How to prove this conjecture？
Conjecture $f$ is Riemann integrable function in$[0,1]$,then $$ \lim_{n\to\infty}\frac{1}{\varphi(n)}\sum_{\substack{1\le k\le n\\(k,n)=1}}f\left(\frac{k}{n}\right)=\int_0^1 f(x)\mathrm{d}x $$
similar to Weyl's criteria？
The following method gives a proof with the assumption that $f(x)$ is of bounded variation in $[0,1]$. It is possible to generalize this to loose conditions.
For convenience, we define:
\begin{aligned} S(x) &=\sum_{\substack{1\le k\le x\\(k,n)=1}}1=\sum_{d|n}\mu(d)\left\lfloor\frac xd\right\rfloor \\ &={\varphi(n)\over n}x+\mathcal O(2^{\omega(n)}) \end{aligned}
This implies that
$$ Q(x)=\sum_{1\le k\le x}\left[S(n)-S(n-1)-{\varphi(n)\over n}\right]\ll2^{\omega(n)} $$
then we have
\begin{aligned} \sum_{\substack{1\le k\le n\\(k,n)=1}}f\left(\frac kn\right) &={\varphi(n)\over n}\sum_{1\le k\le n}f\left(\frac kn\right)+\sum_{1\le k\le n}f\left(\frac kn\right)[Q(k)-Q(k-1)] \\ \end{aligned}
Certainly, the first sum will converge to $\int_0^1f(x)\mathrm dx$ as $n\to\infty$ when its divided by $\varphi(n)$, so the remaining task is to bound the second sum. By summation by parts, we have
\begin{aligned} E &=\sum_{1\le k\le n}f\left(\frac kn\right)[Q(k)-Q(k-1)] \\ &=\sum_{1\le k\le n}f\left(\frac kn\right)Q(k)-\sum_{0\le k\le n-1}f\left(k+1\over n\right)Q(k) \\ &=f(1)Q(n)-f(0)Q(0)-\sum_{0\le k\le n-1}Q(k)\left[f\left(k+1\over n\right)-f\left(\frac kn\right)\right] \\ \end{aligned}
Since bounded variation is assumed, we have
$$ E\ll2^{\omega(n)}[f(0)+f(1)]+2^{\omega(n)}\ll2^{\omega(n)} $$
Consequently, we have
$$ \lim_{n\to\infty}{1\over\varphi(n)}\sum_{\substack{1\le k\le n\\(k,n)=1}}f\left(\frac kn\right)=\int_0^1f(x)\mathrm dx+\lim_{n\to\infty}{\mathcal O(2^{\omega(n)})\over\varphi(n)} $$
Using the property of divisor function, we know that $2^{\omega(n)}\ll\sigma_0(n)\ll_\varepsilon n^\varepsilon$, so the latter limit will certainly vanish.