\circ F_{\mu}^{-1})(y).\end{aligned}$$ Therefore, $a=f ( \check{F}^{-1}_{\mu}(0) )$ and $b=\nabla_{\mu}f ( \check{F}^{-1}_{\mu}(0) )$. This shows $$\begin{aligned}
    f \circ \check{F}_{\mu}^{-1} \in \mathcal{D}^*(\Delta_{\nu \circ F_{\mu}^{-1}, \, \Lambda})
    \quad \text{and} \quad
    \Delta_{\nu \circ {F}_{\mu}^{-1}, \, \Lambda} (f \circ \check{F}_{\mu}^{-1}) = \Delta_{\nu, \, \mu}f\circ \check{F}^{-1}_{\mu}.
    \end{aligned}$$ If $f \circ \check{F}_{\mu}^{-1} \in \mathcal{D}^*_{\alpha,\beta}(\Delta_{\nu \circ F_{\mu}^{-1}, \, \Lambda})$, then, as above, for $x \in {\operatorname{supp}}(\mu)$ with $F_{\mu}(x-\varepsilon)<F_{\mu}(x)$ for all $\varepsilon>0$, $$\begin{aligned}
    f( x )
    =f(\check{F}_{\mu}^{-1}( F_{\mu}( x)) )
    &= c+dF_{\mu}(x) + \int \mathds{1}_{[0, F_{\mu}(x)]}(y)(F_{\mu}(x)-y ) \Delta_{\nu \circ F_{\mu}^{-1}, \Lambda}(f   \circ \check{F}_{\mu}^{-1})(y) \;{\mathrm{d}}(\nu \circ F_{\mu}^{-1})(y)\\
    &=
c+dF_{\mu}(x)+
\int \mathds{1}_{[0, F_{\mu}(x)]}(F_{\mu}(y))(F_{\mu}(x)-F_{\mu}(y) ) \Delta_{\nu \circ F_{\mu}^{-1}, \Lambda}
(f   \circ \check{F}_{\mu}^{-1})
 \circ F_{\mu}(y)\;{\mathrm{d}}\nu (y)  \\ &=
c+dF_{\mu}(x)+
\int \mathds{1}_{[0, x]}(y)(F_{\mu}(x)-F_{\mu}(y) ) \Delta_{\nu \circ F_{\mu}^{-1}, \Lambda}
(f   \circ \check{F}_{\mu}^{-1})
 \circ F_{\mu}(y)\;{\mathrm{d}}\nu (y),\end{aligned}$$ with $c=f (\check{F}^{-1}_{\mu}(0 ))$ and $d =\nabla f (\check{F}^{-1}_{\mu} (0 ) )$. The case $x  \in {\operatorname{supp}}(\mu)$ and $F_{\mu}(x-\varepsilon)=F_{\mu}(x)$ for some $\varepsilon>0$ implies that $x$ lies in an interval of constancy. Notice, $$\begin{aligned}
    \check{F}_{\mu}^{-1}( [0,1])=\{0\} \cup {\operatorname{supp}}(\mu) \cap \{ x \in [0,1] \text{ is not a right endpoint of an interval of constancy}\},       \end{aligned}$$ thus we can consider a modification of $f$ such that $f$ is constant on each interval of constancy of $F_{\mu}$. Thus, we have $c=f(0), \ d=\nabla_{\mu}f(0)$. 