Lebesgue decomposition $\nu =\nu_{cont}+\require{cancel} \cancel{\nu _{sing}}+\nu_{pp}$

Question

While reading about the Lebesgue decomposition theorem I learned that we can decompose a measure (with respect to the Lebesgue measure I guess) $$\,\nu =\nu _{{{\mathrm {cont}}}}+\nu _{{{\mathrm {sing}}}}+\nu _{{{\mathrm {pp}}}}$$ where

• $ν_{cont}$ is the absolutely continuous part ($\nu_{cont}<<\lambda$)
• $ν_{sing}$ is the singular continuous part ("singular" meaning $\nu\perp\lambda$ and "continuous" meaning $\mu\{x\}=0$ for any $x\in\mathbb{R}^d$ (I am not sure about this one))
• $ν_{pp}$ is the pure point part (a discrete measure) (meaning $\nu_{pp}=\sum\limits_{i=1}^{\infty}\delta_{x_i}$)

An example of singular continuous measure is the Cantor measure (the probability measure on the real line whose cumulative distribution function is the Cantor function).

As I understand, singular continuous measures are pathological and I was wondering what condition we could impose on the original measure in order not to have any singular continuous terms in the decomposition. I think that asking that the original measure is Radon is not enough right?

PS : A relevant discussion here

Answer 1

The Radon-Nikodym theorem completely characterizes the measures $\nu \ll \lambda$ which are absolutely continuous w/r/t Lebesgue measure as being those of the form $$\nu(A) := \int_A f \, d\lambda $$ for some nonnegative Lebesgue measurable function $f$, and all Lebesgue measurable sets $A$.
The $f$ inside the integral should be thought of as a density function; it measures how much $\nu$ bunches up or thins out measure compared to Lebesgue measure $\lambda$.
Because $f$ quantifies the local "concentration" of measure at a point, it is usually called the Radon-Nikodym derivative and often denoted $d\nu/d\lambda$.
In fact, by the Lebesgue differentiation theorem, if you have a measure $\nu \ll \lambda$, then $\lambda$-almost everywhere, $$\lim_{h \to 0} \frac{\nu(B_h(x))}{\lambda(B_h(x))} = f(x),$$ where $B_h(x)$ denotes the open ball of radius $h$ centered at $x$ and $f = d\nu/d\lambda$ is the aforementioned Radon-Nikodym derivative.

So to answer the original question, a measure with a decomposition w/r/t Lebesgue measure $\nu = \nu_{cont} + \nu_{sing} + \nu_{pp}$ has $\nu_{sing} = 0$ if and only if the original measure minus its discrete part $\nu - \nu_{pp}$ is a continuous measure with a density function, and furthermore, $\nu - \nu_{pp}$ can be differentiated w/r/t Lebesgue measure to recover its density function at almost every point.