Frist:- I am not sure about what title this question should be.
Suppose the function $f:[0,\frac{1}{2}]\rightarrow \mathbb{R}$ defined by
$$ f(x) = \begin{cases} 0, & \text{if }x=0 \\ x \cos(\frac{\pi}{x}), & \text{if } x\in(0,\frac{1}{2}] \end{cases} $$ I want to prove that there is no Lebesuge integrable function $g$ such that :- $$f(x)=\int_{[0,x]} g \, dm$$
this is the lemma form my note in which I want to use it
Let $f\in L^{1}[a,b]$. Define a new functions $g:[a,b]\rightarrow \mathbb{R}. g(x)=\int_{a}^{x}f(y)\, dy$. Then $g\in AC$.
My attempt:- I try to show that f is not absolutely continuous since in case $f$ is absolutely continuous the function g will exist ( there is a theorem I am not sure if they have name or not ). So I was able to show that f is not Bounded Variation, but this is not enough since family of BV is larger than family AC.
It basically oscillates too fast. We can identify points where the function is large by looking at the extrema of the $\cos(\pi/x)$ part. Specifically, consider a finite sequence $1/2,1/3,1/4,\dots,1/n$, the function values there are $1/2$, $-1/3$, $1/4$, $-1/5$, ..., $(-1)^n/n$. So for $n>2$, the total variation on $[1/n,1/2]$ is at least $\sum_{k=3}^n \left ( \frac{1}{k} + \frac{1}{k-1} \right )$, which diverges as $n \to \infty$. Hence $f$ is not BV, which implies it cannot be AC either.