I had asked this question in MSE. It got lot of upvotes but no answer (except one which was too long to be posted as a comment) hence I am posting it in MO.

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 approach**: It was 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 property of equidistributed sequence.

**Motivation**: There are several identities, limits etc on prime numbers which can be easily proven using this simple formula.