Is a Boolean algebra with an order continuous topology a measure algebra? Assume that $B$ is a complete boolean algebra endowed with a Hausdorff topology, with respect to which all operations on $B$ are continuous, $0$ has a base of full sets (recall that $A\subset B$ is full if $b\le a\in A\Rightarrow b\in A$), and for any net $(b_i)$ in $B$ which decreases to $0$ we have $b_i\to 0$.

What can be said about $B$? Is $B$ a measure algebra? Can existence of such a topology characterized in terms of $B$?

Generally, what is a good references on topologies on Boolean algebras?
 A: It is not true that $B$ is necessarily a measure algebra. The counterexample is due to Michel Talagrand, who constructed a Maharam algebra that is not a measure algebra.

Maharam, D., An algebraic characterization of measure algebras, Ann. Math. (2) 48, 154-167 (1947). ZBL0029.20401.


Talagrand, Michel, Maharam’s problem, Ann. Math. (2) 168, No. 3, 981-1009 (2008). ZBL1185.28002.

The relevant material can also be found in Fremlin's textbook Measure Theory, specifically in Chapter 39.
First I'll give some background, which I think will be helpful for your more general question, as well as relating Talagrand's example to your more specific question. A submeasure on a Boolean algebra $A$ is a function $A \rightarrow [0,\infty]$ that:

*

*$\mu(0) = 0$

*$\mu$ is monotone

*$\mu$ is subadditive: $\mu(a \lor b) \leq \mu(a) + \mu(b)$.

A finite submeasure is one where $\mu(1) < \infty$, and a positive (sometimes called strictly positive) submeasure is one such that $\mu(a) = 0$ implies $a = 0$.
Proofs about finitely-additive measures that only use axioms 1 to 3 without using additivity for disjoint elements go through for submeasures. In particular, if $\mu$ is finite we have a pseudometric on $A$:
$$
d(a,b) = \mu(a \triangle b)
$$
which is a metric iff $\mu$ is positive.
It is not difficult to prove that for all $\mu$, the functions $\lnot, \land, \lor$ and $\mu$ are uniformly continuous with respect the metric $d$. The monotonicity of $\mu$ implies that an $\epsilon$-ball around $0$ is down-closed (a "full set" in your terminology) so $0$ has a base of full sets.
If we try out the submeasure that has $\mu(0) = 0$ and $\mu(a) = 1$ for all $a \neq 0$, then $d$ is the discrete metric with distance $1$ between distinct elements. This shows we need an extra assumption to get your third requirement on the topology on $A$. In fact we only need to add a version of it for sequences:
A submeasure is continuous (called Maharam in Fremlin 393A) if for each non-increasing sequence $(a_n)_{n \in \mathbb{N}}$ with $\bigwedge_{n=0}^\infty a_n = 0$ we have $\lim_{n \to \infty}\mu(a_n) = 0$.
If $A$ is $\sigma$-complete and $\mu$ is finite, strictly positive and continuous, then $A$ is c.c.c. (Fremlin 393C), which is to say a disjoint set of non-zero elements is countable. It follows that $\mu$ preserves infima of downward directed nets (see e.g. Fremlin 316F(c), though try your own proof using Zorn's lemma or the well-ordering theorem). Therefore your third condition is also satisfied, and also $A$ is a complete Boolean algebra.
So a (sigma-)complete Boolean algebra with a continuous positive finite submeasure (known as a Maharam algebra) will satisfy your criteria. By Theorem 1.2 of Talagrand's paper, there are such algebras that are not measure algebras, in fact admitting no non-zero finite measures at all.
You can find many interesting proofs and references to the literature for results characterizing both Maharam algebras and measure algebras in Fremlin's §393, e.g. 393E, as well as in Talagrand's article.
