only if the support of $\nu$ is a finite set.

\[densecm\] For $\gamma=(\alpha,\beta )\in [0,{\pi/2}]^{2} $, set $$\begin{aligned}
    C_{\nu,\Lambda}^{\gamma} 
    \coloneqq \begin{cases}
    \left\{ f \in C_{\nu,\Lambda}  \colon f(0)=0, \ f(1)=0\right\}& \text{ if } \alpha=\beta=0, \\
    \left\{ f \in C_{\nu,\Lambda}  \colon f(0)=0\right\}& \text{ if }  \alpha=0, \ \beta \in (0,\pi/2],\\
    \left\{ f \in C_{\nu,\Lambda}  \colon f(1)=0\right\} &\text{ if } \alpha \in (0,\pi/2], \ \beta=0, \\
     C_{\nu,\Lambda
     }& \text{ if } \alpha,\beta \in (0,\pi/2].
    \end{cases}
    \end{aligned}$$ The set $\mathcal{D}^s_{\gamma}(\Delta_{\nu, \, \Lambda})$ is dense in $\left(C_{\nu,\Lambda}^{\gamma}, \lVert \cdot \rVert_{\infty}\right)$.

The result can be found in [@MR0314125 Behauptung 2.4] without a detailed proof. We will sketch a proof of this result for the case $\gamma=(\alpha,\beta) \in (0,\pi/2) \times [0,\pi/2)$; the other cases follow in a similar fashion. Fix $\Phi \in C_{\mu,\Lambda}' $ such that $$\forall f \in \mathcal{D}^s_{\gamma}(\Delta_{\nu, \, \Lambda}): \
\Phi(f)=\int f(x) \,{\mathrm{d}}\phi(x)=0,$$ where $\phi$ is the signed distribution function representing $\Phi$. By definition $\phi$ is of bounded variation and local constant on the complement of ${\operatorname{supp}}(\nu)$. With $E:=-(1+\tan(\beta))A_{\alpha,\beta} $ it follows from the proof of Lemma \[Surjec\] for all $g \in C_{\nu,\Lambda}^{\gamma}$ $$\begin{aligned}
\int  (E+s\cdot A_{\alpha,\beta})g(s)\,{\mathrm{d}}\nu(s) \cdot \int (\tan(\alpha)+x) \,{\mathrm{d}}\phi(x)&= \int \int  (x-s)g(s) \,{\mathrm{d}}\nu (s)\,{\mathrm{d}}\phi(x) \label{eq:dual1}\\
&= \int  \left(\phi(1)(1-s)- \int_{[s,1]}\phi(s) \,{\mathrm{d}}s \right)g(s)\,{\mathrm{d}}\nu(s).\nonumber\end{aligned}$$ Furher, 