# Lambda system generated by a non-atomic collection

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 $$\lambda$$, $$\mathcal{B}_{\lambda}:= \cup_{\alpha<\lambda}\mathcal{B}_\alpha$$.

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.