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Hi, I am struggling with the following question that is tangentially arising from a paper I'm working on. It is not at all essential for the revision but it would be nice to know if there is a pleasant answer, or references, to the question. The question was asked by me on math.stackexchange.com because I thought that surely it must appear as homework in some graduate class; if so, then apologies for posting here; but nevertheless a hint would be very much appreciated.

Let $(\Omega,\Sigma)$ be a measurable space that need not be countably generated. Let $\mathcal{N}$ be a subset of $\Sigma$. What conditions on $\mathcal{N}$ (and $\Sigma$) guarantees the existence of a (non-atomic) probability measure $\mu: \Sigma\to [0,1]$ such that for any $E\in \Sigma$ if $\mu(E)=0$, then $E\in \mathcal{N}$?

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  • $\begingroup$ The obvious necessary condition is that $\mathcal N$ is a $\sigma$-ideal ($A\subseteq N\in\mathcal N$ implies $A\in\mathcal N$ and $N_n\in\mathcal N$ implies $\bigcup_n N_n\in\mathcal N$). If $\Sigma$ is generated by $\mathcal N$ this is also sufficient ($\mu(N)=0$ if $N\in\mathcal N$ and $\mu(A)=1$ if $\Omega\setminus A\in\mathcal N$). $\endgroup$
    – Jochen Wengenroth
    Commented Oct 29, 2012 at 8:14
  • $\begingroup$ Except that that's not necessary. $\;\;$ $\mathcal{N}$ could just be $\: \Sigma-\{\Omega\} \:$. $\;\;\;\;$ $\endgroup$
    – user5810
    Commented Oct 29, 2012 at 8:21
  • $\begingroup$ Sorry, I misunderstood the question. Being a $\sigma$-ideal is necessary for being included in all $\mu$-null sets. $\endgroup$
    – Jochen Wengenroth
    Commented Oct 29, 2012 at 8:39
  • $\begingroup$ @Jochen The answer that $\mathcal{N}$ contains a $\sigma$-ideal that generates $\Sigma$ is kinda too sufficient a condition because I'm also interested in the "non-atomic" part of the question. $\endgroup$
    – Rabee Tourky
    Commented Oct 29, 2012 at 9:14

1 Answer 1

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We may rephrase the condition saying: $\mathcal{N}$ includes the class $\Sigma_0$ of all $E\in \Sigma$, such that $E$ is a null set for some non-atomic probability measure $\mu_E$ on $\Sigma$, or equivalently, of all $E\in \Sigma$ such that there is a non-atomic probability measure supported on $\Omega\setminus E$. This is a condition only on the set $\Omega\setminus E$ as a measurable space with the $\sigma$-algebra induced by $\Sigma$. Re-naming this measure subspace, the question is:

Which measurable space $(\Omega, > \Sigma)$ can support a non-atomic probability measure $\mu$?

There is a number of papers on this matter, that you can easily find on google, and should check, though mainly addressed to spaces with additional topological or metrc sturctures. However, in the general setting of a measurable spaces $(\Omega,\Sigma)$ I think one can actually give a caracterization for the existence of a non-atomic probability measure in terms of the partial order structure of $\Sigma$ (I'm quite sure somebody here has the right reference handy).

A first necessary condition on the structure of $\Sigma$ as a partially ordered set is given by Sierpiński's theorem : a non-atomic probability measure space $(\Omega,\Sigma,\mu)$ is divisible; precisely: $\Sigma$ contains a monotone family $\{E _ \lambda\} _ {\lambda\in[0,1]}$ such that $\mu(E_\lambda)=\lambda$ (in other words, the measure $\mu$, as a function $\mu:\Sigma\to[0,1]$, admits a monotone section $[0,1]\ni\lambda\mapsto E_\lambda\in\Sigma$). And, of course, every set of positive measure also have this property.

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Summarizing: Denoting by $\mathcal{M^*}$ the set of non-atomic probability measures on $(\Omega,\Sigma)$, and $$\Sigma_0:=\{A\in\Sigma: \exists \mu\in\mathcal{M^*} \, s.t. \mu(E)=0\}\, ,$$

a necessary condition for $E_0\in\Sigma_0$ is that it can be embedded into a family $\{E _ \lambda\} _ {\lambda\in[0,1]}$ of subsets of $\Sigma$ strictly increasing by inclusion. In other words, $E_0$ is a minimum element of a chain of $\Sigma$ that is order-isomorphic to $[0,1]$.

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  • $\begingroup$ Pietro, can you please explain the "equivalently, such that there is a non-atomic probability...." What's $E$ in your $\Omega\setminus E$? $\endgroup$
    – Rabee Tourky
    Commented Oct 29, 2012 at 9:33
  • $\begingroup$ To clarify, digging deeper into my question; what I've wanted is indeed a partial order characterization guaranteeing that $\mathcal{N}$ includes the class of $\Sigma_0$ of all $E\in \Sigma$ that are zero measure sets of a non-atomic probability measure. I really haven't understood your answer. $\endgroup$
    – Rabee Tourky
    Commented Oct 29, 2012 at 9:44
  • $\begingroup$ (to the first question). Just the obvious fact that $\mu(E)=0$ iff $\mu(\Omega\setminus E)=1$; so $\mu$ restricts to a non-atomic probability measure on $\Omega\setminus E$ with the induced $\sigma$-algebra $\Sigma_{|\Omega\setminus E}:=\{A\in\Sigma:A\subset \Omega\setminus E\}$; and any such measure on $\Omega\setminus E$ extends to $\Omega$ with $\mu(E)=0$. $\endgroup$
    – Pietro Majer
    Commented Oct 29, 2012 at 9:53
  • $\begingroup$ I have edited and added a few lines of clarification. I think, that is also a sufficient condition or very close to a sufficient condition. As I was saying, the problem of existence of non-atomic probability measures should be treated in the literature, so I'd wait for somebody with a reference, before we do it ourselves :) $\endgroup$
    – Pietro Majer
    Commented Oct 29, 2012 at 10:14
  • $\begingroup$ Pietro, your $\Sigma_0$ after the edit is not what I understood in my second comment. I thought it was there exists $\mu\in \mathcal{M^*}$ such that $\Sigma_0=\\{E\in \Sigma: \mu(E)=0\\}$. I am now very confused about your answer "\mathcal{N} includes (your) $\Sigma_0$." How do we go from your $\Sigma_0$ to the existence of one $\mu$ for which all sets in $\Sigma_0$ have measure zero? $\endgroup$
    – Rabee Tourky
    Commented Oct 29, 2012 at 10:30

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