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\begin{document}

{\bf Question}

Let ${\cal M}$ be any $\sigma$-algebra of sets.  Show that

\begin{itemize}
\item[i)]
$E\epsilon {\cal M}, \,\, F\epsilon {\cal M} \Rightarrow E\cup
F\epsilon {\cal M}$

\item[ii)]
$E\epsilon {\cal M}, \,\, F\epsilon {\cal M} \Rightarrow E\cap
F\epsilon {\cal M}$

\item[iii)]
$E\epsilon {\cal M}, \,\, F\epsilon {\cal M} \Rightarrow
E-F\epsilon {\cal M}$

\item[iv)]
$\phi\epsilon {\cal M}$

\item[v)]
$\ds E_1,E_2,\cdots E_n\cdots \epsilon {\cal M} \Rightarrow
\bigcap_{n=1}^\infty E_n\epsilon {\cal M}$

\end{itemize}



\vspace{0.25in}

{\bf Answer}

${\cal M}$ is a $\sigma$-algebra of sets. i.e.

\begin{itemize}
\item[a)]
$E\epsilon {\cal M} \Rightarrow E^C\epsilon{\cal M}$

\item[b)]
$\ds E_1,E_2\cdots\epsilon {\cal M} \Rightarrow
\bigcup_{i=1}^\infty E_i\epsilon{\cal M}$
\end{itemize}

\begin{itemize}
\item[i)]
Suppose $E,F\epsilon {\cal M}$.  Let $E_1=E, \,\,
E_2=E_3=\cdots=F$.

Then $\ds \bigcup_{i=1}^\infty E_i=E\cup F\epsilon {\cal M}$.

\item[ii)]
$E\cap F=(E^C\cup F^C)^C\epsilon {\cal M}$

\item[iii)]
$E,F\epsilon{\cal M} \Rightarrow E\cap F^C=E-F\epsilon{\cal M}$

\item[iv)]
$E\epsilon{\cal M} \Rightarrow E^C\epsilon{\cal M} \Rightarrow
E\cap E^C=\phi\epsilon{\cal M}$

\item[v)]
$\ds E_1,E_2\cdots E_n\cdots\epsilon{\cal M} \Rightarrow
E_1^C,E_2^C\cdots E_n^C\cdots\epsilon{\cal M}$

$\ds \Rightarrow
\left(\bigcup_{i=1}^\infty(E_i)^C\right)^C\epsilon{\cal M}
\Rightarrow \bigcap_{i=1}^\infty E_i\epsilon{\cal M}$

\end{itemize}

\end{document}