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We now show that $\mathcal{B} \subsetneq
\mathcal{P}(\RR)$ (in fact,
$\mathcal{M}_{\text{Leb}} \subsetneq
\mathcal{P}(\RR)$). Vitali (1905) first showed
such a set exists.

Consider $E = [0, 1)$ with addition modulo $1$.
Then Lebesgue measure is invariant under this
operation, i.e.
\[ \lambda(B) = \lambda(B + x) \]
where adding is done modulo $1$.

\textbf{Question:} If $B \in \mathcal{B}$, why is
$B + x \in \mathcal{B}$? $\mathcal{G} = \{B \in
\mathcal{B} : B + x \in \mathcal{B}\}$.

For $x, y \in [0, 1)$, define the equivalence
relation $x \sim y$ if $x - y \in \QQ \cap [0, 1)$
($\QQ \cap [0,1)$ a subgroup on $[0, 1)$).
Using the Axiom of Choice (uncountable version),
we can select a representative from each
equivalence class and form the set $S$. We will
show $S \not\in \mathcal{B}$.

The sets $(S + q)_{q \in \QQ \cap [0, 1)}$ are all
disjoint and their union is $[0, 1)$, i.e.
\[ [0, 1) = \bigcup_{q \in \QQ \cap [0, 1)} (S +
q) \]
Now, if $S$ were a Borel set, so would $S + q$ be,
and by translation invariance of $\lambda$ and
countable additivity of $\lambda$,
\[ 1 = \lambda([0, 1]) = \sum_{q \in \QQ \cap [0,
1)} \lambda(S + q) = \sum_{q \in \QQ \cap [0, 1)}
\lambda(S) \]
giving a contradiction. So $S \not\in
\mathcal{B}$.

\begin{remark*}
  Extend this proof to show that $S \not\in
  \mathcal{M}_{\text{Leb}}$.
\end{remark*}

\begin{theorem*}[Banach-Kuratowski 1929]
  Assuming the continuum Hypothesis, there does
  not exist a measure $\mu$ on $\mathcal{P}([0,
  1])$ such that
  \[ \mu([0, 1]) = 1 \qquad \text{and} \qquad
  \mu(\{x\}) = 0 ~\forall x \]
\end{theorem*}

\begin{proof}
  Dudley (appendix).
\end{proof}

Henceforth, whenever we are on a metric space $E$,
we will work with $\mathcal{B}(E)$, which will be
perfectly satisfactory.

\subsection{Probability Measures}

We usually use $(\Omega, \mathcal{F}, \PP)$ to
denote a probability space.
\begin{itemize}
  \item $\Omega$ is the set of possible outcomes
    of the experiment / sample space.
  \item $\mathcal{F}$ is the set of events.
  \item $\PP(\Omega) = 1$, $\PP(A)$ for $A \in
    \mathcal{F}$ is the probability that the event
    occurs.
\end{itemize}
Kolmogorov (1933) set the axioms of probability
theory.

Axioms:
\begin{enumerate}[(1)]
  \item $\PP(\Omega) = 1$, $\PP(\emptyset) = 0$.
  \item $\PP(E) \ge 0 ~\forall E \in \mathcal{F}$.
  \item $\PP \left( \bigcup_n A_n \right) = \sum_n
    \PP(A_n)$ for all $A_n \in \mathcal{F}$
    disjoint, i.e. $\PP$ is a measure.
\end{enumerate}

\begin{remark*}
  \phantom{}
  \begin{itemize}
    \item $\PP \left( \bigcup_n A_n \right) \le
      \sum_n \PP(A_n)$.
    \item $A_n \uparrow A \implies \PP(A_n)
      \uparrow \PP(A)$.
    \item $A_n \downarrow A \implies \PP(A_n)
      \downarrow \PP(A)$.
  \end{itemize}
\end{remark*}

\begin{flashcard}[independent-sigma-algebras-defn]
\begin{definition*}[Independence]
  \glsadjdefn{indep_event}{independent}{events}
  \glsadjdefn{indep_sigma}{independent}{$\sigma$-algebras}
  \cloze{
  We say $(A_i, i \in I)$, $A_i \in \mathcal{F}$,
  are \emph{independent}, if for all finite sets
  $J \subseteq I$, we have
  \[ \PP \left( \bigcap_{j \in J} A_j \right) =
  \prod_{j \in J} \PP(A_j) \]
  Say that the $\sigma$-algebras $(\mathcal{A}_i,
  i \in I)$, $\mathcal{A}_i \in \mathcal{F}$ for
  all $i$, are \emph{independent}, if $(A_i, i \in
  I)$ is independent for all $A_i \in
  \mathcal{A}_i$.
  }
\end{definition*}
\end{flashcard}

\begin{theorem*}
  Let $\mathcal{A}_1$, $\mathcal{A}_2$ be
  $\pi$-systems contained in $\mathcal{F}$ such
  that
  \[ \PP(A_1 \cap A_2) = \PP(A_1) \PP(A_2) \qquad
  \forall A_1 \in \mathcal{A}_1, A_2 \in
  \mathcal{A}_2 \]
\end{theorem*}

\begin{proof}
  Fix $A_1 \in \mathcal{A}_1$, and define for $A
  \in \sigma(\mathcal{A}_2)$:
  \[ \mu(A) = \PP(A_1 \cap A), \qquad \nu(A) =
  \PP(A_1) \PP(A) .\]
  Then $\mu$ and $\nu$ are finite measures, and
  they agree on the $\pi$-system $\mathcal{A}_2$.
  Hence, by \nameref{uniqueness_thm},
  \[ \label{5_133_star} \mu(A) = \nu(A) ~\forall A \in
  \sigma(\mathcal{A}_2) \tag{$*$} \]
  Repeat the same argument, now by fixing $A_2 \in
  \sigma(\mathcal{A}_2)$.
  \[ \mu'(A) = \PP(A \cap A_2), \qquad \nu'(A) =
  \PP(A) \PP(A_2) \qquad \forall A \in
  \sigma(\mathcal{A}_1) \]
  By \eqref{5_133_star}, $\mu' = \nu'$ on
  $\mathcal{A}_1$, hence by
  \nameref{uniqueness_thm}, $\mu' = \nu'$ on
  $\sigma(\mathcal{A}_1)$.
\end{proof}

\subsection{Borel-Cantelli Lemmas}

Given a sequence of events $(A_n, n \in \NN)$,
we may ask for the probability that infinitely
many of them occur.

\begin{flashcard}[liminf-limsup-defn]
\begin{definition*}
  \glsnoundefn{io}{i.o.}{N/A}
  \glsnoundefn{eventually}{eventually}{N/A}
  For $A_n \in \mathcal{F} ~\forall n$, define
  \[ \limsup A_n
  = \cloze{\bigcap_{n = 1}^\infty \bigcup_{m \ge n} A_n
  = \{\text{$A_n$ infinitely often}\}} \]
  \[ \liminf A_n
  = \cloze{\bigcup_{n = 1}^\infty \bigcap_{m \ge n} A_n
  = \{\text{$A_n$ eventually}\}} \]
\end{definition*}
\end{flashcard}

\setcounter{customlemma}{0}
\begin{lemma*}[Borel-Cantelli Lemma 1]
  \refstepcounter{customlemma}
  \label{borel_cantelli_1}
  If $\sum \PP(A_n) < \infty$, then
  $\PP(\text{$A_n$ infinitely often}) = 0$.
\end{lemma*}

\begin{proof}
  Fix any $n \in \NN$. Then
  \[ 0 \le \PP(\text{$A_n$ infinitely often}) \le
  \PP \left( \bigcup_{m \ge n} A_m \right) \le
  \sum_{m \ge n} \PP(A_m) \]
  Take limit as $n \to \infty$.
\end{proof}