Atoms in a measure space and Sierpinski theorem for non-atomic measures I have two questions concerning atoms in a measure space.
My first question is about two different definitions I've encountered and I'm not sure if they are equivalent.
Definition 1: Given a measure space $(\Omega, \Sigma, \mu)$ we say that an element $A \in \Sigma$ is an atom of $\mu$ if it satisfies that $\mu(A) >0$ and for every $B \in \Sigma$ such that $B \subset A$ either $\mu(B)=0$ or $\mu(A \setminus B)=0$.
Definition 2: Given a measure space $(\Omega, \Sigma, \mu)$ we say that an element $A \in \Sigma$ is an atom of $\mu$ if it satisfies that $\mu(A) >0$ and for every $B \in \Sigma$ such that $B \subset A$ either $\mu(B)=0$ or $\mu(B)=\mu(A)$.
So I know that if $\mu(\Omega)$ is finite then both definitions are indeed equivalent but, are they still equivalent if the measure is not finite?
Now, concerning atoms of a measure I'm trying to prove the following theorem, apparently due to Sierpinski:
Theorem: If $(\Omega, \Sigma, \mu)$ is a measure space with no atoms, then for every $t \in [0, \mu(\Omega)]$ there exists an $A \in \Sigma$ such that $\mu(A)=t$.
Following Wikipedia's article on atoms, I'm trying to fully proof the sketch of the proof at the end of the page.
So I want to prove there is a funciton $S:[0, \mu(\Omega)] \longrightarrow \Sigma$ satisfying:

*

*$S$ is monotone, that is, if $t \leq t'$ then $S(t) \subset S(t')$

*$\mu(S(t))=t$ for every $t \in [0, \mu(\Omega)]$
because this would directly imply the theorem.
To this end, we define
$$\Gamma= \left\{  S:D \longrightarrow \Sigma: D \subset [0, \mu(\Omega)], S \textrm{ is monotone}, \mu(S(t))=t, \forall t \in D \right\}$$
Then $\Gamma$ is not empty beacause we can define $S_0:\{0\} \longrightarrow \Sigma$ given by $S(0)=\emptyset$ and is clearly in $\Gamma$, and we can make it a partially ordered set by establishing that $S_1 \leq S_2$ if, and only if, $\operatorname{graph}(S_1) \subset \operatorname{graph}(S_2)$.
Now, if $\Gamma_c$ is a chain in $\Gamma$ then we define $S_c: D_c \longrightarrow \Sigma$ where
$$D_c=\bigcup_{S \in \Gamma_c} \operatorname{dom}(S)$$
and given $t \in D_c$, we choose any $S \in \Gamma_c$ such that $t \in \operatorname{dom}(S)$, and define $S_c(t)=S(t)$.
First, $S_c$ is well defined because if we have that given $t \in D_c$, there are $S_1, S_2 \in \Gamma_c$ such that $t \in \operatorname{dom}(S_1) \cap \operatorname{dom}(S_2)$, then as $\Gamma_c$ is a chain, we can assume wlog that $S_1 \leq S_2$ and then, $(t,S_1(t)) \in \operatorname{graph}(S_2)$ so $S_2(t)=S_1(t)$ as we wanted to show.
Now, we will prove that $S_c \in \Gamma$.
First, observe that given $t_1, t_2 \in D_c$ there exists $S_1, S_2 \in \Gamma_c$ such that $t_i \in \operatorname{dom}(S_i)$, so as $\Gamma_c$ is a chain, we may assume $S_1 \leq S_2$ and so $\operatorname{dom}(S_1) \subset \operatorname{dom}(S_2)$, getting that $t_1, t_2 \in \operatorname{dom}(S_2)$. So, if $t_1 \leq t_2$ as $S_2 \in \Gamma$ we will have that
$$S_c(t_1)=S_2(t_1) \subset S_2(t_2) = S_c (t_2$$
so $S_c$ is monotone.
On the other hand, it is clear that given $t \in D_c$ if it is $S \in \Gamma_c$ such that $t \in \operatorname{dom}(S)$, then
$$\mu(S_c(t))=\mu(S(t))=t$$
So we have proven that $S_c \in \Gamma$ and it is clear by its construction that it is an upper bound of $\Gamma_c$.
By Zorn's lemma, there exits then some $S: D \longrightarrow \Sigma$ maximal in $\Gamma$, and the claim is that this function is the one we are looking for, and it suffices to prove that $D=[0, \mu(\Omega)]$.
This last part is the one I cannot prove, and if everything I've done is correct I suppose here is where you should use that $\mu$ is non-atomic. If tried to prove that $D$ is closed and open in $[0, \mu(\Omega)]$, and as it is conected, we'll have that $D=[0, \mu(\Omega)]$ as we want.
First, I've shown by using the maximality of $S$ that $\{0, \mu(\Omega)\} \subset D$, for example if we suppose $0 \not \in D$ (the other case is analogue), then defining $\bar S: D \cup \{0\} \longrightarrow \Sigma$ as $\bar S(t)=S(t)$ if $t \in D$ and $\bar S(0)=\emptyset$, it is clear that $\bar S \in \Gamma$ and $S < \bar S$ contradiction with its maximality.
Now, to show it is closed, I've taken a sequence $\{t_n\} \subset D$ converging to some $t_0 \in [0, \mu(\Omega)]$ and I've supposed that $t \not \in D$.
Then, we have that $t_0 \in (0, \mu(\Omega))$ so in particular is finite, and we know then there must exists a monotone subsequence of $\{t_n\}$ converging to $t_0$, so we can suppose that $\{t_n\}$ is monotone.
If it is increasing, defining $A_n=S(t_n)$ and $\bar S: D \cup \{t_0\} \longrightarrow \Sigma$ as $\bar S(t)= S(t)$ if $t \in D$ and $\bar S(t_0)= \cup A_n$ we have:

*

*If $t <t_0$ then there exists some natural number $n$ such that $t \leq t_n$ and so
$$\bar S(t)=s(t) \subset S(t_n) = A_n \subset \bar S(t_0)$$

*If $t_0 < t$ then $t_n < t$ for every $n \geq 1$ so
$$A_n=S(t_n) \subset S(t), \forall n \geq 1$$
and then
$$\bar S(t_0) \subset S(t)$$

*Since $\{t_n\}$ is increasing, by the monotonity of $S$, $\{S(t_n)\}$ is also increasing and we then have
$$\mu\left(\bar S(t_0)\right)=\mu\left( \bigcup_{n=1}^\infty S(t_n) \right)=\lim_{n \rightarrow \infty} \mu\left( S(t_n)\right) = \lim_{n \rightarrow \infty} t_n = t_0$$
So, this three facts imply that $\bar S \in \Gamma$ but $S < \bar S$ so we have a contradiction with the maximality.
Now, if $\{t_n\}$ is decreasing, we do the same thing but taking $\bar S(t_0)=\cap A_n$ and, as $t$ is finite, we can assume that $\mu(A_1)=\mu(S(t_1))=t_1$ is finite, so we can argue as before arriving at the same contradiction.
If everything I've write here is correct, then to show $D$ is open I must use the fact that $\mu$ is non-atomic because I haven't used it yet, so my doubt concerging this part are:

*

*If the definitions I gave at the beggining are not equivalent, can this last part be shown with both definitions?

*In case it is two difficult to show $D$ is open, is there a simple proof that $D=[0,\mu(\Omega)]$? And, in that case, can it be shown with both definitions of atoms if they are not equivalent?

 A: About the definitions:
The definitions $1$ and $2$ are not equivalent in the case of infinite measures.
The definition $1$ is more adequate in some cases, but it is not the most common one.  On the other hand definition $2$ may lead to pathological examples when applied to infinite measures. Consider $(\mathbb{N}, 2^{\mathbb{N}}, \mu)$ where $\mu(\emptyset)=0$ and, for all $E \subseteq \mathbb{N}$ such that $E\neq \emptyset$ , $\mu(E)=\infty$. According to definition $1$,  only the singletons will be atoms. According to definiton $2$, all non-empty subsets of $\mathbb{N}$ (so you have atoms being proper subsets of other atoms, for instance). However definition $2$ is more common and it is more useful when we want to talk about atom-free measures.
About your proof:
Yes, by Zorn's lemma, there exits then some $S: D \longrightarrow \Sigma$ maximal in $\Gamma$. You have already proved that $D$ is closed.
To complete your proof, it is enough to proof that if $t_1, t_2 \in D$ and $t_1<t_2$, there there is $t\in D$ such that  $t_1<t<t_2$.
Proof: Suppose $t_1, t_2 \in D$ and $t_1<t_2$. Suppose  there is no $t\in D$ such that  $t_1<t<t_2$ (in other words, for all $t\in(t_1,t_2), t\notin D$). Let us show that this contradicts the maximality of $S: D \longrightarrow \Sigma$.
Since $t_1<t_2$, $S(t_1)\subseteq S(t_2)$. But clearly $S(t_1)= S(t_2)$ is not possible because $\mu(S(t_1))=t_1$ and $\mu(S(t_2))=t_2$. So $S(t_1)\subsetneq S(t_2)$.
Let $B= S(t_2) \setminus S(t_1)$, since $t_1<t_2$, $t_1$ is finite. So
$\mu(B) = t_2-t_1 >0$. As $\mu$ has no atoms (definition $2$). There is $C\subsetneq B$
such that $0< \mu(C) < \mu(B) = t_2-t_1$. Let $\delta = \mu(C)$.
It is immediate that $t_1<t_1+\delta<t_2$,  $S(t_1)\subsetneq S(t_1) \cup C \subsetneq  S(t_2)$ and $\mu(S(t_1) \cup C) = t_1+\delta$. So  $S: D \longrightarrow \Sigma$ can be extended to include the pair $(t_1+\delta, S(t_1) \cup C)$. Contradiction the maximality of $S: D \longrightarrow \Sigma$.
Since $D$ is closed, $0 \in D$ and $\mu(\Omega) \in D$, we can conclude that  $D=[0,\mu(\Omega)]$.
In fact, suppose there is $x\in [0,\mu(\Omega)]$ such that $x \notin D$ since $D$ is closed, $0 \in D$ and $\mu(\Omega) \in D$, there is an open interval $(a,b)$ such that $x\in (a,b)\subset [0,\mu(\Omega)]$ and $D\cap (a,b) =\emptyset$.
Now note that $0\in \{y\in D: y<a\}$  and $\mu(\Omega) \in \{y\in D: y>b\}$.
So both set are not empty and we can define
$$ a_1 = \sup \{y\in D: y<a\} $$
and
$$ b_1 = \inf  \{y\in D: y>b\}$$
It is easy to see that $(a,b) \subseteq (a_1,b_1)$ and $D \cap (a_1,b_1)=\emptyset$. But $D$ is closed, so $a_1, b_1 \in D$. But then, by what we have proved before, there is $t\in (a_1,b_1)$ such that $t \in D$. Contradiction to  $D \cap (a_1,b_1)=\emptyset$. So for all $x\in [0,\mu(\Omega)]$, $x\in D$. So  $D=[0,\mu(\Omega)]$
