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{\bf Question}

Suppose ${\cal M}_1$ and ${\cal M}_2$ are two $\sigma$-algebras of
sets

\begin{itemize}
\item[i)]
Is ${\cal M}_1\cup{\cal M}_2$ necessarily a $\sigma$-algebra?

\item[ii)]
Is ${\cal M}_1\cap{\cal M}_2$ necessarily a $\sigma$-algebra?

\item[iii)]
Is ${\cal M}_1-{\cal M}_2$ necessarily a $\sigma$-algebra?
\end{itemize}


\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[i)]
Let ${\cal M}_1$ be the collection of countable subsets of
$[0,1]$, together with their complements.  Then ${\cal M}_1$ is a
$\sigma$-algebra.  Let ${\cal M}_2$ be the collection of countable
subsets of $[1,2]$ together with their complements.  Then ${\cal
M}_2$ is a $\sigma$-algebra, ${\cal M}_1\cup{\cal M}_2$ is not a
$\sigma$-algebra, since it is not closed under unions.

\item[ii)]
${\cal M}_1\cap{\cal M}_2$ is a $\sigma$-algebra.

$E\epsilon{\cal M}_1\cap{\cal M}_2 \Rightarrow E\epsilon{\cal
M}_1\wedge E\epsilon{\cal M}_2 \Rightarrow  E^C\epsilon{\cal
M}_1\wedge E^C\epsilon{\cal M}_2 \Rightarrow E^C\epsilon {\cal
M}_1\cap{\cal M}_2$

$E_i\epsilon{\cal M}_1\cap{\cal M}_2 \Rightarrow E_i\epsilon{\cal
M}_1$ and $E_i\epsilon{\cal M}_2 \Rightarrow \bigcup E_i\epsilon
{\cal M}_1\cap{\cal M}_2$

\item[iii)]
Let ${\cal M}_1={\cal P}(S)$ then ${\cal M}_1$ is a
$\sigma$-algebra.

Let ${\cal M}_2=\{\phi,S\}$ then ${\cal M}_2$ is a
$\sigma$-algebra.

But ${\cal M}_1-{\cal M}_2$ is not a $\sigma$-algebra since it
does not contain $\phi$.

\end{itemize}

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