Measure theory and probability theory are deeply connected through the interpretation of subset measures on the sample space as probabilities of events.
A major (and somewhat disturbing) result from measure theory is the existence of non-measurable sets, even in concrete cases such as that of the Lebesgue measure on $\mathbb R$. What is even more disturbing is that the existence of such sets appears conditional on us believing in the Axiom of Choice (see Solovay model).
Probability theory usually avoids this problem by choosing event algebras that ensure measurability (such as the Borel algebra if the sample space is a subset of $\mathbb R^n$). Does, then, the existence of non-measurable sets have any implications on probability at all?
I'm especially interested in the following question: Assuming a model of set theory that does not include the Axiom of Choice (where it is not possible to prove the existence of non-measurable subsets of the reals), we should be able to explore probability spaces $(\mathbb R,2^{\mathbb R},P)$ instead of being limited to just $(\mathbb R,B(\mathbb R),P)$, that is, use the full power set as an event space in place of the Borel algebra. From a probabilistic point of view, what would that mean?
Well, from a probabilistic point of view, you might not be able to have countable additivity at all. Since it might be that $\Bbb R$ is the countable union of countable sets, then the only $\sigma$-additive measure which gives $0$ to singletons, is the trivial measure.
In that case you can still do some measure theory or probability, but you need to reduce to a strict subset of $2^\Bbb R$, which is the sets which have Borel codes. This alleviates some problems but comes with the cost of having to chase after codes all the time.
Of course, you can assume instead that $\sf DC$ or some sufficient choice principle holds that you can still develop basic measure theory, and basic probability. Then you can do most of what you can do in the usual aspects of measure theory, but you cannot use Hahn-Banach or functional analytic tools. If you don't use those very much, then it shouldn't matter.
But such universes come with their own price. In this sort of universe you can partition $\Bbb R$ into pairwise disjoint non-empty sets, and have strictly more sets than elements of $\Bbb R$. And that's just strange. Even if it doesn't have many measure theoretic implications.
(It should be added, by the way, that the additional cost of having measure theory behave nicely while all sets are measurable is that you need to increase in consistency strength, namely the theory you need to believe is consistent is now strictly stronger than just $\sf ZFC$. It's not a very large jump compared to other similar jumps, but it's one nonetheless, and if people have trouble with the consistency of set theory, asking them to believe even stronger theories are consistent is a different philosophical problem.)