Expressing equal probability on an infinite line with probability axioms Is there any way using the usual (Kolmogorov) axioms of probability to describe/model the following situation :
A value $v \in \mathbb{R}$ has an equal probability of being measured anywhere in the interval $(-\infty, +\infty)$. What we want to express is that :


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*The probability of getting some value, in other words, some $v \in (-\infty, \infty)$, is 1.

*If there are two intervals $A$ and $B$, and $|A|=k\ |B|$, then the probability of $v$ being in $A$ is $k$ times the probability of $v$ being in $B$.
When attempting to model it with probability axioms, since the interval of all possible values for $v$ is infinite, each finite interval in $\mathbb{R}$ will have probability $0$. Countably adding the probability of finite intervals to get the entire $\mathbb{R}$ will then still yield $0$, which is not compatible with the axiom that $P(\mathbb{R})=1$. Also, there seems to be no way to express 2 above.
 A: Your argument is solid. In fact, given any infinite set $X$ for which there is a countably-infinite partition $\mathscr P$ into sets of equal cardinality, there is no uniform probability measure on $X$ making all $\mathscr P$-sets measurable.
A: Depending upon the source, there are two sets of axioms that are sometimes called "Kolmogorov's axioms."


*

*One set of axioms requires that probability be finitely additive, i.e., for any finite collection of disjoint events $A_1, \ldots, A_n$, the probability of the union $P(\bigcup_{i \leq n} A_i)$ is equal to the sum $\sum_{i \leq n} P(A_i)$.

*Often times, however, Kolmogorov's axioms are said to require countable additivity, i.e., for any countable collection of disjoint events $\{A_n: n \in \mathbb{N}\}$, the probability of the union $P(\bigcup_{n \in \mathbb{N}} A_n)$ is equal to the sum $\sum_{n \in \mathbb{N}} P(A_i)$.
The latter condition is obviously strictly stronger than the former.
Your two conditions are possible to fulfil if only finite additivity is required of a probability measure, but they are impossible to satisfy if a probability measure is required to be countably additive.
To see that they cannot be satisfied for a countably additive measure,
note that the intervals $\{[z,z+1): z \in \mathbb{Z}\}$ cover the real number line.  By countable additivity, at least one of those intervals must have positive probability if the entire real line has positive probability.  Now if any interval $A$ has positive probability $r$, then the interval $[0,1]$ has probability $1/|A|$ by your second condition.  By finite additivity (which is entailed by countable additivity), it follows that the intervals $[0,1), [1,2), \ldots [M, M+1)$ have probability greater than $1$, where $M$ is the ceiling of $(|A|)+1$.  That's obviously impossible.
If probability is only required to be finitely additive, however, it is possible to define a measure that satisfies your two conditions.  The reason is that one can assign every interval of finite length probability zero, and so your second condition is immediately satisfied.  A good source to consult is Savage's Foundation of Statistics, which contains a discussion of the importance of finitely additive probability measures.
