In my lecture on measure theory, we wanted to derive a refined decomposition of the Lebesgue decomposition.
Let $\nu$ be a locally finite measure on $(\mathbb R, \mathcal B(\mathbb R))$. I already know that the Lebesgue decomposition $\nu = \nu_a + \nu_s$ wrt. to the Lebesgue measure $\lambda^1$ exists and is unique, where $\nu_a \ll \lambda^1$ is absolute continuous and $\nu_s \perp \lambda^1$ is singular.
Furthermore, I know of the decomposition $\nu = \nu_c + \nu_d$, where $\nu_d$ is a discrete measure and $\nu_c$ has a continous distribution function (meaning $\nu_c(\{x\}) = 0$ for all $x \in \mathbb R$, i.e. $\nu_c$ has no atoms). This decomposition is also unique.
Claim: $\nu_d \le \nu_s$ and $\nu_c \ge \nu_a$.
How can I derive these relations? Preferably from first principles, i.e. only using general properties of measures or distribution functions.
We needed this to further prove $\nu_a = (\nu_c)_a$ and $\nu_d = (\nu_s)_d$ and thus $\nu = \nu_a + \nu_{sc} + \nu_d$ where $\nu_{sc} = (\nu_s)_c = (\nu_c)_s$ is the singular continuous part.