A $\sigma$-algebra that is complete as a Boolean algebra?

This MO thread has several examples of complete Boolean algebras that are not isomorphic to $\sigma$-algebra of sets. But what is a non-trivial example of a complete Boolean algebra that is isomorphic to a $\sigma$-algebra of set as Boolean algebras? Equivalently, is there a $\sigma$-algebra of set that is complete as a Boolean algebra?

A trivial example is the power set algebra $\mathcal{P}(X)$, or any algebra of set that is atomic and complete as Boolean algebra. Also, from the above post it seems a non-trivial example is probably not ccc. I am wondering if the tree algebra on some tree such as $2^{<\aleph_1}$ can help, but the tree algebra itself doesn't seem complete, or even $\sigma$. Maybe we can consider its completion?

Clarification: By a complete Boolean algebra I mean a Boolean algebra whose any subset has supremum. An algebra of set (on $X$) is a nonempty subset of $\mathcal{P}(X)$ closed under union and complementation. An algebra of set can be viewed as a Boolean algebra in an obvious way.

Edit: Now I have the feeling that the Boolean completion of the poset $(2^{<\aleph_1},\supseteq)$ should does it. This is the poset of partial map from $\aleph_1$ to $\{0,1\}$ with countable domain, under inverse inclusion (the largest element being the empty map). Every branch in $2^{<\aleph_1}$, equivalently every element in $2^{\aleph_1}$, determines an $\sigma$-complete ultrafilter in the Boolean completion because the branch has uncountable cofinality. The Boolean completion seems to be the regular-open algebra on $2^{\aleph_1}$ with countable support product topology. I'm not sure if it is extremally disconnected; probably not.

Here is a partial answer. In particular, I will show it is consistent with ZFC that no nontrivial examples exist, i.e. that any $\sigma$-algebra of sets which is complete as a Boolean algebra is isomorphic to a power set.

First, some general remarks on representing Boolean algebras as algebras of sets. Suppose $B$ is a Boolean algebra. Then to give a homomorphism $f:B\to\mathcal{P}(X)$ is the same thing as giving, for each $x\in X$, an ultrafilter $U_x$ on $B$ (and then $f$ is defined by $f(A)=\{x\in X:A\in U_x\}$). The homomorphism $f$ is injective iff these ultrafilters separate points of $B$, i.e. iff every nonzero element of $B$ is contained in some $U_x$. If $\kappa$ is a cardinal and $B$ is $\kappa$-complete, then $f$ is a $\kappa$-complete homomorphism iff each $U_x$ is a $\kappa$-complete ultrafilter. In particular, in the case $\kappa=\aleph_1$, this says that a Boolean $\sigma$-algebra $B$ is isomorphic to a $\sigma$-algebra of sets iff every nonzero element of $B$ is contained in some countably complete ultrafilter.

Now here's the key fact which makes it difficult for this to happen in a complete Boolean algebra.

Theorem: Let $B$ be a complete Boolean algebra and let $U$ be a nonprincipal countably complete ultrafilter on $B$. Then there exists a measurable cardinal $\kappa$ such that $U$ is $\kappa$-complete.

In the special case where $B$ is a power set, this is a well-known fact in the basic theory of measurable cardinals. The proof for a general complete Boolean algebra is similar.

Proof of Theorem: Let $\kappa$ be the least cardinal such that $U$ is not $\kappa^+$-complete (such a cardinal exists since $U$ is not principal; in particular $U$ is not $|U|^+$-complete since if $\bigwedge U$ were in $U$ then it would generate all of $U$). Let $(x_\alpha)_{\alpha<\kappa}$ be a sequence of elements of $U$ such that $\bigwedge_{\alpha<\kappa} x_\alpha\not\in U$. Adding $\neg\bigwedge x_\alpha$ as one more term at the start of the sequence, we may assume that in fact $\bigwedge_{\alpha<\kappa} x_\alpha=0$. By minimality of $\kappa$, for each $\alpha<\kappa$, $\bigwedge_{\beta<\alpha} x_\beta\in U$. So, replacing each $x_\alpha$ with $\bigwedge_{\beta<\alpha} x_\beta$, we may assume the sequence $(x_\alpha)$ is decreasing and continuous.

Now let $y_\alpha=x_{\alpha+1}\wedge \neg x_\alpha$. Then the $y_\alpha$ are pairwise disjoint, each $y_\alpha\not\in U$, and $\bigvee_{\beta<\alpha} y_\beta=\neg x_\alpha$. In particular, since $\bigwedge_{\alpha<\kappa}x_\alpha=0$, this means $\bigvee_{\alpha<\kappa}y_\alpha=1$. There is then a is a complete homomorphism $f:\mathcal{P}(\kappa)\to B$ defined by $f(S)=\bigvee_{\alpha\in S} y_\alpha$. Now consider the pullback ultrafilter $f^{-1}(U)$ on $\mathcal{P}(\kappa)$. Since $f$ is a complete homomorphism and $U$ is $\kappa$-complete, $f^{-1}(U)$ is also $\kappa$-complete. Since $y_\alpha\not\in U$ for each $\alpha$, $f^{-1}(U)$ is nonprincipal. Thus $\kappa$ is measurable. $\blacksquare$

Now suppose $B$ is a $\sigma$-algebra of sets that is complete as a Boolean algebra and is not isomorphic to a power set. In particular, then, $B$ is not atomic, so there exists some nonzero $a\in B$ which contains no atom. Since $B$ is a $\sigma$-algebra of sets, $a$ is contained in some countably complete ultrafilter $U$, which must be nonprincipal since there is no atom below $a$. By the Theorem, this implies the existence of a measurable cardinal.

So no such $B$ can exist if measurable cardinals do not exist. I don't know whether it is possible for such a $B$ to exist if measurable cardinals do exist.