Is it possible to construct probability theory which is not based on measure theory but on logic? A probability space is a triplet $(\Omega, F, P)$, where $\Omega$ is a set, $F$ is a $\sigma$-algebra on $\Omega$ and $P$ is a probability measure on $(\Omega, F)$.

However, do we need $\Omega$?

What if we define as follows:


A probability set $G$ is a set of propositions which satisfies


$\bot \in G$
If $p \in G$, then $\lnot p \in G$
If $p_1, p_2, \dots \in G$, then $\lor_{i=1}^\infty p_i \in G$

A probalitiy function $P$ is a function from $G$ to $[0, 1]$ which satisfies


$P(\top)=1$
$P(\lor_{i=1}^\infty p_i) = \sum_{i=1}^\infty P(p_i)$ if $p_i \land p_j$ is false when $i \neq j$
Yes.  If you think about it, you need $\Omega$ and $P$ first. Or else for a given $p\in\Omega$, $\neg p$ does not make sense.  Since $$\neg p = \{q\in\Omega | q \neq p\}$$
The $\sigma$-algebra $F$ is in some sense "What events in $\Omega$ satisfy conditions of $P$"
Yes.  If you think about it, you need $\Omega$ and $P$ first. Or else for a given $p\in\Omega$, $\neg p$ does not make sense.  Since $$\neg p = \{q\in\Omega | q \neq p\}$$
The $\sigma$-algebra $F$ is in some sense "What events in $\Omega$ satisfy conditions of $P$"