Can you have perfect secrecy with countable message/key spaces by dropping countable additivity? This classic paper by Chor and Kushilevitz shows that if the key space and the message space are both countably infinite, then it is impossible to have a perfectly secure private-key encryption scheme.  Their proof relies on the fact there exists no uniform probability measure on the set of natural numbers, which in turn relies on the fact that probability measures are countably additive.
But there’s a generalized notion of measure that only requires finite additivity and not countable additivity.  In particular this paper talks about how the asymptotic density of a set of natural numbers constitutes a finitely additive probability measure that is “uniform” in the sense of translation invariance.
I’m wondering whether that could be used to recover some notion of perfect security.  So let the message space, key space, and ciphertext space all equal $\mathbb{N}$, let the family of measurable sets be the family $F$ of all subsets of $\mathbb{N}$ which have a well-defined asymptotic density, let the key be chosen using the asymptotic density measure, and let the adversary have some finitely additive probability measure $P$ over the message space.  Then my question is, does there exist an encryption scheme such that for all $X,Y\in F$ for which $P(C\in Y)\neq 0$, we have $P(M\in X | C\in Y) = P(M\in X)$? (Note that I’m using the same letter $P$ for bother he probability measure over the message space and the probability measure of the ciphertext.)
Of course dropping countable additivity may make this all unrealistic, but I’m just asking a theoretical question.
Let the key, message, and ciphertext spaces all be $\mathbb{Z}$, which is in bijection with $\mathbb N$. I'm going to pick a specific bijection by making the non-negative integers map to evens and the negative integers map to odds, and let $\mu$ be the asymptotic density measure applied to $\mathbb{Z}$ using this bijection. We can then construct a one-time pad by setting $c = m + k$. For any (possibly finitely-additive) probability measure $\mu_M$ chosen by the adversary on $m$, and any set $Y \in F$, $(\mu_M \times \mu)(\{(m, k) \in \mathbb Z^2 \mid m + k \in Y\}) = \mu(Y)$. That is, the resulting measure on the ciphertext is uniform, no matter what measure $m$ comes from. I am stating it this way so as to avoid the division inherent in the $P(m \in X | c \in Y)$ expression in your question.
Details: First, we need to show that $\mu$ is translation invariant on $\mathbb Z$. $\mu(A)$ could equivalently be defined as $\lim_{n \to \infty} \frac{A \cap [-n, n]}{2n + 1}$. To see that translation invariance holds, notice that $|[-n, n] \cap A|$ and $|[-n, n] \cap (A - x)|$ only differ by at most $2 x$, since $[-n, n]$ and $[-n + x, n + x]$ overlap everywhere except for $x$ points at either end. This becomes irrelevant in the limit as $n \to \infty$.
Next, we need to evaluate the product measure. Unfortunately product measures do not seem to be very well defined for finitely additive measures. I will pick one product measure, but in general it is not unique.
\begin{align*}
(\mu_M \times \mu)(\{(m, k) \in \mathbb Z^2 \mid m + k \in Y\})
&amp;= \int_{\mathbb Z} \mu(\{k \in \mathbb Z \mid m + k \in Y\}) d\mu_M(m) \\
&amp;= \int_{\mathbb Z} \mu(Y - m) d\mu_M(m) \\
&amp;= \int_{\mathbb Z} \mu(Y) d\mu_M(m) \\
&amp;= \mu(Y) \\
\end{align*}
Here we have used that $\mu$ is translation invariant on $\mathbb Z$ and that $\mu_M$ is a probability measure, so the integral (i.e., expected value) of a constant is that constant.
However, none of this is physically realizable, at least with a digital computer. Any way of generating a random key is going to give a distribution that has countable additivity, since it can always be broken down into discrete choices, with each outcome having paths leading to it with only have a finite number of choices. I don't think it is physically possible in general, but I have less justification for that.