# absolutely continuous function

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 $\mathbbmss{R}$. Then a function $f\colon[a,b]\to\mathbbmss{C}$ is 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 ().

Let $f\colon[a,b]\to\mathbbmss{C}$ 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 $g\colon(a,b)\to\mathbbmss{C}$ 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 differentiable almost everywhere and $f^{\prime}=g$ almost everywhere. (Above, both integrals are Lebesgue integrals.)

See [2, 3] for proof.