TITLE: Are there sigma-algebras of cardinality $\kappa>2^{\aleph_0}$ with countable cofinality?
QUESTION [8 upvotes]: A standard homework in measure theory textbooks asks the student to prove that there are not countably infinite $\sigma$-algebras. The only proof that I know is via a contradiction argument which yields no estimate on the minimum cardinality of an infinite $\sigma$-algebra.
Given an a set $X$ of infinite cardinality $\kappa$, the $\sigma$-algebra of all co-countable subsets of $X$ is of cardinality $2^\kappa$ $\kappa^{\aleph_0}$. This example doesn't tell me whether there are $\sigma$-algebras of cardinality below $2^{\aleph_0}$, if I don't assume the Continuum Hypothesis.
My question is as the title says: Are there $\sigma$-algebras of every uncountable cardinality? 
Edit: The combined answer with Stephen, Matthew proves that the cardinality of a $\sigma$-algebra is necessarily at least $2^{\aleph_0}$. Further, for each cardinality $\kappa\ge 2^{\aleph_0}$ with uncountable cofinality, the $\sigma$-algebra of countable (or cocountable) subsets of a set $X$ with cardinality $\kappa$, is of cardinality $\kappa$.
What is left is whether for $\kappa\ge 2^{\aleph_0}$ with $cf(\kappa)=\aleph_0$ are there $\sigma$-algebras of cardinality $\kappa$. (I changed the title to reflect this.)
Thanks Stephen, Matthew, Apollo, for the combined work!

REPLY [9 votes]: A Boolean algebra is $\sigma$-complete if every countable subset has a least upper bound
and a greatest lower bound.  Every $\sigma$-algebra is a $\sigma$-complete Boolean algebra.
Every (infinite) $\sigma$-complete Boolean algebra $B$ satisfies $|B|^{\aleph_0}=|B|$.
(I am almost certain that a proof of this fact is in the Handbook of Boolean Algebras, Volume 1.)
Since for every infinite cardinal $\kappa$, $\kappa<\kappa^{\text{cf}(\kappa)}$,
there is no infinite $\sigma$-algebra of some size $\kappa$ of cofinality $\aleph_0$.  
Something stronger is actually true:
By a result of Koppelberg [Boolean algebras as unions of chains of subalgebras, Algebra Universalis, Vol. 7 (1977), 195-203], no $\sigma$-complete Boolean algebra is the union of
a countable increasing chain of proper subalgebras.  This also implies that the
size of a $\sigma$-algebra cannot be of countable cofinality.