Return to Revisions

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.