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%  Lecture 4
%  Primary source: handwritten notes (Mathpix mmd)
%  Fallback: kashlak.pdf (only for OCR/notation/curriculum clarity)
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\section[Lecture 4 -- Lebesgue Measure; Non-Measurable Sets]{Lecture 4 \textemdash{} Lebesgue Measure; Non-Measurable Sets; Product Measures and Independence}
\label{sec:lec04}

The Carath\'eodory and Dynkin machinery of Lecture~3 is now put to work.
We build Lebesgue measure on \((0,1]\) (and on \(\R\)), exhibit a Vitali
set that no translation-invariant measure can size, then sketch how the
same toolkit yields product measures and independence of
\(\sigma\)-fields.

\subsection{Lebesgue measure on \texorpdfstring{$(0,1]$}{(0,1]} and \texorpdfstring{$\R$}{R}}

Take \(\Omega=(0,1]\) (or \(\Omega=\R\)) and let \(\Acal\) be the
collection of all finite unions of half-open intervals \((a,b]\),
together with \(\emptyset\). The target is a set function with
\[
\lambda\bigl((a,b]\bigr) \;=\; b-a,
\]
extended by additivity to \(\Acal\).

\begin{proposition}{$\Acal$ is a $\pi$-system and a ring}{halfopen-ring}
The collection \(\Acal\) of finite (disjoint) unions of half-open
intervals \((a,b]\subseteq(0,1]\), together with \(\emptyset\), is a
\(\pi\)-system: intersections of half-open intervals are again half-open
intervals,
\[
(a,b]\cap(c,d] \;=\;
\begin{cases}
\emptyset & \text{if } b\le c,\\
(c,b]     & \text{if } a\le c<b\le d,
\end{cases}
\]
and unions of finitely many such intersections remain in \(\Acal\). The
class \(\Acal\) is also a ring: it contains \(\emptyset=(a,a]\), is
closed under set difference \((a,b]\setminus(c,d]\), and closed under
finite unions by definition.
\end{proposition}

\begin{proposition}{$\lambda$ is a pre-measure on $\Acal$}{lambda-premeasure}
The set function \(\lambda\colon\Acal\to[0,\infty]\) defined by
\(\lambda((a,b])=b-a\) and extended additively to finite disjoint unions
is a pre-measure on the ring \(\Acal\).
\end{proposition}

Combining the two propositions with the Carath\'eodory and
\(\pi\)-\(\lambda\) theorems of Lecture~3 produces the construction.

\begin{theorem}{Lebesgue measure}{lebesgue}
Let \(\Bcal=\sigma(\Acal)\) be the Borel \(\sigma\)-field of \((0,1]\)
(equivalently, the \(\sigma\)-field generated by the open sets). There
exists a unique measure \(\lambda\) on \(\Bcal\) with
\(\lambda((a,b])=b-a\). It is obtained by Carath\'eodory extension
(Step~2) of the pre-measure on \(\Acal\) (Step~1); uniqueness on
\(\sigma(\Acal)\) follows because \(\Acal\) is a \(\pi\)-system
(Step~3). Writing \(\Mcal_\lambda\) for the \(\sigma\)-field of
\emph{Lebesgue measurable} sets,
\[
\Acal \;\subset\; \Bcal \;=\; \sigma(\Acal)
       \;\subsetneq\; \Mcal_\lambda \;\subsetneq\; \Pcal\bigl((0,1]\bigr),
\]
where \(\Mcal_\lambda\) is the completion of \(\Bcal\) with respect to
the \(\lambda\)-null sets \(\Ncal_\lambda\).
\end{theorem}

\begin{remark}
Lebesgue measure is the \emph{only} translation-invariant measure on
\(\R\) (up to a multiplicative constant) assigning length \(b-a\) to
\((a,b]\); the same statement holds on \(\R^n\). The strict inclusion
\(\Mcal_\lambda\subsetneq\Pcal((0,1])\) is the content of the next
subsection.
\end{remark}

\subsection{Non-measurable sets: the Vitali construction}

We exhibit a set \(H\subseteq(0,1]\) outside \(\Mcal_\lambda\). The
construction uses the Axiom of Choice and addition modulo~\(1\), which
wraps \((0,1]\) into a circle.

\begin{definition}{Addition modulo $1$}{add-mod-one}
For \(x,y\in(0,1]\) define
\[
x \oplus y \;=\;
\begin{cases}
x+y     & \text{if } x+y\le 1,\\
x+y-1   & \text{if } x+y>1.
\end{cases}
\]
For \(A\subseteq(0,1]\) and \(x\in(0,1]\), set
\(A+x=\{y\in(0,1]: y-x\in A\}\), with subtraction taken mod~\(1\).
\end{definition}

\begin{figure}[h]
\centering
\begin{tikzpicture}[>=Stealth]
  \draw[thick, deepnavy] (0,0) circle (1.6);
  \node[above] at (0,1.6) {\small \(\tfrac12\)};
  \node[below] at (0,-1.6) {\small \(0\equiv 1\)};
  \draw[->, thick, exampleblue]
    (1.6,0) arc[start angle=0, end angle=120, radius=1.6];
  \node at (-0.95,1.05) {\small \(\bullet\)};
  \node at (-1.35,1.30) {\small \(x\oplus y\)};
  \node[right=2pt] at (1.6,0) {\small \(\tfrac14\)};
\end{tikzpicture}
\caption{Addition mod \(1\) wraps \((0,1]\) onto a circle: shifts
\(A\mapsto A+x\) become rotations.}
\label{fig:circle-mod-one}
\end{figure}

\begin{lemma}{Translation invariance on Borel sets}{translation-invariance}
Let
\(\Lcal=\{A\in\Mcal_\lambda:\lambda(A+x)=\lambda(A)\text{ for all }x\in(0,1]\}\).
Then \(\Lcal\) is a \(\lambda\)-system, and
\(\Acal\subseteq\Lcal\) since
\(\lambda((a,b]+x)=\lambda((a+x,b+x])=b-a\). By Dynkin's
\(\pi\)-\(\lambda\) theorem,
\(\Bcal=\sigma(\Acal)\subseteq\Lcal\); equivalently, \(\lambda\) is
translation invariant on every Borel set.
\end{lemma}

\begin{example}[Vitali set]
Define an equivalence relation on \((0,1]\) by
\(x\sim y \iff x-y\in\Q\); for instance \(\tfrac1{\sqrt2}\sim
\tfrac1{\sqrt2}+\tfrac1{100}\). By the Axiom of Choice pick a set
\(H\subseteq(0,1]\) containing exactly one representative from each
equivalence class. Then for distinct \(r_1,r_2\in\Q\cap(0,1]\) we have
\((H+r_1)\cap(H+r_2)=\emptyset\), and
\[
(0,1] \;=\; \bigsqcup_{r\in\Q\cap(0,1]}(H+r).
\]
If \(H\) were Lebesgue measurable, countable additivity together with
\cref{lem:translation-invariance} would give
\[
1 \;=\; \lambda\bigl((0,1]\bigr)
   \;=\; \sum_{r\in\Q\cap(0,1]}\lambda(H+r)
   \;=\; \sum_{r\in\Q\cap(0,1]}\lambda(H),
\]
which is \(0\) if \(\lambda(H)=0\) and \(\infty\) if
\(\lambda(H)>0\)\,---\,either way a contradiction. Hence
\(H\notin\Mcal_\lambda\).
\end{example}

\begin{remark}
This shows the strict inclusion \(\Mcal_\lambda\subsetneq\Pcal((0,1])\).
A related fun fact: there is no infinite-dimensional analogue of
Lebesgue measure: the only locally finite, translation-invariant Borel
measure on an infinite-dimensional separable Banach space is the trivial
one.
\end{remark}

\subsection{Product measures, briefly}

The half-open construction generalises directly to \(\R^p\) using
half-open rectangles.

\begin{definition}{Lebesgue measure on $\R^p$}{lebesgue-rp}
On \(\R^p\), define
\[
\lambda^{(p)}\bigl((a_1,b_1]\times\cdots\times(a_p,b_p]\bigr)
\;=\; \prod_{i=1}^{p}\lambda\bigl((a_i,b_i]\bigr)
\;=\; \prod_{i=1}^{p}(b_i-a_i).
\]
The collection of half-open rectangles is a \(\pi\)-system, and the
extension theorem gives a unique measure on \(\Bcal(\R^p)\). For
\(p=2\),
\[
\lambda^{(2)}\bigl((a,b]\times(c,d]\bigr)
   \;=\; (b-a)(d-c)
   \;=\; \lambda\bigl((a,b]\bigr)\,\lambda\bigl((c,d]\bigr).
\]
\end{definition}

\begin{definition}{Product measure}{product-measure}
Given two \(\sigma\)-finite measure spaces
\((\Xset,\Xcal,\mu)\) and \((\Yset,\Ycal,\nu)\), the
\emph{product measure space} is
\((\Xset\times\Yset,\,\Xcal\times\Ycal,\,\pi)\), where the
product \(\sigma\)-field is
\[
\Xcal\times\Ycal
   \;=\; \sigma\bigl(\{A\times B:A\in\Xcal,\,B\in\Ycal\}\bigr),
\]
and \(\pi\) is uniquely determined by
\[
\pi(A\times B)\;=\;\mu(A)\,\nu(B),\qquad A\in\Xcal,\;B\in\Ycal.
\]
\end{definition}

\begin{remark}
The product of two Borel \(\sigma\)-fields satisfies
\(\Bcal(\Xset)\times\Bcal(\Yset)\subseteq\Bcal(\Xset\times\Yset)\), and
equality holds in ``nice'' (e.g.\ second-countable) settings, including
\(\Xset=\Yset=\R\).
\end{remark}

\subsection{Independence}

Switch perspective from measure theory to probability: let
\((\Omega,\Fcal,\mu)\) be a probability space.

\begin{definition}{Independence for sets}{indep-sets}
A countable collection \(\{A_i\}_{i\in I}\subseteq\Fcal\) is
\emph{independent} if, for every finite \(J\subseteq I\),
\[
\mu\!\left(\bigcap_{j\in J}A_j\right)
   \;=\; \prod_{j\in J}\mu(A_j).
\]
\end{definition}

\begin{example}[Standard 52-card deck]
Draw one card uniformly at random and let
\(A_1=\{\text{red}\}\), \(A_2=\{\text{heart or club}\}\),
\(A_3=\{\text{queen}\}\). Then
\[
\mu(A_1)=\tfrac12,\quad \mu(A_2)=\tfrac12,\quad \mu(A_3)=\tfrac1{13},
\]
and one checks
\(\mu(A_1\cap A_2)=\tfrac14\),
\(\mu(A_1\cap A_3)=\tfrac1{26}\),
\(\mu(A_2\cap A_3)=\tfrac1{26}\),
\(\mu(A_1\cap A_2\cap A_3)=\tfrac1{52}\), so the three events are
independent.
\end{example}

\begin{definition}{Independence for $\sigma$-fields}{indep-sigma-fields}
A countable collection \(\{\Fcal_i\}_{i\in I}\) of sub-\(\sigma\)-fields
of \(\Fcal\) is \emph{independent} if every selection
\(\{A_i\in\Fcal_i:i\in I\}\) is an independent collection of sets in the
sense of \cref{def:indep-sets}.
\end{definition}

The next theorem is the workhorse result: independence on a generating
\(\pi\)-system already forces independence of the generated
\(\sigma\)-fields.

\begin{theorem}{Independence from $\pi$-systems}{indep-pi}
Let \(\Acal_1,\Acal_2\subseteq\Fcal\) be \(\pi\)-systems. If
\[
\mu(A_1\cap A_2) \;=\; \mu(A_1)\,\mu(A_2)
\quad\text{for all } A_1\in\Acal_1,\;A_2\in\Acal_2,
\]
then \(\sigma(\Acal_1)\) and \(\sigma(\Acal_2)\) are independent.
\end{theorem}