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

1 Answer 1


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 $D=[0,\mu(\Omega)]$

  • $\begingroup$ 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 $D$? $\endgroup$
    – Eparoh
    Commented Nov 23, 2020 at 19:15
  • $\begingroup$ 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 $t_1<t<t_2$). $\endgroup$
    – Ramiro
    Commented Nov 23, 2020 at 20:24
  • $\begingroup$ 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. $\endgroup$
    – Eparoh
    Commented Nov 23, 2020 at 20:54
  • $\begingroup$ 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 $D=[0,\mu(\Omega)]$. $\endgroup$
    – Ramiro
    Commented Nov 23, 2020 at 23:38
  • $\begingroup$ 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. $\endgroup$
    – Eparoh
    Commented Nov 24, 2020 at 7:11

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