Let $g: \mathbb R \to \mathbb R$ be a function of locally bounded variation, and $f$ a locally integrable function with respect to $dg$, the Lebesgue–Stieltjes measure associated with $g$.


Let $\eta$ be a smooth, compactly supported function. Define


$$F(x) := \int\_0^x f(s) \, dg(s).$$


**Question:** Is it true that


$$(F \ast \eta)(x) = \int\_0^x f \ast \eta (s) \, dg(s)?$$


*Remark: This is true for ordinary integrals, i.e. when $g$ is the identity.*



Does convolution commute with Lebesgue–Stieltjes integration?
If (say) $g$ is absolutely continuous (with an almost-everywhere derivative $g'$), then the left-hand side of your identity is
$$L(\eta):=(F\*\eta)(x)=\int dy\,\eta(x-y)\int\_0^y ds\,g'(s) f(s)$$
and its right-hand side is
$$R(\eta):=\int\_0^x dg(s)\,(f\*\eta)(s)=\int\_0^x ds\,g'(s)\,\int dt\,f(t)\eta(s-t).$$
We see that the identity cannot hold in general, because the arguments of $g'$ and $f$ are the same on the left but different on the right.




---


More specifically, suppose that for all natural $n$ we have $L(\eta\_n)=R(\eta\_n)$, where $f(s)=g'(s)=e^{-s^2}\,1(s>0)$ for real $s$, and the $\eta\_n$'s are smooth compactly supported positive functions such that $\eta\_n\to1$ monotonically pointwise. Then for all real $x>0$ we have $\infty=L(1)=R(1)<\infty$, a contradiction. $\quad\Box$


