Measure decomposition 
*

*By Lebesgue's decomposition theorem we have for $\sigma$-finite measures $\nu$ and $\mu$ that


$\nu=\nu_{sing}+\nu_{ac}$,
where $\nu_{ac}<<\mu$ and $\nu_{sing}\perp\mu$. Now, (under certain assumptions) I can decompose the singular part into a pure points $\nu_{pp}$ and a continuous $\nu_{sc}$ part (continuous means that $\nu_{sc}(\{x\})=0$ for all $x\in X$). Therefore I get together
$\nu=\nu_{sc}+\nu_{pp}+\nu_{ac}$
Do we have $\nu_{sc}\perp \mu$ and $\nu_{pp}\perp\mu$?


*More generally, can I conclude from $\nu \perp \mu$ and $\nu=\nu_1+\nu_2$ that $\nu_1\perp \mu$ and $\nu_2\perp\mu$?

 A: You have that $\nu \perp \mu$ if there exists $E \in \mathcal B$ (the underlying $\sigma$-algebra) such that 
$$
|\nu|(E) = 0 \text{ and } |\mu|(E^c) = 0,
$$
where $|\nu|$, is the total variation of $\nu$.
If I am not mistaken, then you want to set $v_{pp}(A) := \nu_{sing}(A \cap P)$, where 
$$
P:= \{ x :  \nu_{sing}(x) \neq 0 \}.
$$
Notice that this definition requires that $\{x\}$ is in $\mathcal{B}$ for all $x$ in the underlying space.
Now since $\nu_{ac} \perp \nu_{sing}$ there is $\tilde{E}$ such that 
$$
|\nu_{ac}|(\tilde{E}^c) = 0 \text{ and }|\nu_{sing}|(\tilde{E}) = 0.
$$
Now, obviously for $P$ as above we know that $P\cap \tilde{E} = \emptyset$, as otherwise it would be easy to show that $|\nu_{sing}|(\tilde{E}) > 0$.
Hence $|\nu_{pp}|(\tilde{E}) = |\nu_{sing}|(\tilde{E} \cap P)= |\nu_{sing}|(\emptyset) = 0$ and consequently $\nu_{pp} \perp \nu_{ac}$. As a consequence $\nu_{sc}(\tilde{E}) = \nu_{sing}(\tilde{E}) - \nu_{pp}(\tilde{E}) = 0$. Thus, $\nu_{sc}\perp \nu_{ac}$. 
The second question can be answered negatively by setting $\mu := \nu_1 := -\nu_2$.
