TITLE: Riemann sum formula for definite integral using prime numbers
QUESTION [11 upvotes]: 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.

REPLY [19 votes]: The statement follows from the prime number theorem. By an approximation argument, we can assume that $f$ is continuously differentiable on $[0,1]$. Then,
$$\sum_{r=1}^{n}f\left(\frac{p_r}{p_n}\right)=\int_0^1 f(x)\,d\pi(p_n x)=nf(1)-\int_0^1 f'(x)\pi(p_n x)\,dx.$$
We estimate the last integral for fixed $f$:
$$\int_0^1 f'(x)\pi(p_n x)\,dx=\int_{\frac{1}{\log n}}^1 f'(x)\pi(p_n x)\,dx+o(n).$$
In the last integral, we have by the prime number theorem,
$$\pi(p_n x)\sim\frac{p_n x}{\log(p_n x)}\sim\frac{p_n x}{\log n}\sim nx,$$
so that
\begin{align*}\int_0^1 f'(x)\pi(p_n x)\,dx
&=\int_{\frac{1}{\log n}}^1 f'(x)\bigl(nx+o(nx)\bigr)\,dx+o(n)\\
&=\int_{0}^1 f'(x)\bigl(nx+o(nx)\bigr)\,dx+o(n)\\
&=n\int_0^1 f'(x)x\,dx+o(n)\\
&=nf(1)-n\int_0^1 f(x)\,dx + o(n).
\end{align*}
Putting everything together,
$$\sum_{r=1}^{n}f\left(\frac{p_r}{p_n}\right)=n\int_0^1 f(x)\,dx + o(n),$$
that is,
$$\frac{1}{n}\sum_{r=1}^{n}f\left(\frac{p_r}{p_n}\right)=\int_0^1 f(x)\,dx + o(1).$$