we have $$\int (\tan(\alpha)+x) \,{\mathrm{d}}\phi(x)=\tan(\alpha)\left( \phi(1)-\phi(0)\right)+\phi(1)-\int  \phi(s) \,{\mathrm{d}}s\eqqcolon B_{\phi}.$$ Combining these identities gives $$\begin{aligned}
\int  \left(E \cdot B_{\phi} +s \cdot  A_{\alpha,\beta} \cdot B_{\phi} -\phi(1)(1-s)+ \int_{[s,1]}\phi(s) \,{\mathrm{d}}s\right)g(s) \,{\mathrm{d}}\nu(s)=0\end{aligned}$$ If we consider $g(s):=E \cdot B_{\phi} +s \cdot  A_{\alpha,\beta} \cdot B_{\phi} -\phi(1)(1-s)+ \int_{[s,1]}\phi(x) \,{\mathrm{d}}x \in \mathcal{C}_{\nu,\Lambda}^{\gamma}$ it follows, that for all $s \in [0,1]$ $$\int_{[s,1]}\phi(x) \,{\mathrm{d}}x=\phi(1)-E \cdot B_{\phi} -s(   A_{\alpha,\beta} \cdot B_{\phi}+\phi(1)),$$

which is only possible if $\phi(s)=A_{\alpha,\beta} B_{\phi}+\phi(1)$, for all $s \in [0,1]$. Therefore $\Phi$ is a Dirac-measure in $\{0\}$ with weight $\phi(0)$ and we have by , or all $f \in \mathcal{C}_{\nu,\Lambda}^{\gamma}$, $$\phi(0) \ \tan(\alpha) \ \int\left(E+A_{\alpha,\beta} \cdot s\right)f(s) \,{\mathrm{d}}\nu(s)=0$$ For the particular choice $f \in \mathcal{C}_{\nu,\Lambda}^{\gamma}$ given by $f:s\mapsto E+A_{\alpha,\beta} \cdot s$ the above integral is positive and hence $
\phi(0)=0$. Consequently, $\Phi=0$ and we have shown that the annihilator of $  \mathcal{D}^s_{\gamma}(\Delta_{\nu, \Lambda}
)$ is trivial and therefore $  \mathcal{D}^s_{\gamma}(\Delta_{\nu, \, \Lambda})$ is dense in $ \mathcal{C}_{\nu,\Lambda}^{\gamma}$.

Generalized Kreĭn-Feller operators and transformations of measure spaces
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Let us first state two basic key observations. three key observation.

\[identity\] The function $\check{F}^{-1}_{\mu} \circ F_{\mu}$ equals the identity $\nu$-almost everywhere.

Note, $\check{F}^{-1}_{\mu}( F_{\mu}(x) )\neq x$ if and only if there exists $\varepsilon>0$ with $F_{\mu}(x-\varepsilon)=F_{\mu}(x)$. This means, if $\check{F}^{-1}_{\mu}( F_{\mu}(x) )\neq x$, 