How probability-rich is the $\sigma$-algebra generated by a sequence of sets? (Sierpiński's theorem on non-atomic measures without using the AoC.)

Question

$\newcommand\F{\mathcal F}\newcommand\si{\sigma}\newcommand\Om{\Omega}\newcommand\ep{\varepsilon}$Let $p\in(0,1)$ and let $(\Om,\F,P)$ be a probability space. Let $(A_n)$ be a sequence in $\F$ such that $P(A_n)\to p$ as $n\to\infty$.

Without using the axiom of choice, can it then be shown that,

necessarily, for each real $\ep>0$ there is some set $B_\ep\in\F$ such that $P(B_\ep)\in[p,p+\ep)$?

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One may note that, without loss of generality, the $\si$-algebra $\F$ is generated by the $A_n$'s.

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Motivation: A positive answer to this question would provide a proof of Sierpiński's theorem on non-atomic measures without using the axiom of choice.

(On the other hand, it is not hard to deduce the highlighted statement from Sierpiński's theorem.)

Answer 1

Here is a proof that probability measures have closed range under DC, summarized from the discussion with Iosif Pinelis. I try to be as explicit as possible about any choice arguments used.

For any probability measure $(\Omega,\mathcal{F},P)$, there exists $A\in\mathcal{F}$ such that $P$ is purely atomic on $A$ and purely nonatomic on $A^C$; the decomposition is usually not unique. To construct $A$, let $A_1$ be an atom of maximal probability. Given $A_1,\ldots,A_n$ are defined, let $A_{n+1}$ be an atom in $\Omega\setminus(A_1\cup\cdots\cup A_n)$ of maximal probability. This gives us by DC a sequence of atoms $\langle A_n\rangle$ such that there are no atoms in $\Omega\setminus A$ for $A=\bigcup_n A_n$. We can assume by the usual trick that all atoms in the sequence are disjoint. We can now define two finite measures $P^{\text{a}}$ and $P^{\text{na}}$ on $(\Omega,\mathcal{F})$ such that $P^{\text{a}}(E)=P^{\text{a}}(E\cap A)$ and $P^{\text{na}}(E)=(E\cap A^C)$. Then $P^{\text{a}}$ and $P^{\text{na}}$ are purely atomic and nonatomic, respectively, and $P=P^{\text{a}}+P^{\text{na}}$.

We know from this answer by Eric Wofsey that under DC, the range of a finite nonatomic measure is a closed and bounded interval. To show that every probability measure has a closed range under DC, it suffices, therefore, to show that $P^{\text{a}}$ has a closed range. So we can assume for the remainder that $P=P^{\text{a}}$. Let $\alpha$ be in the closure of the range of $P$. For every $m$, there exists some $E_m\in\mathcal{F}$ such that $|P(E_m)-\alpha|<1/m$. DC implies the countable axiom of choice, so we can select a sequence $\langle E_m\rangle$ with this property. In particular, $\alpha=\lim_m P(E_m)$.

Since $P$ is purely atomic we can, for our original sequence of atoms $\langle A_n\rangle$ and each $E_n$ canonically specify a number $c_{nm}$ such that $c_{nm}=1$ if $P(A_n\cap E_m)=P(A_n)$ and $c_{nm}=0$ if $P(A_n\cap E_m)=0$. Write $c_m$ for the sequence $\langle c_{nm}\rangle$. Viewing $P$ is a function on $\{0,1\}^\mathbb{N}$ given by $P(c_1,c_2,\ldots)=\sum c_n P(A_n)$, we see that $P$ is continuous in the product topology.

We construct a limit point in the product topology of $\langle c_m\rangle$ by a diagonal argument. Either, $(c_{11},c_{12}, c_{13},\rangle)$ is infinitely often $0$ or infinitely often $1$. In the first case, take the subsequence with the $0$s, otherwise, the one with the $1$s. Continue this way for every coordinate to get a sequence of subsequences. Picking the diagonal elements gives you the desired limit point $\langle c_n^*\rangle$. No choice was needed for this argument. By the continuity of $P$, we have $\sum_n c_n^* P(A_n)=\alpha$. Since the $A_n$ are disjoint, $$P\bigg(\bigcup_{n:c_n^*=1} A_n\bigg)=\alpha.$$