Borel Measures: Lebesgue Decomposition

Question

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Context

Given a measure space $\Omega$ with sigma-finite measure $\lambda$.

Consider a finite measure $\kappa<\infty$.

Then there exists a decomposition: $$\kappa=\kappa_{ac}+\kappa_s\quad(\kappa_{ac}\ll\lambda,\kappa_s\perp\lambda)$$

Moreover the singular measure splits into: $$\kappa_s=\kappa_{sc}+\kappa_{pp}$$

Question

Is the last decomposition w.r.t. singletons or w.r.t. atoms?
(See the wikipedia articles Discrete Measure and Atom.)

Moreover, can it make a really important difference?
(I guess so but cannot imagine what...)

Remark

The absolutely continuous measures are characterized precisely by the measurable functions: $$\kappa_{ac}(E)=\int_Ehd\lambda\quad(h\geq0)$$

$\mu(\Omega\setminus { \omega \in \Omega : \mu{\omega}\ne 0 })=0$. Discrete doesn't make a lot of sense unless you consider the measure of a singleton set. — Disintegrating By Parts, Oct 14 '14 at 21:03

score 0 · Accepted Answer · edited Apr 13 '17 at 12:19

0

It is really into discrete and continuous!
(See thread on Atomic vs. Discrete and Atomic Decomposition.)

The reason is that one is tempted to compare the point spectrum to the eigenspaces and that happens to come out fine only if one restricts to measures concentrated on countable sets.

edited Apr 13 '17 at 12:19

Community

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answered Nov 06 '14 at 20:23

freishahiri

17,045

Borel Measures: Lebesgue Decomposition

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