Measures: Atom Definitions 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:
$$(1)\quad\forall E\subseteq A:\quad\mu(E)&lt;\mu(A)\implies\mu(E)=0\quad(\mu(A)&gt;0)$$
and according to the paper by Johnson it is an atom if:
$$(2)\quad\forall E\in\Sigma:\quad\mu(E\cap A)=0\lor\mu(E^c\cap A)=0\quad(\mu(A)&gt;0)$$

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 $0&lt;\mu(F)&lt;\infty$.)

So far I checked that:
$$(2)\implies(1):\quad\mu(E)&lt;\mu(A)\implies\mu(A\setminus E)&gt;0\implies\mu(E)=0$$
but what about the other direction?
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 $\mu(A)&lt;\infty$ 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 $\mu(A_n)&lt;\infty$.
By continuity, there exists a natural $n_0\in\mathbb{N}$ with $\mu(A_{n_0})&gt;0$.
By defintion of an atom, this implies $\mu(A_n)=\mu(A)=\infty$.
Contradiction!

Concluding that $\mu(A)&lt;\infty$.
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 $\mu(A)&lt;\infty$ 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 $\mu(A_n)&lt;\infty$.
By continuity, there exists a natural $n_0\in\mathbb{N}$ with $\mu(A_{n_0})&gt;0$.
By defintion of an atom, this implies $\mu(A_n)=\mu(A)=\infty$.
Contradiction!

Concluding that $\mu(A)&lt;\infty$.