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Let $\Omega$ be a measure space with measure $\mu$.
(Here, a measure is only meant to be countable additive!)

Consider a subset $A\in\Sigma$.

Then according to the wikipedia article it is an atom if: $$(1)\quad\forall E\subseteq A:\quad\mu(E)<\mu(A)\implies\mu(E)=0\quad(\mu(A)>0)$$ and according to the paper by Johnson it is an atom if: $$(2)\quad\forall E\in\Sigma:\quad\mu(E\cap A)=0\lor\mu(E^c\cap A)=0\quad(\mu(A)>0)$$

Now, these definitions agree for the really trivial cases: $$\mu\equiv0$$ $$\mu\equiv\infty$$ but they differ for the less trivial case: $$\quad\mu(E\neq\varnothing):=\infty,\,\mu(\varnothing):=0$$ namely the atoms are w.r.t. (1) all nonempty subsets whereas w.r.t. (2) only the singletons.

Excluding this pathological case, are the definitions equivalent?
(More precisely, assume there exists a measurable subset $0<\mu(F)<\infty$.)

So far I checked that: $$(2)\implies(1):\quad\mu(E)<\mu(A)\implies\mu(A\setminus E)>0\implies\mu(E)=0$$ but what about the other direction?

freishahiri
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    The second definition definitely has a typo. – Asaf Karagila Oct 14 '14 at 22:54
  • Yepp, a complement was missing, thanks, corrected! – freishahiri Oct 14 '14 at 22:55
  • The pathological case you want to exclude is $\mu(A)=\infty$. Just the fact that there is a set of finite measure somewhere tells you nothing at all about $A$. – tomasz Oct 15 '14 at 00:28
  • @tomasz: Riiight one could just adjoin a finite measure space to the pathological one $\Omega_{pathological}\sqcup\Omega_{finite}$ getting again the same problem... – freishahiri Oct 15 '14 at 00:37
  • @tomasz: But what about the sigma-finite case then as studied in the paper? – freishahiri Oct 15 '14 at 00:38
  • @Freeze_S: If the measure is $\sigma$-finite, there can be no atoms of infinite measure, as any set can be partitioned into (countably many) sets of finite measure. – tomasz Oct 15 '14 at 07:10

1 Answers1

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First, as mentioned in the comments excluding the pathological case is not enough:
(Thanks alot @tomasz!!)

Consider the measure space: $$\mu:\mathcal{P}(\mathbb{N}\cup{\infty})\to\overline{\mathbb{R}}_+:\quad\mu(\varnothing):=0,\;\mu(\{\infty\}):=1,\;\mu(E):=\infty\text{ else}$$ Then the atoms are w.r.t. (1) all nonempty subsets whereas w.r.t. (2) only the singletons
and for both (1) and (2) additionally the point at infinity.

Next, for atoms of finite mass the proof goes as follows:
(Thanks alot @asaf karagila!!)

Assume, it holds (1). Let $E\in\Sigma$ arbitrary.

If $\mu(A\cap E)=0$ we're done.
Otherwise $\mu(A\cap E)=\mu(A)$ by (1). But then $\mu(A\cap E^c)=0$ as $\mu(A)<\infty$ and we're done, too.

Concluding that (2) holds.

Finally, for the sigma-finite case atoms have finite mass and the preceding applies:

Assume $A$ is an atom w.r.t. (1) with $\mu(A)=\infty$.

By sigma-finiteness, there exists a sequence $A_n\uparrow A$ with $\mu(A_n)<\infty$.
By continuity, there exists a natural $n_0\in\mathbb{N}$ with $\mu(A_{n_0})>0$.
By defintion of an atom, this implies $\mu(A_n)=\mu(A)=\infty$.
Contradiction!

Concluding that $\mu(A)<\infty$.

freishahiri
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