I think the usual way is to define a sequence $f_n$ of probability measures whose limit has whichever property you consider to be the essence of uniformity.

For example, let $f_n(A) = |A \cap \{1,2,\dots,n\} | / n$. Each $f_n$ is a probability measure, and the limit $\lim f_n(A)$ (if it exists) is called the natural density of $A$. Another example, let $f_n(A) = \zeta(1+1/n)^{-1} \sum_{k\in A} k^{-1-1/n}$. The limit $\lim f_n(A)$ (if it exists) is the logarithmic density of $A$.

Another approach is to take an unlimited hyperinteger (as in, nonstandard analysis) $N$, and to take the measure $f(A)$ to be the standard part of $|{}^\ast A\cap \{1,2,\dots,N\}|/N$. This has a tendency to be meaningless, however, unless you can prove a result that works for all unlimited hyperintegers. Essentially, this is the same difficulty that arises in Denis Serre's answer.

I think the usual way is to define a sequence $f_n$ of probability measures whose limit has whichever property you consider to be the essence of uniformity (invariance under translation, the evens have probability 1/2, finite additivitiy, et cetera).

For example, let $f_n(A) = |A \cap \{1,2,\dots,n\} | / n$. Each $f_n$ is a probability measure, and the limit $\lim f_n(A)$ (if it exists) is called the natural density of $A$. Another example, let $f_n(A) = \zeta(1+1/n)^{-1} \sum_{k\in A} k^{-1-1/n}$. The limit $\lim f_n(A)$ (if it exists) is the logarithmic density of $A$.

Another approach is to take an unlimited hyperinteger (as in, nonstandard analysis) $N$, and to take the measure $f(A)$ to be the standard part of $|{}^\ast A\cap \{1,2,\dots,N\}|/N$. This has a tendency to be meaningless, however, unless you can prove a result that works for all unlimited hyperintegers. Essentially, this is the same difficulty that arises in Denis Serre's answer.