Let $f:[a,b]\to\mathbb R$ be a Henstock-Kurzweil-integrable function (short: HK-integrable). 



> 
> Can $f$ always be written as a sum of a Lebesgue-integrable function and a function which has a classical primitive, i.e. are there $f\_1\in L^1([a,b])$ and an everywhere differentiable function $F:[a,b]\to\mathbb R$ such that $f=f\_1+F'$?
> 
> 
> 


Since every integral function of HK-integrable functions is $ACG\_\*$, the following question is equivalent:



> 
> Can every $ACG\_\*$-function be written as a sum of an absolutely continuous function and an everywhere differentiable function?
> 
> 
> 


For me it is sufficient to (dis)prove this "only" for HK-integrable functions $f$ which are in $L^1([c,b])$ for each $c\in(a,b)$.


Since HK-integration is brand new to me, I don't really have a feeling whether this is true or not. By reading a couple of books I've gathered a bunch of nice properties of HK-integrable functions, but none of them helped. I would be very grateful if anyone who is more familiar with this type of integration theory can at least have a look at this problem. Thank you very much in advance! 



Decomposition of Henstock-Kurzweil-integrable functions
The answer is negative. Consider the space $\Delta'=\Delta'([a,b])$ of functions, which are classical derivatives of functions $[a,b]\to\mathbb R$. Assume the claim was true. Then any HK-integrable $f:[a,b]\to\mathbb R$ had a representation $f=f\_1+f\_2$, where $f\_1\in L^1([a,b])$ and $f\_2\in\Delta'$. Let $g$ be a multiplier for $\Delta'$, i.e. a function $[a,b]\to\mathbb R$ with the property that $hg\in\Delta'$ whenever $h\in\Delta'$. Then it is easy to see that $g$ is bounded and hence belongs to $L^\infty([a,b])$. Then $f\_1g\in L^1([a,b])$ and hence $f\_1g$ is HK-integrable, and $f\_2g\in\Delta'$ and thus $f\_2g$ is HK-integrable. But then $fg$ where HK-integrable. Since $f$ was arbitrary, $fg$ is HK-integrable whenever $f$ is HK-integrable. But then it is well-known that $g$ must agree almost everywhere with a function of bounded variation, and since $g$ has a primitive, it can be shown that $g$ must be continuous and of bounded variation. This would imply that every multiplier for $\Delta'$ is continuous and of bounded variation, which is not true in general. There are examples of such multipliers which are neither.


