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.

Every $\sigma$-complete Boolean algebra is isomorphic to a quotient $\mathcal{M}/I$ where $(X,\mathcal{M})$ is a $\sigma$-algebra and $I$ is a $\sigma$-complete ideal (See the Handbook of Boolean Algebras for a proof or look here). Here by a $\sigma$-complete Boolean algebra, we mean a Boolean algebra where every countable subset has a least upper bound. A $\sigma$-complete Boolean algebra $B$ is isomorphic to an algebra of sets if and only if each element in $B\setminus\{0\}$ is contained in some $\sigma$-complete ultrafilter.

$\textbf{Proposition}$ Suppose that $B$ is an atomless $\sigma$-complete Boolean algebra that satisfies the countable chain condition. Then $B$ has no $\sigma$-complete ultrafilters. In particular, $B$ is not representable as an algebra of sets.

$\textbf{Proof}$ Suppose to the contrary that $\mathcal{U}$ is a $\sigma$-complete ultrafilter. Then since $B$ is atomless, $\mathcal{U}$ is non-principal. Let $c\subseteq B$ be a maximal family under inclusion such that $c\cap\mathcal{U}=\emptyset$ and $a\wedge b=0$ for each $a,b\in c,a\neq b$. Such a set $c$ exists by Zorn's lemma. By maximality, $c$ is in fact a partition of $B$. By the countable chain condition, the set $c$ must be countable, so since $c\cap\mathcal{U}=\emptyset$, we have $1=\bigvee c\not\in\mathcal{U}$ since $\mathcal{U}$ is $\sigma$-complete. $\mathbf{QED}$

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...