absolutely continuous function

is the precise condition one needs toimpose in order for the fundamental theorem of calculusto hold for the Lebesgue integral.

DefinitionSuppose $[a,b]$ be a closed bounded interval of $\\mathbbmss{R}$ .Then a function $f\\colon[a,b]\\to\\mathbbmss{C}$ isabsolutely continuous on $[a,b]$ ,if for any $\\varepsilon>0$ , there is a $\\delta>0$ such that the followingcondition holds:

( $\\ast$ )
If $(a_{1},b_{1}),\\ldots,(a_{n},b_{n})$ is a finitecollection of disjoint open intervals in $[a,b]$ such that
$\\sum_{i=1}^{n}(b_{i}-a_{i})<\\delta,$
then
$\\sum_{i=1}^{n}|f(b_{i})-f(a_{i})|<\\varepsilon.$

Theorem 1 ().

Let $f\\colon[a,b]\\to\\mathbbmss{C}$ be afunction. Then $f$ is absolutely continuous if and only ifthere is a function $g\\in L^{1}(a,b)$ (i.e. a $g\\colon(a,b)\\to\\mathbbmss{C}$ with $\\displaystyle\\int_{a}^{b}|g|<\\infty$ ), such that

f(x)=f(a)+\\int_{a}^{x}g(t)dt

for all $x\\in[a,b]$ .What is more, if $f$ and $g$ are as above, then $f$ is differentiablealmost everywhere and $f^{\\prime}=g$ almost everywhere. (Above, both integrals are Lebesgue integrals.)

See [2, 3] for proof.

See also [1], and [4] for a discussionabout different proofs.

References

1 Wikipedia, entry onhttp://en.wikipedia.org/wiki/Absolute_continuityAbsolute continuity.
2 F. Jones, Lebesgue Integration on Euclidean Spaces,Jones and Barlett Publishers, 1993.
3 C.D. Aliprantis, O. Burkinshaw, Principles of Real Analysis,2nd ed., Academic Press, 1990.
4 D. B’arcenas,The Fundamental Theorem ofCalculus for Lebesgue Integral,Divulgaciones Matemáticas, Vol. 8, No. 1, 2000, pp. 75-85.