TITLE: Complete Boolean algebra not isomorphic to a $\sigma$-algebra
QUESTION [18 upvotes]: 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?

REPLY [10 votes]: Here is yet another way to see this. In a $\sigma$-algebra of sets, countable intersections an unions are computed setwise, which means that we always have $$\bigwedge_{n<\omega} \bigvee_{m < \omega} A_{m,n} = \bigcap_{m<\omega} \bigcup_{n < \omega} A_{m,n} = \bigcup_{f:\omega\to\omega} \bigcap_{m<\omega} A_{m,f(m)},$$ where the right hand side is to be considered in a purely set theoretic basis. In any complete boolean algebra, we have $$\bigwedge_{n<\omega} \bigvee_{m < \omega} A_{m,n} \geq \bigvee_{f:\omega\to\omega} \bigwedge_{m<\omega} A_{m,f(m)}.$$ In the case of a $\sigma$-algebra of sets, we have $$\bigvee_{m<\omega} A_{m,f(m)} = \bigcap_{m<\omega} A_{m,f(m)}$$ for every $f:\omega\to\omega$, hence the set theoretic identity above shows that in a complete $\sigma$-algebra we must have $$\bigwedge_{n<\omega} \bigvee_{m < \omega} A_{m,n} = \bigvee_{f:\omega\to\omega} \bigwedge_{m<\omega} A_{m,f(m)}.$$ There are plenty of complete Boolean algebras that do not satisfy this distributive identity. In fact, from the perspective of forcing, this distributive property for a complete atomless Boolean algebra is equivalent to not adding new reals, so any complete atomless Boolean algebra that adds a new real is not isomorphic to any $\sigma$-algebra of sets. This includes Cohen forcing as Joel pointed out, Random forcing as Simon pointed out, and a whole lot more...