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Revision difference : absolutely continuous function
Version 2 Version 1
Absolutele continuity is the precise condition one needs to Absolutele continuity is the precise condition one needs to
impose in order for the fundamental theorem of calculus impose in order for the fundamental theorem of calculus
should hold for the Lebesgue integral. should hold for the Lebesgue integral.
\PMlinkescapeword{absolutely continuous} \PMlinkescapeword{absolutely continuous}
\PMlinkescapeword{property} \PMlinkescapeword{property}
{\bf Definition} {\bf Definition}
Suppose $[a,b]$ be a closed bounded interval of $\R$. Suppose $[a,b]$ be a closed bounded interval of $\R$.
Then a function $f\colon [a,b]\to\C$ is Then a function $f\colon [a,b]\to\C$ is
{\bf absolutely continuous} on $[a,b]$, {\bf absolutely continuous} on $[a,b]$,
if for any $\varepsilon>0$, there is a $\delta>0$ such that the following if for any $\varepsilon>0$, there is a $\delta>0$ such that the following
condition holds: condition holds:
\begin{itemize} \begin{itemize}
\item[($\ast$)] If $(a_1,b_1), \ldots, (a_n,b_n)$ is a finite \item[($\ast$)] If $(a_1,b_1), \ldots, (a_n,b_n)$ is a finite
collection of disjoint open intervals in $[a,b]$ collection of disjoint open intervals in $[a,b]$
such that such that
$$ $$
\sum_{i=1}^n (b_i-a_i)< \delta, \sum_{i=1}^n (b_i-a_i)< \delta,
$$ $$
then then
$$ $$
\sum_{i=1}^n |f(b_i)-f(a_i)|< \varepsilon. \sum_{i=1}^n |f(b_i)-f(a_i)|< \varepsilon.
$$ $$
\end{itemize} \end{itemize}
\begin{thm}[Fundamental theorem of calculus for the Lebesgue integral \begin{thm}[Fundamental theorem of calculus for the Lebesgue integral
\cite{jones,aliprantis}] \cite{jones,aliprantis}]
Let $f\colon [a,b] \to \C$ be a Let $f\colon [a,b] \to \C$ be a
function. Then $f$ is absolutely continuous if and only if function. Then $f$ is absolutely continuous if and only if
there is a function $g\in L^1(a,b)$ (i.e. a $g\colon(a,b)\to \C$ with there is a function $g\in L^1(a,b)$ (i.e. a $g\colon(a,b)\to \C$ with
$\int_{(a,b)} |g|< \infty$), such that $\int_{(a,b)} |g|< \infty$), such that
$$ $$
f(x) = f(a) + \int_a^x g(t) dt f(x) = f(a) + \int_a^x g(t) dt
$$ $$
for all $x\in[a,b]$. for all $x\in[a,b]$.
What is more, if $f$ and $g$ are as above, then $f$ is differentiable What is more, if $f$ and $g$ are as above, then $f$ is differentiable
almost everywhere and $f'=g$ almost everywhere and $f'=g$
almost everywhere. (Above, both integrals are Lebesgue integrals.) almost everywhere. (Above, both integrals are Lebesgue integrals.)
\end{thm} \end{thm}
See also \cite{wikiabs}. See also \cite{wikiabs}.
\begin{thebibliography}{9} \begin{thebibliography}{9}
\bibitem{wikiabs} Wikipedia, entry on \bibitem{wikiabs} Wikipedia, entry on
\PMlinkexternal{Absolute continuity}{http://en.wikipedia.org/wiki/Absolute_continuity}. \PMlinkexternal{Absolute continuity}{http://en.wikipedia.org/wiki/Absolute_continuity}.
\bibitem{jones} \bibitem{jones}
F. Jones, \emph{Lebesgue Integration on Euclidean Spaces}, F. Jones, \emph{Lebesgue Integration on Euclidean Spaces},
Jones and Barlett Publishers, 1993. Jones and Barlett Publishers, 1993.
\bibitem{aliprantis} \bibitem{aliprantis}
C.D. Aliprantis, O. Burkinshaw, \emph{Principles of Real Analysis}, C.D. Aliprantis, O. Burkinshaw, \emph{Principles of Real Analysis},
2nd ed., Academic Press, 1990. 2nd ed., Academic Press, 1990.
\end{thebibliography} \end{thebibliography}