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FUNCTIONAL ANALYSIS

INTEGRATION-FUNCTIONAL APPROACH

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\begin{description}

\item[]
The starting-point is a vector lattice $L$ of functions defined on
a space $X$.

We have a functional $I$ defined on $L$ with the following
properties

\begin{description}

\item[(i)]
$I:L\to R$

\item[(ii)]
$I$ is linear on $L$

\item[(iii)]
$f\geq0\Rightarrow I(f)\geq0$

\item[(iv)]
If $f_n \downarrow0\ I(f_n)\to 0$

\end{description}

we now extend $I$ to a wider class of functions
$U=\{g:X\to\overline{R}:f_n\uparrow g$ for some sequence.\}

Suppose $f_n\uparrow g$ and $g_n\uparrow g$ where
$\{f_b\}\{g_n\}\in L$.

For every $n_0$

$f_n\geq f_n\wedge g_{n_0}\uparrow g_{n_0}$ as $n\to\infty$
therefore

$$I(g_{n_0})=\lim I(f_n\wedge g_{n_0})\leq I(f_n)$$

Hence $\lim I(g_n)\leq\lim I(f_n)$.

Similarly $\lim I(g_n)\geq \lim I(f_n)$. Therefore

$$\lim I(g_n)=\lim I(f_n)=^{df}I_h$$

$U$ is no longer a vector lattice, but it has the following
properties:

\begin{description}

\item[(i)]
$f,g\in U\Rightarrow f+g\in U$

\item[(ii)]
$f\in U\ c\geq 0\Rightarrow cf\in U$

\item[(iii)]
$f,g\in U\Rightarrow f\vee g\ f\wedge g\in U$.

\end{description}

We have the following results.

If $g_n\geq0$ and $g_n\subset U$ and if $\sum g_n=g$ then
$g\subset U$ and $I(g)=\sum Ig_n$, for $g\geq0\ g\subset
U\Leftrightarrow \exists\{f_n\}\subset L$ such that $f_n\geq0$ and
$g=\sum_1^\infty f_n$. In this case $I(g)=\sum I f_n$.

If $g_n\geq0$ and $g_n\subset U$ and if $g=\lim g_n$ then
$g\subset U$ and $I(g)=\lim I(g_n)$.

We now extend the definition of $I$ to a different class of
functions.

Define

\begin{eqnarray*}
\overline{I}(h)&=&\inf\{I(g)\ g\in U\ g\geq h\}\\
\underline{I}(h)&=&-\overline{I}(-h)
\end{eqnarray*}

\begin{description}

\item[(i)]
$\overline{I}(f+g)\leq\overline{I}(f)+\overline{I}(g)$

\item[(ii)]
$\overline{I}(cf)=c\overline{I}(f)\ c\geq0$

\item[(iii)]
$f\leq g\Rightarrow \overline{I}(f)\leq\overline{I}(g)$ and
$\underline{I}(f)\leq\underline{I}(g)$

\item[(iv)]
$\underline{I}(f)\leq\overline{I}(f)$,

\end{description}

for
$0=\overline{I}(0)=\overline{I}(f-f)\leq\overline{I}(f)+\overline{I}(-f)$
 therefore $I(f)=-\overline{I}(-f)\leq\overline{I}(f)$.

\item[Theorem]
If $f\in U\ \underline{I}(f)=\overline{I}(f)=I(f)$.

\item[Proof]
If $f\in U$ clearly $\overline{I}(f)=I(f)$.

$\exists\{f_n\}\subset L$ such that $f_n\uparrow f$ therfore
$-f_n\downarrow-f$

\begin{eqnarray*}
I(f)&=&\lim I(f_n)\\ -I(f)&=&\lim
I(-f_n)\geq\overline{I}(-f)=-\underline{I}(f)\\ \textrm{ therefore
}\underline{I}(f)&\geq &I(f)
\end{eqnarray*}

Hence the result.

\item[Theorem]
If $f_n$ is a sequence of non-negative functions and $f=\sum f_n$

$$\overline{I}(f)\leq\sum\overline{I}(f_n).$$

\item[Proof]
Suppose without loss of generality $\sum\overline{I}(f_n)<\infty$.

Given $\varepsilon>0$, for each $n$ we can find $g_n\in U$ such
that $g_n\geq f_n$ and
$I(g_n)<\overline{I}(f_n)+\frac{\varepsilon}{2^n}$.

If $g=\sum g_n,\ g\in U\ g\geq f$ so

\begin{eqnarray*}
I(g)&\leq&\sum\overline{I}(f_n)+\varepsilon\\ \textrm{therefore
}\overline{I}(f)&\leq&\sum\overline{I}(f_n)+\varepsilon\\
\textrm{therefore }\overline{I}(f)&\leq&\sum \overline{I}(f_n)
\end{eqnarray*}

Define $L'=\{f:\overline{I}(f)=\underline{I}(f)<\infty\}$.

$L'$ contains all functions of $U$ on which $I$ is finite.
$L'\subset L$.

If $f\in L'$ define $I(f)=\overline{I}(f)=\underline{I}(f)$.

$L'$ is a vector lattice.

Let $\circ=+\ \wedge$ or $\vee$.

Let $f,g\in L',\ \varepsilon>0$

$\exists f_1\ f_2:f_1\in U\ f_2\in -U\ f_2\leq f\leq f_1$ and
$I(f_1)+I(-f_2)<\varepsilon$.

$\exists g_1\ g_2:g_1\in U\ g_2\in -U\ g_2\leq g\leq g_1$, and
$I(g_1)+I(-g_2)<\varepsilon$

\begin{eqnarray*}
&&f_2\circ g\leq f\circ g\leq f_1\circ g\ f_1\circ g_1\in U\
f_2\circ g_2\in -U\\ &&I(f_1\circ g_1)+I(-f_2\circ g_2)\leq
I(f_1-f_2)+I(g_1-g_2)<2\varepsilon
\end{eqnarray*}

Therefore $f\circ g\in L'$

Scalar multiplication is trivial.

\item[Theorem]
If $\{f_n\}$ is an increasing sequence of functions in $L'$, if
$\lim I(f_n)<\infty$ and if $=\lim f_n$ then $f\in L'$ and
$I(f)=\lim (f_n)$.

\item[Proof]
We may suppose that $f_1=0\ f=\sum_1^\infty(f_{n+1}-f_n)$
therefore

$$\overline{I}(f)\leq\sum_1^\infty I(f_{n+1}-f_n)=\lim I(f_n)$$

Since $f\geq f_n$ for every $n$

$$\underline{I}(f)\geq \underline{I}(f_n)=I(f_n)$$

Therefore $\underline{I}(f)\geq\lim I(f_n)$ hence the result.

\item[Theorem (Fatou's Lemma)]
Let $f_n$ be a sequence of non-negative integrable $(\in L_1)$
functions. Then $\inf f_n\in L'$. If
$\underline{\lim}I(f_n)<\infty$ then $\underline{\lim}f_n\in L'$
and

$$I(\underline{\lim}f_n)\leq\underline{\lim}I(f_n)$$

\item[Proof]
Define $g_n=f_1\wedge f_2\wedge\ldots\wedge f_n$

$g_n\in L'$ and $g_n\downarrow\inf f_n$ therefore $-g_n\uparrow
-\inf f_n$ therefore $-\inf f_n\in L'$ therefore $\inf f_n\in L'$.

Define $h+n=\inf_{r\geq n}f_r\ \ h_n\in L'$

$h_n\uparrow\underline{\lim}f_n$ therefore provided
$\underline{\lim}I(f_n)<\infty\ \underline{\lim}f_n\in L'$

\begin{eqnarray*}
I(h_n)&\leq&I(f_r)\ r\geq n\\ \textrm{therefore
}I(h_n)&\leq&\underline{\lim}I(f_n)\\ \textrm{therefore
}I(\underline{\lim}h_n)&\leq&\underline{\lim}I(f_n)
\end{eqnarray*}

\item[Theorem (Dominated Convergence)]
If $\{f_n\}$ is a sequence of integrable functions such that
$|f_n|\leq g$ for some $g\in L'$, for every $n$ and if $f_n\to f$
as $n\to\infty$ then $f\in L'$ and $I(f_n)\to I(f)$ as
$n\to\infty$.

\item[Proof]
$o\leq g+f_n\leq2g$ therefore applying Fatou's Lemma to this
sequence

\begin{eqnarray*}
I(\underline{\lim}|g+ f_n)&\leq&\underline{\lim}I(g+ f_n)\\
\textrm{ therefore }I(g+f)&\leq&I(g)+\underline{\lim}I(f_n)\ f\in
L'\\ \textrm{ therefore }I(f)&\leq&\underline{\lim}I(f_n)\\
\textrm{ therefore }I(-f)&\leq&\underline{\lim}I(-f_n)\\
&=&-\overline{\lim}I(f_n)\\ \textrm{ therefore
}I(f)&\geq&\overline{\lim}I(f_n)
\end{eqnarray*}

Hence the result.

this approach ties up with the measure approach in the following
sort of way.

$f\geq 0\ f:X\to R$. $f$ is said to be measurable if $f\wedge g\in
L'$ for every $g\in L'$.

The measurable functions constitute a vector lattice in which
$\lim f_n$ is measurable. A subset $Y$ of $x$ is called a
measurable set if $X_Y$ is a measurable function. The measurable
sets constitute a $\sigma$-algebra of sets. If we assume that
$f\wedge 1\in L'$ then $\{x:f(x)>a\}$ is measurable. If we define
$\mu(Y)=I(X_Y)$ then for $f\in L'\ \ \int f\,d\mu=I(f)$.

\end{description}

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