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Consider a probability space $(X,\Sigma,P)$. Let say that a collection $\mathcal{B}\subseteq\Sigma$ is non-atomic if for every $E\in\mathcal{B}$ and $\alpha\in(0,P(E))$ there exists $F\in\mathcal{B}$ such that $F\subseteq E$ and $P(F)=\alpha$.

Suppose $\Sigma$ non-atomic. Is the $\lambda$-system generated by the collection of all the $2^n$-fold uniform partitions, non-atomic ?

An idea to prove it could be to use transfinite induction since for any collection $\mathcal{B}\subseteq\Sigma$ we have (I think) $\lambda(\mathcal{B})=\cup_{\alpha<\omega_1}\mathcal{B}_\alpha$ with,

  • $\mathcal{B}_0:= \mathcal{B}$;
  • For every ordinal $\alpha$, $\mathcal{B}_{\alpha+1}:= \{\sqcup_{n\in\mathbb{N}}E_n : \forall n\in\mathbb{N}\ E_n\in \mathcal{B}_\alpha\} \cup \{E^c : E\in \mathcal{B}_\alpha\}$ ;
  • For every ordinal limit $\lambda$, $\mathcal{B}_{\lambda}:= \cup_{\alpha<\lambda}\mathcal{B}_\alpha$.
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Take the probability space $([0,1],\mathcal{B},\lambda)$, where $\lambda$ denotes the Lebesgue measure. $\mathcal{B}$ is non-atomic. However, let $X=\{A\in\mathcal{B}\;|\;A\subseteq[0,1/2]\}$. $X$ is non-atomic itself. The $\lambda$-system generated by $X$ is $$Y:=X\cup\{(1/2,1]\cup B\;|\;B\in X\}\cup\{[0,1]\}$$ since the above set is closed unter complements (complements of elements $B\in X$ are of the form $(1/2,1]\cup C$ for $C=[0,1/2]\smallsetminus B$, which is in $X$ and complements of elements $(1/2,1]\cup B$ are $[0,1/2]\smallsetminus B$, which is in $X$) and unions of pairwise disjoint sets (since $X$ is closed under pairwise disjoint unions). Therefore, the above system is a $\lambda$-system and also minimal, since $(1/2,1]$ has to be in the $\lambda$-System generated by $X$. But $Y$ is not non-atomic, since $Y$ does not contain any subset of $[1/2,1]$ of positive measure.

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