\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\parindent=0pt
\begin{document}


{\bf Question}

Define what is meant by a measurable set, and what is meant by a
measurable function.

Show that if $\{A_n\}$ is a sequence of measurable sets with the
properties $A_{n+1} \subseteq A_n$ for$n = 1, 2, \cdots$ and
$$m(A_1) < \infty$$ then $$m \left( \bigcap_{n=1}^\infty A_n
\right) = \lim_{n \to \infty} m(A_n).$$ Suppose $f$ is a
measurable function defined on [0,1].  Define the function $g(x)$
by $$g(x) = m(\{y:f(y) \geq x\}).$$ Show that for each real number
$a$, $$\lim_{x \to a-}g(x)= g(a).$$ Is it true $$\lim_{x\to a+}
g(x) = g(a)?$$  Justify your assertion.

\vspace{.25in}

{\bf Answer}


A set $E$ is said to be measurable if for each set $S$, we have
$$m^*(S) = m^*(S-E) + m^*(S \cap E)$$  A function $F$ is said to
be measurable  if for each number $c$, the set $$\{x|f(x) \geq c\}
$$ is measurable.

${}$

${}$

We first prove that if $A_1, \, A_2$ are all measurable ad
$A-1\subseteq A_2 \subseteq \cdots$ then \begin{eqnarray*}m\left(
\bigcup_{n=1}^\infty A_n\right) & = &  \lim_{n\to \infty} m(A_n)
\\ \bigcup_{n=1}^\infty A_n & = & A_1 \cup (A_2 - A_1 ) \cup (A_3
- A_2) \cup \cdots \\ & = & A_1 \cup \bigcup_{n=1}^\infty (A_{n+1}
- A_n) \end{eqnarray*} So, by additivity , \begin{eqnarray*}
m\left( \bigcup_{n=1} ^\infty\right) & = & m(A_i) +
\sum_{n-1}^\infty m(A_{n+1} - A_n) \\ & = & \lim_{n\to \infty}
[m(A_1) + (A_{2} - A_n) + \cdots + (A_{n+1} - A_n)] \\ & = &
\lim_{n\to \infty} m(A_n) \end{eqnarray*}  Now let $B_i = A_{1} -
A_i$

$\phi = B_1\subseteq B_2 \subseteq\cdots$ so $\ds m\left(
\bigcup_{i=1}^\infty B_i\right) = \lim_{n\to \infty} m(B_n)$

Thus $\ds m\left(A_{1} - \bigcap_{i=1}^\infty A_i\right) =
\lim_{n\to \infty}m(A_{1} - A_n)$

Therefore $\ds m(A_1) - m \left(\bigcap_{n=1}^\infty A_n\right) =
\lim_{n\to\infty} (m(A-1) - m(A-n))$

\hspace{2in} using $m(A_i) < \infty$

Therefore $\ds m \left( \bigcap_{n=1}^\infty A_n\right) =
\lim_{n\to \infty} m(A_n)$

Let $\ds g(x) = m(\{y | f(y) \geq x\})$

Let $a_1, a_2, \ldots$ be an arbitary sequence with the properties
that $ a_1 \leq a_2 \leq \ldots$ and $ a_n \to a$

Let $\ds A_n = \{ y| f(y) \geq a_n\}$

Let $\ds A = \{ y| f(y) \geq a\}$

We first prove that $\ds A = \bigcap_{n=1}^\infty A_n$

Suppose $y \epsilon A$ then $f(y) \geq a$ and so fo all $n$, $f(y)
\geq a_n$.  Hence $\ds y \epsilon \bigcap_{n=1}^\infty A_n$
conversley if $\ds y \epsilon \bigcap_{n=1}^\infty A_n$ then for
all n, $f(y) \geq a_n$ and so $f(y) \geq a_{n+1} \geq a_n$ and so
$y\epsilon A_n$.

Also , since $A_1 \subseteq [0,1]$, we have $\ds m(A_1) \leq 1
\leq \infty$.

Hence $\ds m \left( \bigcap_{n=1}^\infty A_n \right) = \lim_{n\to
\infty} m(A_n)$

or $\ds g(a)  =  \lim_{n\to \infty} g(a_n)$

But $a_n$ is an arbitary increasing sequence $\to a$ and so $\ds
g(a) = \lim_{x \to a-} g(x)$

It is not true that $\ds\lim_{x \to a+} g(x) = g(a)$ as the
following example shows:

Let $f(x) = 1$ for all $x \epsilon[0,1]$

$\ds g(1) = m(\{ y | f(y) \geq 1 \}) = 1$

If $x>1 \, \, g(x) = m(\{y|f(y) \geq x >1\}) = 0$

So $\ds \lim_{x \to 1+} g(x) = 0 \not=g(1)$







\end{document}
