# 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:

1. $$S$$ is monotone, that is, if $$t \leq t'$$ then $$S(t) \subset S(t')$$
2. $$\mu(S(t))=t$$ 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:

1. If $$t then there exists some natural number $$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)$$
2. If $$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)$$
3. Since $$\{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:

1. If the definitions I gave at the beggining are not equivalent, can this last part be shown with both definitions?
2. 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?

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.

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 $$t_1, there there is $$t\in D$$ such that $$t_1.

Proof: Suppose $$t_1, t_2 \in D$$ and $$t_1. Suppose there is no $$t\in D$$ such that $$t_1 (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 $$t_1, $$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_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 $$t_1, $$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 $$0\in \{y\in D: y and $$\mu(\Omega) \in \{y\in D: y>b\}$$. So both set are not empty and we can define

$$a_1 = \sup \{y\in D: y and $$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 $$D=[0,\mu(\Omega)]$$

• Thank you, the definitions are clear now, but I don't get the proof. I suposse at the beggining there is a typo and you mean $t \not \in D$, but at the end what you get is that $t_1+\delta$ must be in $D$, but how does that contradicts the fact that the fixed $t$ at the beggining is not in $D$? Commented Nov 23, 2020 at 19:15
• Thanks. There was a typo (a missing "no"). I have corrected it. We start with the fact that $S: D \longrightarrow \Sigma$ is maximal and we suppose that there is no $t\in D$ such that $t_1<t<t_2$. In the end we find $t_1+\delta$ such that $t_1<t_1+\delta<t_2$ and the pair $(t_1+\delta, S(t_1) \cup C)$ can be used to extend $S: D \longrightarrow \Sigma$, which contradicts the maximality of $S$. (Or if you prefer: since $S$ is maximal, such $t+\delta$ must be in $D$ which contradicts that there is no $t\in D$ such that $t_1<t<t_2$). Commented Nov 23, 2020 at 20:24
• Okay, that's clear now, but I don't see why the density of $D$ is obvious from that. I mean, I know how to prove from what have been done, that $D$ must be dense, but is not straightforward to me and I just want to know if there is a simple argument to conclude it. Commented Nov 23, 2020 at 20:54
• Since $D$ is closed, to prove it is dense is almost the same effort to prove directly that $D=[0,\mu(\Omega)]$. I have changed the end of my proof to porve directly that $D=[0,\mu(\Omega)]$. Commented Nov 23, 2020 at 23:38
• Thank you, that was the type of argument with suppremum and infimum I was thinking about but I thought maybe I was missing something and was more obvious. Commented Nov 24, 2020 at 7:11