TITLE: Peculiar examples to the Stone Representation Theorem
QUESTION [6 upvotes]: The Stone Representation theorem states that every Boolean algebra is isomorphic to a field of sets. That is, a Boolean algebra whose elements are sets, and sums, products, negation are union, intersection and complements respectively.
But the theorem doesn't say that we preserve any more than the basic Boolean structure. One of my undergrad teachers does in fact work a lot around Boolean algebras. So once I asked him,

If we have a $\sigma$-algebra (or a complete Boolean algebra) is the field of sets from Stone's theorem closed under countable unions (or arbitrary unions)?

His answer was no. But he never gave me an example.
Question. Was my professor right (probably yes), and if so, what is an example of a $\sigma$-algebra (or even a complete Boolean algebra) which is not isomorphic to any $\sigma$-field of sets (or a field of set closed under arbitrary unions)?

REPLY [2 votes]: Your professor is right.
As you said, Stone's representation theorem asserts that every Boolean algebra is isomorphic to a field of set. However, not every Boolean $\sigma$-algebra is $\sigma$-isomorphic to a $\sigma$-field of sets.
A typical example is the quotient of the algebra of Lebesgue measurable subsets of the unit interval by the ideal of sets of measure zero. It is a Boolean $\sigma$-algebra which is not $\sigma$-isomorphic to any $\sigma$-field of sets (a result proved, if I am not mistaken, by Marczewski (1946): Mesure dans les corps de Boole).