Does the map $f \mapsto \mu_f$ (BV to Lebesgue-Stieltjes measure) behave nicely under function concatenation? Consider two continuous functions $f,g : [0,1]\rightarrow\mathbb{R}$ of bounded variation, and let $\mu_f, \mu_g : \mathcal{B}([0,1])\rightarrow\mathbb{R}$ be their associated Lebesgue-Stieltjes (signed) measures.
Let further $h:= f\star g : [0,1]\rightarrow\mathbb{R}$ be the [$x=1/2$ centered] concatenation of $f$ and $g$ (that is: $h(x) = f(2x)$ for $x\in[0,1/2]$, and $h(x)= g(2x-1) + (f(1)-g(0))$ for $x\in(1/2,1]$).
Is there an algebraic (or otherwise explicit or insightful) relation between the measures $\mu_h$ and $\mu_f, \mu_g$?
 A: Not an answer, but an informed guess that is too long for a comment.
Let $A: [0, \frac{1}{2}] \to  [0, 1]$ be the function $x \to  2x$ and $B: (\frac{1}{2}, 1] \to (0, 1]$ the function $x \to 2x - 1$.
I would guess that that $\mu_h$ is given by
$$\mu_h (E) = \mu_f \left(A^{-1}\big (E \cap [0, \frac{1}{2}] \big )\right) + \mu_g \left(B^{-1}\big (E \cap (\frac{1}{2}, 1] \big )\right)$$
More might be able to be said.
Let $\mu_f = \mu^{ac}_f + \mu^{s}_f$ be the Lebesgue decomposition of $\mu_f$ into absolutely continuous and singular part, and similarly for $\mu_g, \mu_h$.
We should then further have that the Lebesgue decomposition of $\mu_h$ is given by
$$\mu^{ac}_h (E) =\mu^{ac}_f \left(A^{-1}\big (E \cap [0, \frac{1}{2}] \big )\right) +  \mu^{ac}_g \left(B^{-1}\big (E \cap (\frac{1}{2}, 1] \big )\right)$$
$$\mu^{s}_h (E) = \mu^{s}_f \left(A^{-1}\big (E \cap [0, \frac{1}{2}] \big )\right) + \mu^{s}_g \left(B^{-1}\big (E \cap (\frac{1}{2}, 1] \big )\right).$$
Below follows some heuristics as to why I think the guess is correct.
The Lebesgue Stiltjes measure is such that it depends only on the increments of the inducing function - thus the Lebesgue Stiltjes measure associated to $h$ should depend only on $f$ on $[0, \frac{1}{2}]$, and only on $g$ on $(\frac{1}{2}, 1]$. Since on each of these intervals is just a scaled version of the orginal functions, we expect the Lebesgue Stiltjes measure to be merely scaled as well.
Since $g$ is assumed continuous, the slight asymmetry at the point $\frac{1}{2}$, i.e. not including $g(0)$ in the definition of $h$ should not matter much, apart from possibly introducing an atom at $\frac{1}{2}$, but the way you defined concatenation means that there is no jump corresponding to the difference between $f(1)$ and $g(0)$.
Some ideas to prove this rigorously - the first claim may possibly be directly verified by examining the increments of $h$.
The decomposition may be provable by looking at the Lebesgue decomposition of each of $f$ and $g$, recalling that the absolutely continuous part of the Lebesgue Stiltjes measure is given by the integral of the (a.e. defined) derivative of the inducing function, and then noting that concatenation doubles the derivative.
A: $\newcommand{\B}{\mathcal B}\renewcommand{\S}{\mathcal S}$Note that for any $a$ and $b$ such that $0\le a\le b\le1$
\begin{equation}
    \mu_h((a,b])=h(b)-h(a)=
    \left\{
    \begin{alignedat}{2}
&   \mu_f(2(a,b])&&\text{ if }(a,b]\subseteq[0,1/2], \\ 
&   \mu_g(2(a,b]-1)&&\text{ if }(a,b]\subseteq(1/2,1],
    \end{alignedat}
    \right.
\end{equation}
where $2A:=\{2x\colon x\in A\}$ and $2A-1:=\{2x-1\colon x\in A\}$ for any $A\subseteq\mathbb R$.
So, for any left-open subinterval $(a,b]$ of the interval $[0,1]$,
\begin{equation}
\begin{aligned}
        \mu_h((a,b])&=  \mu_h((a,b]\cap[0,1/2])+\mu_h((a,b]\cap(1/2,1]) \\ 
        &=\tilde\mu_h((a,b]) \\ 
        &:=\mu_f(2((a,b]\cap[0,1/2]))+\mu_g(2((a,b]\cap(1/2,1])-1).  
\end{aligned}       
\end{equation}
The function $\tilde\mu_h$ is a finite signed measure on the semiring (say $\S$) of all left-open subintervals of the interval $[0,1]$, and $\sigma(\S)=\B([0,1])$. So, by the uniqueness of the measure extension (cf. e.g. Proposition 13 of Kisil - Introduction to functional analysis),
\begin{equation}
\begin{aligned}
\mu_h(B)&=\mu_f(2(B\cap[0,1/2]))+\mu_g(2(B\cap(1/2,1])-1)   \\ 
&=\mu_f(2B\cap[0,1])+\mu_g((2B-1)\cap(0,1])  
\end{aligned}
\end{equation}
for all $B\in\B([0,1])$.
