Does there exist a complete Boolean algebra that is not isomorphic to any $\sigma$-algebra? If so, what is an easy or canonical example or construction?



Complete Boolean algebra not isomorphic to a $\sigma$-algebra
Let $\Sigma\_0$ be the $\sigma$-algebra of Lebesgue measurable sets on the real interval $[0,1]$. Define $\Sigma$ to be $\Sigma\_0$ quotiented by the relation $U \simeq V$ if $U$ and $V$ differ by a Lebesgue negligible set.


$\Sigma$ is a complete boolean algebra, but $\Sigma$ is not a $\sigma$-algebra.


Indeed, assume $\Sigma$ is identified with a $\sigma$-algebra in $\mathcal{P}(X)$ for some set $X$, and let $x \in X$. Then for each integer $k$, $x$ has to belong to a set of the form $[a/k,(a+1)/k]$ but the countable intersection of a family of such set is always empty in $\Sigma$ (it has zero measure), hence $x$ belong to the empty set, which yields a contradiction.


