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{\bf Question}
Give an outline of the development necessary for a definition of
the Lebesgue intergral of measurable function $f:{\bf R}^n \to
{\bf R}.$ Any theorems concerning measurable functions should be
stated but not proved.
Show that if $f$ and $g$ are two integrable functions then the
function $min(f,g)$ is integratable, and that $$f \, min(f,g) \leq
min \left(\int f, \int g\right).$$ Discuss the case when equality
occurs.
\vspace{.25in}
{\bf Answer}
We first consider the so called simple function. A function $f:
\Omega \rightarrow R^*$ is called a simple function if it can be
expressed in the form \begin{eqnarray}f(x) & = & \sum_{i=1}^n c_i
X_{Ei}(x)\end{eqnarray} where $e_i \epsilon R^*$, $\{E_1, E_2,
\ldots E_n\}$ is a partition of $Omega$ into disjoint measurable
sets, and $X_A(x)$ is the characteristic function of the set $A$.
We then introduce the concept of a measurable function.. A
function $f; \Omega \to R^*$ is said to be measurable if and onlt
if for all $c \epsilon R^*$, $\{ x| f(x) \leq c\}$ is a measurable
set.
We the prove the fundemental result stating that any
non-negitative measurable function can be expressed as the limit
if monotonic increasing sequence of simple functions of the form
(1) by $$\int f = \sum_{i=1}^n c_i \cdot m(E_i),$$ proving htat
$\ds \int f$ is independent of the representation of $f$ in the
form (1).
We then define the integral of a non-negative measurable function
$f$ by expressing $f$ as \begin{eqnarray} f = \lim_{n\to
\infty}f_n \end{eqnarray} where $\{ f_n\}$ is an increasing
sequence of simple functions, and by defining $$\int f =
\lim_{n\to \infty} \int f_n,$$ proving also that $\ds \int f$
independent of the representation of $f$ in the form (2)
We extend the definition to measurable functions which are
positive or negative by adding the functions $f_+, \, f_-:$
\begin{eqnarray*} f_+(x) & = & \left\{ \begin{array}{rcl} f(x) & {\rm if} & f(x) >
0 \\ 0 & {\rm if} & f(x) \leq 0 \end{array} \right. \\ & & \\
F_-(x) & = & \left\{ \begin{array}{rcl} -f(x) & {\rm if} & f(x) <
0 \\ 0 & {\rm if} & f(x) \geq 0 \end{array} \right.
\end{eqnarray*} Then for all $x$ we have that $f(x) = f_+(x) -
f_-(x)$. After proving that for $f$, measurable, $f_+$ and $f_-$
are also measurable, we define $$\int f = \int f_+ - \int f_- $$
whenever R.H.S. is meaningful (i.e.excluding $\infty - \infty$)
${}$
${}$
We now prove that $$ min(f,g) = f - (f-g) _+$$ Suppose $\ds f(x)
\leq g(x)$ then $\ds f(x) - g(x) \leq 0$ So $\ds (f-g) _+(x) = 0$
and $\ds min(f(x),g(x)) = f(x)$ If $\ds f(x) >g(x)$ then $\ds f(x)
- g(x) >0$. Thus $\ds f(x) - (f-g)_+(x) = g(x) = min
(f(x),g(x)).$
Now the difference of two integrable function is integrable, and
so $$min(f,g) = f- (f-g)_+$$ is integrable. Also $$\int min(f,g)
= \int f- \int (f-g)_+ \leq \int f$$ since $\ds (f-g)_+ \geq 0$
and so $\int (f-g)_+ \geq 0$
Similarly $$\int min(f,g) \leq \int g,$$and so $$\int min(f,g)
\leq min\left(\int f \, \int g \right).$$
Suppose $\ds min \left( \int f \, \, \int g \right) = \int f$ and
that $\ds \int min(f,g) = \int f$
then $\ds \int (f-g)_+ = 0$. But $(f-g)_+ \geq 0$ so $(f-g) _+ =
0$ a.e.
Therefore $\ds f(x) \leq g(x)$ a.e.
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