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.
Best Answer
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. $\quad\Box$