Lebesgue decomposition from a sum-measure density ($B=\{h=1\},f=h/(1-h)\text{ on }{B}^c$)

Use the Hilbert-space construction of dominated measure densities with $\rho =\mu +\nu$. The set $B=\{h=1\}$ is a null set for $\mu$. Put $f=0$ on $B$ and $f=h/(1-h)$ elsewhere. Then $\int fd\mu =\nu (B^c)<\infty$ and $\nu (A)={\int }_{A}fd\mu +\nu (A\cap B)$. This is the Lebesgue decomposition into a density part and a part concentrated on a $\mu$-null set.