For what it's worth, $F^\infty$ cannot be complete in any metric, by the Baire category theorem. (It is the union of countably many finite-dimensional subspaces, which are closed sets of empty interior).

@MohammadGolshani Thanks, good to know.

I don't know enough about forcing to know how products of random real forcings of different cardinals differ from ordinary random real forcing.

@MohammadGolshani The measure algebras of probability spaces are not necessarily Maharam homogeneous. They are countable products of Maharam homogeneous ones. This is described in Maraham's article, Theorems 1 and 2: ncbi.nlm.nih.gov/pmc/articles/PMC1078424 Fremlin's version is 332B.

I particularly like Grothendieck's version of this. He proves that unconditional convergence is the same as absolute convergence for a locally convex space if and only if it is nuclear, i.e. every continuous linear map to a normed space is a nuclear map. The falsity of this belief then follows from the fact that a nuclear normed space is finite dimensional.

In case this is not commonly known, the weight $w(X)$ of a topological space $X$ is the smallest cardinality of a base for the topology.

@YCor The Haar measure on a compact group can be taken to be a probability measure, so Borel sets modulo ~ is a $\sigma$-complete Boolean algebra with the c.c.c., and therefore a complete Boolean algebra.

@MohammadGolshani This poster is saying that the complete Boolean algebra you get from the measure algebra (i.e. Borel sets modulo nullsets) of $G$ is isomorphic to the measure algebra of $2^{w(G)}$, with its usual coin-flipping measure. This is the complete Boolean algebra for adding $w(G)$ random reals. A good reference for the statements I have just made is Fremlin's Measure Theory, volume 3. The wikipedia article on Maharam's theorem is not so good, I've been meaning to fix it when I have the time.

@DenisNardin The condition is given in Bourbaki's General Topology, Chapter 1, section 1.2 and called $\mathrm{V}_\mathrm{IV}$. It is that if $V$ is a neighbourhood of $x$, there exists a neighbourhood $W$ of $x$ such that for all $y \in W$, $V$ is a neighbourhood of $y$.

Is there anything to stop us from just calling it a "Lipschitz space"?

How does this equivalence work? The obvious one does not seem to do, for the following reason: Every $\mathcal{G}$ algebra admits $\sigma$-convex combinations, but not every convex space does. For example, the usual structure of a convex space on $\mathbb{R}$ cannot be extended to $\sigma$-convex combinations, by a variation of the argument $0 = (1 - 1) + (1 - 1) + ... = 1 - (1 - 1) - (1 - 1) + ... = 1$ using $\sigma$-convex combinations to express the sums. So we cannot obtain the usual convex structure on $\mathbb{R}$ as the underlying convex space of a $\mathcal{G}$-algebra.

@SamHopkins I suspect you mean "critical" rather than "crucial". Although some senses of these words overlap in meaning, I have never heard of "crucial" being used in percolation theory.

I think you are looking for the notion of a state on an orthomodular lattice. But this is not defined in quite this way, probably because of the "common false belief" mentioned by Pietro Majer. Instead it is that $\mu(A \cup B) = \mu(A) + \mu(B)$ if $A \cap B = \bot$.

Functors are defined on the second page of that paper (though only for the category of groups).