Suppose we fix a quantity $x\geqslant 0$, and consider the set
$$
P(x) = \left\{\frac{p}{\pi(x)} : p\leqslant x \text{ prime}\right\} \subset[0,1] .
$$
As $x\to \infty$, it seems like the $P(x)$ are converging to a measure. Here are a couple histograms:
(Primes $\leqslant 10^5$.)
(Primes $\leqslant 10^{10}$.)
Is the limiting distribution known? That is, can these pictures be made rigorous in a reasonable way and proven (possibly assuming the Riemann Hypothesis?)
Edit: here is a more precise statement.
For a real number $x$, let $P_x = \frac{1}{\pi(x)} \sum_{p\leqslant x} \delta_{p/x}$. This is a discrete measure supported on the interval $[0,1]$. I claim that $$ P_x \approx \frac{x}{\mathrm{Li}(x)\log(t x)}\, \mathrm{d} t $$ in the sense that if $L_x$ denotes the second measure and $$ D(P_x,L_x) = \sup_{0\leqslant t \leqslant 1} \left| P_x[0,t] - L_x[0,t]\right| = \sup_{0\leqslant t \leqslant 1} \left| \frac{\# \{p\leqslant x : p\leqslant t x\}}{\pi(x)} - \frac{\mathrm{Li}(t x)}{\mathrm{Li}(x)}\right| $$ is the discrepancy (Kolmogorov-Smirnov statistic), then $D(P_x,L_x) \ll x^{-\frac 1 2+\epsilon}$.
Is this claim known to be equivalent to the Riemann Hypothesis?