Consider the following measure:
$$f(a, b)=\int_{[a,b]} \frac{1}{dist(x)} d\mu(x) $$
Where dist(x) is the distance from x to the closest prime and $\mu$ is the Lebesgue measure. Some subsets of $\mathbb{R}$ are not measurable using this function, such as [1, 2]. I want to know how many of these subsets are measurable, compared to the Lebesgue measurable sets. This seems like something you would use a measure for. What are some measures on the set of all subsets of $\Bbb{R}$?
Note: I spoke with someone about this, and they came up with the following measure: The unique measure that satisfies $\lambda(\{ s \text{ if } \mu(s) \in [a, b] \} )=b-a$ (Couldn't figure out the {...}), but I don't like this one because intuitively, there are more subsets of $\Bbb{R}$ with measure in [3, 4] than in [2, 3]
When $g$ is any measurable and almost everywhere non-negative real function, $\lambda_g(A) = \int_A f(x) \, d\mu(x)$ is a measure on the Borel subsets of $\mathbb{R}$. It means that any Borel subset of $\mathbb{R}$ is measurable for $\lambda = \lambda_{1/\mathrm{dist}}$.
Your question is more about : which ones have a finite measure ? In this case, indeed, $\lambda([1,2]) = +\infty$ because $\frac{1}{\mathrm{dist}(x)} = \frac{1}{2 - x}$ on $[1,2]$ is not integrable. In general, if you are close enough to a prime number $p$, $\frac{1}{\mathrm{dist}(x)} = \frac{1}{|p - x|}$ is not integrable in a neighborhood of $p$.
It is hard to say exactly which borel subsets of $\mathbb{R}$ have a finite $\lambda$-measure but you can give an answer for intervals. Indeed, since $\frac{1}{\mathbb{dist}(x)}$ is locally integrable everywhere except around prime numbers, a relatively compact interval $I = [a,b]$, $]a,b]$, $[a,b[$ or $]a,b[$ has finite measure if and only if $[a,b] \cap \mathbb{P} = \overline{I} \cap \mathbb{P} = \emptyset$.
For infinite intervals, $[a,+\infty[$, $]a,+\infty[$ and $]-\infty,+\infty[$ obviosuly have infinite measure and, $$ \lambda(]-\infty,a[) = \lambda(]-\infty,a]) = \int_a^\infty \frac{dx}{\mathrm{dist}(x)} \geqslant \int_{\min(a,2)}^\infty \frac{dx}{2 - x} = +\infty. $$ Thus, all infinite intervals have infinite $\lambda$-measure. Notice that if you replace $\frac{1}{\mathrm{dist}}$ by $\frac{1}{\mathrm{dist}^\alpha}$ for any $\alpha < 1$, then for any interval $I$ of $\mathbb{R}$, $\mu(I) < +\infty \Leftrightarrow \lambda_{1/\mathrm{dist}^\alpha}(I) < +\infty$ because $\frac{1}{\mathrm{dist}^\alpha}$ is locally (and not globally) integrable.