1
$\begingroup$

Disclaimer: The question here has been solved, now:

(For jeapardy it is stated below, anyway. Have fun! ;) )

Summary: This is a summary of the discussions:

Reference: Further results are accessible from:


Definitions

Call a measure discrete if it is discretely concentrated: $$\mu(\Omega\setminus\Delta)=0\quad(\#\Delta\leq\aleph_0)$$ and continuous if any discrete has no mass: $$\#\Delta\leq\aleph_0\implies\mu(\Delta)=0$$ (Note that I do not require it to decompose into singletons for reason.)

Call a measure atomic if is made out of atoms only: $$\mu(\Omega\setminus\bigcup_{A\in\mathcal{A}} A)=0\quad(\mathcal{A}:\text{ atoms})$$ and atomless if it has no atoms at all: $$\mathcal{A}\text{ empty}\quad(\mathcal{A}:\text{ atoms})$$ (I have chosen these slightly different formulations for more clarity.)

Discussion

Now, consider a sigma-finite Borel measure $\lambda:\mathcal{B}(\Omega)\to\mathbb{R}_+$.

First, let it be discrete.

Then, by Zorn's lemma it splits into a finest measurable partition.
(See Proof to Finest Measurable Partition.)

These are all atoms and therefore it is atomic.

Conversely, let it be atomic.

Then, by sigma-finiteness atoms cannot have infinite mass.
(See Last Part of Proof to Measure Atoms: Definitions.)

Also, there are at most countably many disjoint ones.

However, they do not need to be countable themselve.
(See Example on Atoms vs. Point Masses.)

Luckily, for separable metric spaces every atom even shrinks to a point mass.
(See Proof by Nate Eldredge.)

Thus, it is concentrated on countable set and therefore discrete.

This completes the equivalence!

Remarks

Besides, note that the only measure that is both discrete and continuous is the trivial measure.

Similarly, the only measure that is both atomic and atomless is the trivial measure.

Problem

Is the first claim of the discussion really true?

Namely, every countable measure space admits a finest measurable partition: $$\#\Omega\leq\aleph_0:\quad\Omega=\bigsqcup_{k=1}^\infty A_k\quad(A_k\in\Sigma)$$

$\endgroup$
3
  • $\begingroup$ You could try something along the lines of considering the equivalence classes under the equivalence relation $x\sim y$ iff for any $A\in\Sigma$, $x\in A\Longleftrightarrow y\in A$. If this works then the equivalence classes are the smallest measurable sets, and define the smallest measurable partition. $\endgroup$ Oct 29 '14 at 22:24
  • $\begingroup$ @OlivierBégassat: Nice idea - seems this works for arbitrary spaces and without Zorn... $\endgroup$ Oct 29 '14 at 22:29
  • $\begingroup$ It relies on the underlying set being numerable to prove that the equivalence class is in $\Sigma$, and some form of countable choice to represent the equivalence class as a countable intersection of measurble sets. $\endgroup$ Oct 29 '14 at 23:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.