While answering another question in MSE, I had used the following result which I thought was a trivial consequence of the prime number theorem and equidistribution. However, I realized from the comments that many people thought that this was not either true or counter intuitive. Hence I am posting this as a question looking for a proof or disproof.
Let $p_k$ be the $k$-th prime and $f$ be a continuous function
Riemann integrable in $(0,1)$ such that$$\lim_{n \to \infty}\frac{1}{n}\sum_{r = 1}^{n}f\Big(\frac{r}{n}\Big)
= \int_{0}^{1}f(x)dx. $$Then, $$ \lim_{n \to \infty}\frac{1}{n}\sum_{r =
1}^{n}f\Big(\frac{p_r}{p_n}\Big) = \int_{0}^{1}f(x)dx. $$
My proof was based on showing that as $n \to \infty$, the ratios $p_r/p_n$ approached equidistribution in $(0,1)$ hence the integral follows as a trivial property of equidistributed sequence.
Motivation: There are several identities, limits etc on prime numbers which can be easily proven using this simple formula, including all answers to all three questions on the arithmetic, geometric and harmonic means of primes mentioned in the above link.
Best Answer
Too long for a comment
Divide $[0,1]$ into $$[0,p_1/p_n],[p_1/p_n,p_2/p_n],\cdots,[p_{n-1}/p_n,1]$$
Then, using Riemann sum, we have $$I:=\int^1_0f(x)dx=\lim_{n\to\infty}\sum^n_{k=1}f\left(\frac{p_k}{p_n}\right)\frac{p_{k+1}-p_k}{p_n}$$
If we assume that $p_j=j\ln j$, $$I=\lim_{n\to\infty}\sum^n_{k=1}f\left(\frac{p_k}{p_n}\right)h(k,n)+\lim_{n\to\infty}\frac1n \sum^n_{k=1}f\left(\frac{p_k}{p_n}\right) \qquad{(1)}$$ where $$h(k,n)=\frac{(k+1)\ln(k+1)-k\ln k}{n\ln n}-\frac1n$$
It can be shown that $$h(k,n)\le h(n,n)=O(\frac1{n\ln n})$$
Therefore, the absolute value of the first term in $(1)$ is upper bounded by $$h(n,n)\cdot nM\to 0$$ where $M$ is a positive constant. This leads us to our desired result.
I am not sure if this argument can be made rigorous. I will review it when I have leisure time.