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{\bf Question}

Let $f$ be a positive bounded measurable function with domain
$[0,1]$.  Using the definition of the Lebesgue integral of $f$
over $[0,1]$, in terms of sequences of simple functions, prove
that

$$\int_0^1 f=m_2(E),$$

where $m_2$ denotes Lebesgue measure in the plane, and $E$ is the
region under the graph of $f$, i.e.

$$E=\{(x,y)|0\leq x\leq 1;0\leq y\leq f(x)\}$$




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{\bf Answer}

In defining $\ds \int_0^1 f$ we prove that a positive measurable
function $f$ can be expressed as a limit of a monotonic increasing
sequence of simple functions.  A simple function is one taking a
finite number of values on measurable subsets of $[0,1]$.  If
$\{Ei\}_{i=1}^n$ is a measurable partition of $[0,1]$ then a
simple function is one of the form

$\ds g(x)=\sum_{i=1}^n c_iX_{Ei}(x), \,$ and we define

$\ds \int_0^1 g=\sum_{i=1}^n c_i m(Ei)$

We then prove that if $\{f_n\}$ and $\{g_n\}$ are two increasing
sequences of functions converging to $f$ then,

$\ds \lim_{n\to\infty}\int_0^1 f_n=\lim_{n\to\infty}\int_0^1 g_n$

We thus define $\ds \int_0^1 f=\lim_{n\to\infty}\int_0^1 f_n$ for
any sequence $\{f_n\}$ of simple functions converging to $f$.

We now use this to prove the stated result.

${}$

Let $Q_{p,s}=\{x\epsilon[0,1)|\frac{p-1}{2^s}\leq
f(x)<\frac{p}{2^s}\} \,\,\,\, p=1,2,\cdots, 2^{2s}$

Since $f$ is bounded, if $s$ is large enough then
$\{Q_{p,s}\}_{p=1}^{2^{2s}}$ forms a partition of $[0,1)$.

We consider a sequence of simple functions defined for such $s$.

$f_s(x)=\frac{p-1}{2^s}, \,\,\,$ for $x\epsilon Q_{p,s} \,\,\,\,
p=1,2,\cdots 2^{2s}$

Clearly $f_s$ is a simple function.  It can be shown that
$f_s\rightarrow f$.

Let $S_{p,s}=\{(x,y)|x\epsilon Q_{p,s}$ and $0\leq y\leq
\frac{p}{2^s}\}$

Then $\ds E\subseteq\bigcup_p S_{p,s}$ for all $s$.  Thus $\ds
E\subseteq\bigcap_s\bigcup_p S_{p,s}$

Now let $\ds (x,y)\epsilon\bigcap_s\bigcup_p S_{p,s}$ then

for all $s$, there exists $p$, such that $x\epsilon Q_{p,s}$ and
$0\leq y\leq \frac{p}{2^s}$

for all $s$, there exists $p$, such that $\frac{p-1}{2^s}\leq
f(x)<\frac{p}{2^s}$ and $0\leq y\leq \frac{p}{2^s}$

for all $s, \,\,\, 0\leq y\leq f(x)+\frac{1}{2^s} \,\,\,$ i.e.
$y\leq f(x)$

Thus $\ds\bigcap_s\bigcup_p S_{p,s}=E$

Now let $T_{p,s}=\{(x,y)|x\epsilon Q_{p,s}$ and $0\leq
y\leq\frac{p-1}{2^s}\}$

Then $\ds m_2\left(\bigcup_p
T_{p,s}\right)=\sum_p\frac{p-1}{2^s}m(Q_{p,s})=\int_0^1 f_s$

$\ds \bigcup_p S_{p,s}-\bigcup_pT_{p,s}=\bigcup_p\{(x,y)|x\epsilon
Q_{p,s}$ and $\frac{p-1}{2^s}<y\leq\frac{p}{2^s}\}$

Thus $\ds m_2\left(\bigcup_p S_{p,s}-\bigcup_p
T_{p,s}\right)=\sum_p\frac{m(Q_{p,s})}{2^s}=\frac{1}{2^s}$

Thus $\ds m_2\left(\bigcup_p S_{p,s}\right)=m_2\left(\bigcup_p
T_{p,s}\right)+\frac{1}{2^s}$

$\ds E=\bigcap_s\bigcup_p S_{p,s}$ and $\bigcup_p S_{p,s}$ has
finite measure as $f$ is bounded, also $\ds \{\bigcup_p
S_{p,s}\}_s$ is a decreasing sequence of sets.

Thus $\ds m_2(E)=\lim_s m_2(\bigcup S_{p,s})$

$\ds =\lim_s\left[m_2\left(\bigcup_p
T_{p,s}\right)+\frac{1}{2^s}\right]=\lim_s m_2\left(\bigcup_p
T_{p,s}\right)$

$\ds =\lim_s\int_0^1 f_s=\int_0^1 f$

Hence the result.


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