By an indefinite integral, I don't mean a primitive (even though the concepts are essentially equivalent, an equivalence easily proven in the 'classical' case) because the equivalence in a more general case is part of what is to be proven in my context. An indefinite integral of a (Lebesgue) integrable function $f:[a,b]\rightarrow\mathbf R$ is by def. any function of the form $x \mapsto \int_a^x f(t)dt + C$ where C is a real constant (I put this constant instead of allowing other lower bounds of the integral to avoid complications due to missing values). I saw that some people define in an unusual manner "absolutely continous" by being such an integral function; of course I don't want my Q. to become trivial in this way. $F:[a,b]\rightarrow\mathbf R$ is absolutely continuous by def. when: for every $\epsilon > 0$ there is a $\delta > 0$ so that for all numbers $a_i, b_i$ with $i = 1, ..., n$ (where $n$ is any integer $> 0$) such that $a \le a_i < b_i \le b$ and $\sum_{i=1}^n(b_i - a_i) < \delta$, the open intervals $]a_i,b_i[$ being disjoint, one has $\sum_{i=1}^n|f(b_i) - f(a_i)| < \epsilon$.
So how does one prove that any increasing such function $F$ is an indefinite integral of an $L^1$ positive real function $f$. (The assumptions increasing / positive - to be understood in non-strict way - are there to simplify matters. Generalizing to the same theorem without them, and treating the converse should be easy, I think.)
Motivation and received hint: this is essentially an exercise in Halmos' Measure Theory but he considers functions defined on $\mathbf R$ while I need the case of a compact interval. He hints (without using the expression) to the (Lebesgue-) Stieltjes measure / integral defined by $F$. To obtain $f$, I realize that $f$ has to be some kind of derivative of $F$, the existence and $L^1$ - property of which must be proven and then that it satisfies the condition with its indefinite integral.
I need a rather simple proof without a lot of measure theory because I don't use that of Halmos & Co (but Bourbaki's) - so no Radon-Nikodym and similar things ...
