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{\bf Question}

Deduce as many of the following properties as you can from the
additive law

\begin{itemize}
\item[i)]
$m(\phi)=0$

\item[ii)]
$m(A-B)=m(A)-m(A\cap B)$

\item[iii)]
$B\subseteq A \Rightarrow m(A-B)=m(A)-m(B)$

\item[iv)]
$m(A\cup B)=m(A)+m(B)-m(A\cap B)$

\item[v)]
$m(A)\geq0$

\item[vi)]
$A\subseteq B \Rightarrow m(A)\leq m(B)$
\end{itemize}



\vspace{0.25in}

{\bf Answer}

The additive law says $A\cap B=\phi \Rightarrow m(A\cup
B)=m(A)+m(B)$

\begin{itemize}
\item[i)]
Put $A=B=\phi$, then $A\cap B=\phi$, $A\cup B=\phi$.

$m(\phi)=m(\phi)+m(\phi)$, therefore $m(\phi)=0$

\item[ii)]
$A=(A\cap B)\cup(A-B) \hspace{0.5in} (A\cap B)\cap(A-B)=\phi$

Therefore $m(A)=m(A\cap B)+m(A-B)$

Therefore $m(A-B)=m(A)-m(A\cap B)$

\item[iii)]
$B\subseteq A \Rightarrow A\cap B=B$

Therefore $m(A-B)=m(A)-m(B)$

\item[iv)]
$A\cup B=A\cup(B-A) \hspace{0.2in} A\cap(B-A)=\phi$

Therefore $m(A\cup B)=m(A)+m(B-A)$

$\hspace{0.2in} =m(A)+m(B)-m(A\cap B)$

\item[vi)]
$A\subseteq B \Rightarrow m(B)=m(A)+m(B-A)$

\item[v)]
We cannot prove $m(A)\geq0$, as the following example shows.

$S=\{a,b\}, \hspace{0.1in} m(\phi)=0 \hspace{0.1in} m(\{a\})=1
\hspace{0.1in} m(\{b\})=-1$

$m(S)=0$

Then for all $A,B\subset S \hspace{0.2in} A\cap B=\phi \Rightarrow
m(A\cup B)=m(A)+m(B)$

But $m(A)\not\geq0$ for all $A$.
\end{itemize}


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