Measure Theory 2

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\title{Lecture 2: Measures}
\author{}
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\maketitle
\section{Measures : Sigma Algebra and its properties}
\begin{lem} 
	If $\mathcal{A}\subseteq\sigma(\mathcal{F})$,then $\sigma(\mathcal{A})\subseteq\sigma(\mathcal{F})$.

\end{lem}
\begin{proof}
	$\sigma(\mathcal{A})$ denotes the intersection of sigma-algebras containing $ \mathcal{A} $. Since $\sigma(\mathcal{F})$ is a $\sigma$-algebra containing $ \mathcal{A} $, $\sigma(\mathcal{A})$is contained in $\sigma(\mathcal{F})  $. Thus $\sigma(\mathcal{A})\subseteq\sigma(\mathcal{F})$.
\end{proof}
\begin{defn}
	Let $ (X,\mathcal{T}) $ be a topological space. Then the $sigma $-Algbera generated by all open sets is called the "Borel $\sigma$-Algebra" on $ (X,\mathcal{T}) $ and is denoted by $ \mathcal{B}_x $.The members of $\mathcal{B}_x $ are called Borel Sets.
\end{defn}

\begin{note}
\begin{enumerate}
		$\mathcal{B}_x $ thus includes open sets,closed sets,countable intersections of open sets,countable intersection of closed sets and so forth.
\end{enumerate}
\end{note}
\begin{defn}
	A countable intersection of open sets is called a $ G_\delta$ set.A countable union of closed sets is called an $ F_\sigma $ set. A countable union of $ G_delta$ sets is called $ G_{\delta \sigma}$ set and finally a countable intersection of $F _\sigma $ sets is called an $ F_{\sigma \delta}$ set.
\end{defn}
\begin{exmp}
	Let X be a non-empty set.
	\begin{enumerate}
		\item $ \mathcal{P}(X) $ is $ \sigma $-algebra.
		\item $ \mathcal{A}=\{X,\emptyset\} $ is the smallest $\sigma$-algbera on X and is called trivial $\sigma$-algebra.
		\item $ \mathcal{A}=\{A\subseteq X \mid A\;or \;X \setminus A\;is\;countable\} $ is a $\sigma$-algebra and is called "countbale cocountable sigma $\sigma$-algebra".It is the smallest $\sigma$-algebra containing all singletons.Hence $\{x\}\in \mathcal{A}\;,\forall x\in X$.
		\begin{proof}
			By definition a cocountable subset of X is a subset Y whose complement in X is a countbale set.The set of all subsets of X which are either countbale or cocountbale form a "countbale cocountable sigma $\sigma$-algebra".\newline
			Since $\emptyset$ is countbale, $\emptyset \in \mathcal{A}$. By definition ,if $A\in\mathcal{A}$, then either A or $A^c$ is countable.If A is countable, then $A^c$ is cocountable and vice versa. Hence for both cases, we have $A^c \in \mathcal{A}$. Thus $\mathcal{A}$ is closed under complementation.\newline
			Let $\{A_j\;\mid\;j\in\mathbb{N}\}\in\mathcal{A}$. Let A=$\bigcup_{j\in\mathbb{N}} A_j$.The sets $A_j$ may be countable or coucountable.We need to show $A\in\mathcal{A}$.Let us counsider 2 cases.\newline
			\textbf{Case 1:} $A_j$ is countbale $\forall\;j\in\mathbb{N}$.Since the countable union of countable sets is also countable, thus $A\in\mathcal{A}$.\newline
			\textbf{Case 2:} Let us consider, there exists some $j\in\mathbb{N}$ such that $A_j$ is cocountable. Now $A_j\subseteq A$. Hence $A^c \subseteq A_j{^c}$ . Since the set A is cocountbale, the set $A^c$ is countable. Hence the set A is cocountable. So A$\in\mathcal{A}$. Thus the set $\mathcal{A}$ is closed under complements and countbale unions and is a $\sigma$-algebra.\newline
			\textbf{For the second part of the proof}, we will show that $\mathcal{A}$ is the $\sigma$-algebra generated by all singletons of X. Let $\mathcal{N}$ denote the $\sigma$-algebra generated by all singletons of X. We need to show $\mathcal{N}=\mathcal{A}$.\newline
			By our assumption, $\mathcal{N}$ is smallest $\sigma$-algebra containing all singeltons. Since each singleton is coutnable, $\mathcal{N} \subseteq \mathcal{A}$.\newline
			Let $P\subseteq X$ and without any loss of generality, we can assume that P is a non-empty set. If P is countable, then P can be written as a countable union of all its singletons. Each of these singletons is an element in $\mathcal{N}$.Thus $P\in\mathcal{N}$. Since the choice of P was arbitrary, $\mathcal{A} \subseteq \mathcal{N}$.\newline
			If P is cocountable, then $P^c$ is countable, hence $P^c\in\mathcal{N}$.Again, $\mathcal{N}$ is a $\sigma$-algebra, so.So $P\in\mathcal{N}$. Hence $P\in\mathcal{N}$ and so we can infer that  $\mathcal{A} \subseteq \mathcal{N}$.\newline
			Thus we have shown that $\mathcal{A} \subseteq \mathcal{N}$ and $\mathcal{N} \subseteq \mathcal{A}$. Hence $\mathcal{N}=\mathcal{A}$.
			
			
			
		\end{proof}
		\item The "Borel $\sigma$-Algebra" on $\Real$ in usual topology is denoted by $\mathcal{B}_\Real$.
		\begin{prop}
			The $\sigma$-Algebra  denoted by $\mathcal{B}_\Real$ on $\Real$ (in ususal topology) is generated by each of the following:\newline
			\begin{itemize}
				\item Open intervals denoted by $\mathcal{E}_1=\{(a,b)\:\mid\:a<b\}$.
				\item Closed intervals denoted by $\mathcal{E}_2=\{[a,b]\:\mid\:a<b\}$.
				\item Half-Open intervals denoted by $\mathcal{E}_3=\{(a,b]\:\mid\:a<b\}\:and\: \mathcal{E}_4=\{[a,b)\:\mid\:a<b\}$.
				\item Open Rays denoted by $\mathcal{E}_5=\{(a,\infty)\}\:and\:\mathcal{E}_6=\{(-\infty,a)\}$.
				\item Open Rays denoted by $\mathcal{E}_7=\{[a,\infty)\}\:and\:\mathcal{E}_8=\{(-\infty,a]\}$.
			\end{itemize}
			
		\end{prop}
	\end{enumerate}
	\begin{proof}
		Among all the above stated elements, all are either open or closed sets except for  $\mathcal{E}_3\:and\:\mathcal{E}_4$. But $\mathcal{E}_3\:and\:\mathcal{E}_4$ can be expressed as countable intersections of closed intervals. We can write $(a,b]=\bigcap_{n\in \mathbb{N}}(a,b+\frac{1}{n})$ and $[a,b)=\bigcap_{n\in \mathbb{N}}(a-\frac{1}{n},b)$.So all the elements are Borel sets, thus by \textbf{Lemma 1.1}, $\sigma(\mathcal{E}_j)\subseteq \mathcal{B}_\Real$.\newline
		By definiton, Borel $\sigma$-algebra on $\Real$ is the $\sigma$-algebra generated by all open intervals. Any open interval over $\Real$ can be expressed as a countabale union of intervals of the form $(a,b)$. Hence $\mathcal{B}_\Real\subseteq \sigma(\mathcal{E}_1)$.\newline
		To prove that $\mathcal{B}_\Real\subseteq \sigma(\mathcal{E}_j)$ for $j\geq 2$, we will show that the open interval $(a,b)$ can be expressed as a compliment,countable intersection and union of the other elements.Let us consider the following cases:\newline
		\begin{itemize}
			\item$(a,b)=\bigcup_{n\in \mathbb{N}}(a+\frac{1}{n},b-\frac{1}{n})$.
			\item$(a,b)=(a,b]\cap[a,b)$.
			\item$(a,b)=(a,\infty)\cap(-\infty,b)$.
			\item$(a,b)=[b,\infty)^c\cap(-\infty,a]^c$.
	  \end{itemize}
	  Thus for the $\sigma$-algebras generated by the elements $\mathcal{E}_j$ for $j\geq 2$, we have shown that the open-intervals can be expressed in terms of compliments,countable unions and intersections of the elements of $\sigma(\mathcal{E}_j) $for $j\geq 2$. Thus we have shown that $\mathcal{B}_\Real \subseteq \sigma(\mathcal{E}_j) $ for $1\leq j \geq 8$. Hence our propsition is proved.
	\end{proof}
\end{exmp}
\section{Measurable Mapping}
\begin{defn}
	A pair (X,$\mathcal{F}_c$) consisting of a non empty set X and the $\sigma$-Algebra of its subsets is called a "measurable space". The elements of $\mathcal{F}_c$ are called "measurable" in $\mathcal{F}_c$.
\end{defn}
\begin{defn}
	A function $f:X\rightarrow Y$ where (X,$\mathcal{F}$) and (Y,$\mathcal{G}$) are measurable spaces, is ($\mathcal{F},\mathcal{G}$) measurable, if $f^{-1}(B)\in \mathcal{F}$ $\forall\:B\in\mathcal{G}$.
\end{defn}
\begin{prop}
	(A Measurability Criterion)Let (X,$\mathcal{F}$) and (Y,$\mathcal{G}$) be 2 measurable spaces. Let $\mathcal{C}\in\mathcal{G}$ such that $\sigma(\mathcal{C})=\mathcal{G}$.If $f:X\rightarrow Y$ be a function with property that if $f^{-1}(C)\in \mathcal{F}$ $\forall\:C\in\mathcal{G}$. The function f is ($\mathcal{F},\mathcal{G}$) measurable.
\end{prop}
\begin{proof}
	Let us define a set $\mathcal{D}=\{B\subseteq Y\:\mid\:f^{-1}(B)\in\mathcal{F}\}$.\newline
	
	\textbf{We will now prove that $\mathcal{D}$ is a $\sigma$-algebra}.\newline
	\begin{proof}
	Let us assume A be a set such that A$\in\mathcal{D}$.we need to show $A^c\in\mathcal{D}$.By definition, since $\mathcal{F}$ is a sigma algebra, ${f^{-1}(A)}^c\in\mathcal{F}$. So now we need to show $\{f^{-1}(A)\}^c=f^{-1}(A^c)$.\newline
	Now, let $x\in f^{-1}(A^c)$. So there exists an element $x\in X$ such that f(x)$\in A^c$. Thus f(x)$\notin A$. Thus, x$\notin f^{-1}(A)$. Hence x$\in\{f^{-1}(A)\}^c$. From this, we can infer that $f^{-1}(A^c)\subseteq \{f^{-1}(A)\}^c$.\newline
	Let y$\in \{f^{-1}(A)\}^c$.So, y is an elment in the set X such that y$\notin f^{-1}(A)$.As a result, f(y)$\notin A$.Then we can write f(y)$\in A^c$ which in turn implies y$\in f^{-1}(A^c)$.
	So, we have shown $f^{-1}(A^c)\supseteq \{f^{-1}(A)\}^c$. Since the choice of x and y was arbitrary, we can conclude that $\{f^{-1}(A)\}^c=f^{-1}(A^c)$ which proves that $\mathcal{D}$ is closed under complementation.\newline
	We will now show that $\mathcal{D}$ is closed under countable unions.Let ,$\{A_j\;\mid\;j\in\mathbb{N}\}\in\mathcal{D}$. We consider the sets, $f^{-1}\{\bigcup_{n\in\mathbb{N}}(A_n)\}$ and $\bigcup_{n \in \mathbb{N}}f^{-1}(A_n)$.Let x$\in f^{-1}\{\bigcup_{n\in\mathbb{N}}(A_n)\}$. So, f(x) $\in \bigcup_{n\in\mathbb{N}}(A_n)$. Thus f(x)$\in some A_k$ where $k\in\mathbb{N}$. Thus x $\in f^{-1}A_k$ where $k\in\mathbb{N}$.Thus x$\in \bigcup_{n \in \mathbb{N}}f^{-1}(A_n)$. Hence $f^{-1}\{\bigcup_{n\in\mathbb{N}}(A_n)\} \subseteq\bigcup_{n \in \mathbb{N}}f^{-1}(A_n)$.\newline
	Let y $\in \bigcup_{n \in \mathbb{N}}f^{-1}(A_n)$. Thus y $\in f^{-1}(A_p)$ where p $\in N$. Thus f(y) $\in A_p$ $\in N$. So, f(y) $\in \bigcup_{n\in\mathbb{N}}(A_n)$. Hence y$\in f^{-1}\{\bigcup_{n\in\mathbb{N}}(A_n)\}$. So,$f^{-1}\{\bigcup_{n\in\mathbb{N}}(A_n)\} \supseteq\bigcup_{n \in \mathbb{N}}f^{-1}(A_n)$. Since the choice of x and y was arbitrary,$f^{-1}\{\bigcup_{n\in\mathbb{N}}(A_n)\}=\bigcup_{n \in \mathbb{N}}f^{-1}(A_n)$.\newline
	Now, as $\{A_j\;\mid\;j\in\mathbb{N}\}\in\mathcal{D}$ and $\mathcal{F}$ is a sigma algebra, $ f^{-1}\{\bigcup_{n\in\mathbb{N}}(A_n)\}\in \mathcal{F}$ and as $f^{-1}\{\bigcup_{n\in\mathbb{N}}(A_n)\}=\bigcup_{n \in \mathbb{N}}f^{-1}(A_n)$, $\bigcup_{n \in \mathbb{N}}f^{-1}(A_n) \in \mathcal{D}$. Thus $\mathcal{D}$ is closed under countable unions and complemenetation. Hence it is a $\sigma$-algebra.
\end{proof} 
By the construction of $\mathcal{D}$, $\mathcal{C}\subseteq \mathcal{D}$. So $\sigma(\mathcal{C})\subseteq \sigma(\mathcal{D})$. As $\sigma(\mathcal{C})=\mathcal{G}$, $\mathcal{G}\subseteq \sigma(\mathcal{D})$.Hence the function f is ($\mathcal{F},\mathcal{G}$) measurable.
\end{proof}
\begin{prop}
	Let (X,$\mathcal{F}$) and (Y,$\mathcal{G}$) be 2 measurable spaces. Let (X,$\mathcal{T}_x$) and (Y,$\mathcal{T}_y$) be 2 topological spaces and $\mathcal{F}\:and\:\mathcal{G}$ are the corresponding Borel $\sigma$-algebras.Then every continuous function f:$X\rightarrow Y$ is ($\mathcal{F},\mathcal{G}$) measurable.
\end{prop}
\begin{proof}
	Since the function f is continuous, then if V be an open set in (Y,$\mathcal{T}_y$) then $f^{-1}(V)$ is also open in the topological space (X,$\mathcal{T}_x$). Every open set in (X,$\mathcal{T}_x$) and (Y,$\mathcal{T}_y$) is an element of the Borel $\sigma$-algebras   $\mathcal{F}\:and\:\mathcal{G}$ respectively. Thus $V\in \mathcal{G}$ and as the function f is continuous, $f^{-1}(V) \in \mathcal{F}$.This condition is true for all open sets in (X,$\mathcal{T}_x$) and (Y,$\mathcal{T}_y$) as f is continuous. Thus the function f is measurable.
\end{proof}
\begin{prop}
	Let f be function which is defined as f:X$\rightarrow \Real$.Then, f is measurable iff $\{x\in X\:\mid f(x)\leq q\}\:\in\mathcal{F}\:\forall\:q\in\mathbb{Q}$.
\end{prop}
\begin{proof}
	\textbf{Proof of the if part:} In this case, we will assume that the function f is measurable. Now the interval $(-\infty,a]$ is an element of $\mathcal{B}_{\Real}$ where $a\in\Real$.Now $\mathbb{Q}\subseteq \mathbb{R}$. Hence the interval $(-\infty,q]$ , where $q\in\mathbb{Q}$, is an element of the Borel-$\sigma$-algebra on $\Real$. So if x$\in X$ be such an element,that f(x)$\leq q$,then, f(x)$\in (-\infty,q] \in \mathcal{B}_{\Real}$. Since f is measurable, then $f^{-1}(-\infty,q] \in \mathcal{F}$. Hence if f is measurable, then $\{x\in X\:\mid f(x)\leq q\}\:\in\mathcal{F}\:\forall\:q\in\mathbb{Q}$.\newline
	\textbf{Proof of the only if part:} In this case, for an arbitrary element x$\in X$ such that $\{x\in X\:\mid f(x)\leq q\}\:\in\mathcal{F}\:\forall\:q\in\mathbb{Q}$. We will prove that f is measurable. Let a be an element of the real number line, we write, $a\in \Real$. The open interval $(-\infty,a)$ can be written as $(-\infty,a)=\bigcup_{q<a}-(\infty,q]$ Thus we can write,  $f^{-1}(-\infty,a)=f^{-1}(\bigcup_{q<a}-(\infty,q])$.Now the set $f^{-1}(\bigcup_{q<a}-(\infty,q])$ is equal to $\bigcup_{q<a}f^{-1}(\infty,q]$, This fact has been proved in \textbf{Proposition 2.3}. The set of rational numbers is countable, hence the set $f^{-1}(-\infty,a)\subseteq X$ and it can be written as a countable union of the elements of $\mathcal{F}$. Thus the set $f^{-1}(-\infty,a) \in \mathcal{F}$ and so f is measurable.
\end{proof}
\begin{prop}
\textbf{Composition of Measurable Maps} Let (X,$\mathcal{F}$),(Y,$\mathcal{G}$) and (Y,$\mathcal{H}$) be measurable spaces and let f and g be functions which are defined as $f:X\rightarrow Y$ and $g:Y \rightarrow Z$ be measurable functions. Then h=$f\circ g$ is $(\mathcal{F},\mathcal{H})$ measurable.
\end{prop}
\begin{proof}
	By definition, the functions f and g are measurable. So, $f^{-1}(B)\in \mathcal{F}$ $\forall\:B\in\mathcal{G}$ and $f^{-1}(C)\in \mathcal{G}$ $\forall\:C\in\mathcal{H}$. Now,h is a function from the set X to the set Z. By definition, h is represented as h=$f\circ g$. So we can write $h^{-1}=g^{-1}\circ h^{-1}$.Let us consider a set $K\subseteq Z$ such that $K\in \mathcal{H}$. So, from the definitions of the measurable mappings f and g, it is seen that$h^{-1}(K) \in \mathcal{F}$. Hence the function h which is represented as composition of the 2 measurable maps f and g, is measurable.
\end{proof}
\begin{cor}
	\textbf{Composition with a Continuous Map} Let $(X,\mathcal{F}$) be a measurable space,(Y,$\mathcal{T}_y$) be a topological space and $\mathcal{G}$ be the Borel $\sigma$-algebra on Y.Let g:$Y\rightarrow \Real$ be a continuous function.Then the map $g\circ f$ is Borel measurable for each measurable function f:$X\rightarrow Y$.
\end{cor}
\begin{proof}
	Let us define a function h as $h=g\circ f$. So we can write $h^{-1}=f^{-1}\circ g^{-1}$. By definition f is a measurable function on ($\mathcal{F},\mathcal{G}$) and g is a .continuous map from Y to $\Real$.Let P be an open set on $\Real$. Thus, P is a member of the Borel-$\sigma$-algbera on $\Real$. Since the map g is continuous, then $g^{-1}(P)\in\mathcal{T}_y$. Now as per definition, $\mathcal{G}$ is the Borel-$\sigma$-algbera on (Y,$\mathcal{T}_y$). Thus $g^{-1}(P)\in\mathcal{G}$.Again as f is a measurable function, then $f{-1}\{g^{-1}(P)\} \in \mathcal{F}$.Thus $h^{-1}(P)\in\mathcal{F}$, where P is any open set in $\Real$.Hence the map $g\circ f$ is Borel measurable for each measurable function $f:X\rightarrow Y$.
\end{proof}
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