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{\bf Question}

Suppose that $S_t$, $t\epsilon(a,b)$ is a family of sets indexed
by $t$.  Suppose also that $t_1<t_2 \Rightarrow S_{t_1}\subseteq
S_{t_2}$.  Prove that if $S_t$ is measurable for each
$t\epsilon(a,b)$, then $\ds \bigcup_{t\epsilon(a,b)}S_t$ is
measurable, and that $\ds
m\left(\bigcup_{t\epsilon(a,b)}S_t\right)=\lim_{t\to b-}m(S_t)$

Formulate and prove a corresponding result involving
intersections.



\vspace{0.25in}

{\bf Answer}

Let $\{t_n\}$ be an arbitrary increasing sequence converging to
$b-$.

Then $S_{t_1}\subseteq S_{t_2}\subseteq\cdots\subseteq
S_{t_n}\subseteq\cdots$

and so $\ds
\lim_{n\to\infty}m(S_{t_n})=m\left(\bigcup_{n=1}^\infty
S_{t_n}\right)$

Now $\ds \lim_{n\to\infty}m(S_{t_n})=\lim_{t\to b-}m(S_t)$

Since $t_n$ is an arbitrary sequence.

Also $\ds \bigcup_{n=1}^\infty
S_{t_n}=\bigcup_{t\epsilon(a,b)}S_t$, for $\subseteq$ obvious

if $\ds x\epsilon\bigcup_{t\epsilon(a,b)}S_t, \hspace{0.2in}$
there exists $t\epsilon(a,b), \hspace{0.1in} x\epsilon S_t$

there exists $n$, with $t_n>t$, therefore $x\epsilon S_{t_n}$
therefore $x\epsilon\bigcup_{n=1}^\infty S_{t_n}$ therefore
$\supseteq$.

Hence equality

Thus $\ds m\left(\bigcup_{t\epsilon(a,b)}S_t\right)=\lim_{t\to
b-}m(S_t)$

For intersections

$\ds m\left(\bigcap_{t\epsilon(a,b)}S_t\right)=\lim_{t\to
a+}m(S_t)$

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