Is $P(tautology) = 1$? What are the connections between logic and probability? It's well-known that sets are "isomorphic" to logic: if we treat $\varphi(A_1, A_2)$ as a shorthand for $\forall x: \varphi(x \in A_1, x \in A_2)$ then $A \land B \equiv A \cap B$ and $A \rightarrow B \equiv A \subseteq B$ and so on.
I've noticed that a large number of true logical statements become events with probability 1 when interpreted probabilistically. For example, if $A \subseteq B$ ($\equiv A \rightarrow B$) then $\mathbb{P}(B|A) = 1$. If you squint hard enough you should see modus ponens there.
To connect a Boolean algebra with a Boolean ring you set $x \lor y := x + y - xy$, and wouldn't you know it, $\mathbb{P}(A \cup B) = \mathbb{P}(A) + \mathbb{P}(B) - \mathbb{P}(A \cap B)$. That connection can't just (ahem) be a random event, can it? ;-)
If we combine some propositional calculus and/or a Boolean algebra with measure/probability theory, can we get some theorems for free? Is it e.g. the case that if $\varphi$ is some tautology then the set-theoretic interpretation of $\varphi$ always has probability 1? Is there something stronger that's also true?
I also notice that $\mathbf{0}$ and $\mathbf{1}$, by which I mean the empty set and the set of all outcomes, are independent from all other events, and that I run into problems with Huntington's equation when I set $\lnot x := 1 - x$ and try to make a Boolean algebra over $[0, 1] \subseteq \mathbb{R}$, to do particularly with higher-order terms.
What are the theorems I'm grasping at but not quite seeing?
 A: Every $\sigma$-algebra over some $\Omega$ is a Boolean algebra with $\Omega = 1$. The 1-element of every Boolean algebra is unique. Thus, if $φ$ provable using e.g. the sound and complete Russel-Bernays axioms with the associated deduction rule and we uniformly substitute in members of our $\sigma$-algebra for the variables in φ and replace disjunction/negation with union/complement, the result must be the unique 1-element in our $\sigma$-algebra, i.e. $\Omega$. But $P(\Omega) = 1$ for every probability measure $P$ by definition, so every tautology has probability 1 (for this value of tautology).
It's easy to show that every $\sigma$-algebra is a Boolean algebra by using the Huntington axiomatization (the "fourth set" on page 7 in Huntington's PDF article).
I assume for now that my failure to make a (Boolean) ring homomorphism out of a probability measure is because it doesn't work in general.  I'm not sure what to make of the similarity between the union rule for probability and disjunction in Boolean rings. Maybe "$\mathbb{P}$ preserves some of the structure, but at the expense of other parts".
A: "To connect a Boolean algebra with a Boolean ring you set x∨y:=x+y−xy, and wouldn't you know it, P(A∪B)=P(A)+P(B)−P(A∩B). That connection can't just (ahem) be a random event, can it?"
One can maintain that it kind of is.  Setting (x$\lor$ y) := max(x, y) comes as simpler.  But, P(AUB) does not equal max(A, B).  This also kind of dovetails with a discussion about possibility theory, which uses possibility measures instead of probability measures.
"If we combine some propositional calculus and/or a Boolean algebra with measure/probability theory, can we get some theorems for free?"
Plenty of propositional calculi can't get combined with a Boolean algebra and consistency get maintained.  The propositional calculus has to come as a classical one for that to work, and there exist plenty of non-classical propositional calculi.  So, it can't be an arbitrary propositional calculus that you select for this sort of thing.
"Is it e.g. the case that if φ is some tautology then the set-theoretic interpretation of φ always has probability 1?"
Are you asking for Cantorian/classical set theory? Because there exist non-Cantorian/non-classical set theories also.
