Consider the increasing, right-continuous function $$ F(x) = \begin{cases} 0 &x < 0 \\ 1+x &x \geq 0 \end{cases} $$
and let $\nu = \nu_F$ be the associated Borel measure on $\mathbb R$ (so $\nu((a,b]) = F(b)-F(a)$). Find the Lebesgue decomposition of $\nu$ with respect to:
(a) $m$
(b) $\delta$, the Dirac measure at 0
(c) the Cantor measure $\mu$
I think I have a solution to (a), but I am lost as to how to do parts (b) and (c). I'd appreciate any help.
This is my attempt for (a):
We want to find measures $\rho$ and $\lambda$ such that $$\nu = \rho + \lambda$$ with $\rho << m$ and $\lambda \perp m$. By the Radon-Nikodym Theorem, $\exists F'$ such that $$\rho(A) = \int_A F'(x) dx. $$
Since $F(x)$ is increasing then for some interval $(a,b] \in \mathcal B$, $$\rho((a,b]) = \int_a^b F'(x) dx \leq F(b)-F(a) = \nu((a,b]). $$
Clearly, $\rho << m $. Next define $$\lambda = \nu-\rho.$$ To show that $\lambda \perp m$, let $E = \{0\}$ and $F = \mathbb R - \{0\}.$ We want to show that $m(E)=\lambda(F)=0$. It is trivial that $m(E) = 0$. To show $\lambda(F) =0$, notice $\lambda \geq 0$ since $\rho(A) \leq \nu(A) \ \ \forall A \in \mathcal B.$ Hence on $(-\infty,0):$ $$ \lambda((-\infty,0))=\nu(-\infty,0)-\rho(-\infty,0)\geq 0 $$ But since $\nu(-\infty,0)=0$, then $\rho((-\infty,0)=0 \implies \lambda((-\infty,0) = 0$. Finally on $(0, \infty),$ note that $\nu = \rho$. Hence $$ \lambda((0,\infty)) = \nu((0,\infty))-\rho((0,\infty))=0$$