + \end{equation*} +\end{proposition} +\begin{proof} + +\end{proof} + +\end{document} diff --git a/Measure Theory/tex/owmaths.cls b/Measure Theory/tex/owmaths.cls new file mode 100755 index 0000000..2b42329 --- /dev/null +++ b/Measure Theory/tex/owmaths.cls @@ -0,0 +1,72 @@ +\NeedsTeXFormat{LaTeX2e} +\ProvidesClass{unswmaths}[2013/01/01 Latex class for OW Maths documents] + +\RequirePackage{amsmath} +\RequirePackage{amssymb} +\RequirePackage{amsthm} +\RequirePackage[a4paper]{geometry} +\RequirePackage{fancyhdr} +\RequirePackage{cite} +\RequirePackage[pdftex]{graphicx} +\RequirePackage{epstopdf} + +\LoadClass{article} +\setlength\parindent{0pt} +\setlength{\parskip}{5mm plus4mm minus3mm} + +\def\author#1{\def\@author{#1}} +\def\subject#1{\def\@subject{#1}} +\def\title#1{\def\@title{#1}} +\def\studentno#1{\def\@studentno{#1}} +\def\supervisor#1{\def\@supervisor{#1}} + +%This puts headers above and below each page. +\pagestyle{fancy} +\fancyfoot[l]{\@author} +\fancyfoot[r]{\today} +\fancyhead[l]{openware} +\fancyhead[r]{\@subject} + + +\newcommand{\HRule}{\rule{\linewidth}{0.5mm}} + +%arguments in the order, author, student number, subject, title +\newcommand{\unswtitle}{ + \begin{titlepage} + \begin{center} + % Upper part of the page. The '~' is needed because \\ + % only works if a paragraph has started. +% \includegraphics[width=0.5\textwidth]{artwork/unswlogo}~\\[0.5cm] + % Upper part of the page. The '~' is needed because \\ + % only works if a paragraph has started. + \includegraphics[width=0.4\textwidth]{artwork/mathslogo}~\\[1cm] + \textsc{\LARGE SKITYL ENTERPRISES}\\[0.5cm] + \textsc{\large }\\[0.5cm] + % Title + \HRule \\[0.4cm] + { \huge \bfseries \@title}\\[0.2cm] + { \Large \@subject }\\[0.2cm] + \HRule \\[1.5cm] + % Author and supervisor + \begin{minipage}[t]{0.4\textwidth} + \begin{flushleft} \large + \emph{Author:}\\ + \@author\\[0.4cm] + \ifx\@supervisor\undefined + \else + \emph{Supervisor:}\\ + \@supervisor + \fi + \end{flushleft} + \end{minipage} + \begin{minipage}[t]{0.4\textwidth} + \begin{flushright} \large + \emph{Student Number:}\\ + \@studentno + \end{flushright} + \end{minipage} + \vfill + \end{center} + \end{titlepage} +} +\endinput diff --git a/Measure Theory/tex/owshortcuts.sty b/Measure Theory/tex/owshortcuts.sty new file mode 100644 index 0000000..d6b74c6 --- /dev/null +++ b/Measure Theory/tex/owshortcuts.sty @@ -0,0 +1,39 @@ +\NeedsTeXFormat{LaTeX2e} +\ProvidesPackage{unswshortcuts} +\RequirePackage{amsmath} +\RequirePackage{amssymb} +\RequirePackage{bbm} + +%Analysis +\newcommand{\Rl}{\mathbb{R}} +\newcommand{\Cplx}{\mathbb{C}} +\newcommand{\Itgr}{\mathbb{Z}} +\newcommand{\Ntrl}{\mathbb{N}} +\newcommand{\Ind}{\mathbbm{1}} +\newcommand{\Hlbt}{\mathcal{H}} + + +%Algebra +\newcommand{\Grp}{\mathcal{G}} + +%Misc +\newcommand{\lra}{\longrightarrow} +\newcommand{\ra}{\rightarrow} +\newcommand{\lla}{\longleftarrow} +\newcommand{\la}{\leftarrow} + +\newtheorem{lemma}{Lemma} +\newtheorem{proposition}{Proposition} +\newtheorem{theorem}{Theorem} +\newtheorem{definition}{Definition} +\newtheorem{corollary}{Corollary} +\newtheorem{remark}{Remark} +\newtheorem{example}{Example} + + +%Stats \ Prob +\newcommand{\E}[1]{\mathbb{E} \left[ #1 \right]} +\newcommand{\Var}[1]{\operatorname{Var} \left[ #1 \right] } +\newcommand{\Cov}[2]{\operatorname{Cov} \left[ #1, #2 \right] } + +\endinput