Absolute continuity is the precise condition one needs to impose in order for the fundamental theorem of calculus to hold for the Lebesgue integral.

Definition Suppose $[a,b]$ be a closed bounded interval of . Then a function is absolutely continuous on $[a,b]$ , if for any $\varepsilon>0$ , there is a $\delta>0$ such that the following condition holds:

($\ast$ )

If $(a_1,b_1), \ldots, (a_n,b_n)$ is a finite collection 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 (Fundamental theorem of calculus for the Lebesgue integral)   Let be a function. Then $f$ is absolutely continuous if and only if there is a function $g\in L^1(a,b)$ (i.e. a 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 differentiable almost everywhere and $f'=g$ almost everywhere. (Above, both integrals are Lebesgue integrals.)

See [2,3] for proof.

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

Bibliography

1

Wikipedia, entry on Absolute 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 of Calculus for Lebesgue Integral, Divulgaciones Matemáticas, Vol. 8, No. 1, 2000, pp. 75-85.