Let $f:[0,1]\rightarrow \mathbb{R}$ be a Riemann integrable function and let $p$ be a real number with $0<p<1$. Prove that
$$ \frac{1}{n}\,\sum_{j=0}^nf\left(\frac{j}{n}\right)\left(1-p^{j}\right)\rightarrow \int_0^1f\left(t\right)\,dt $$
Comments Of course, by Riemann integrability, we have that
$$ \frac{1}{n}\,\sum_{j=0}^nf\left(\frac{j}{n}\right)\rightarrow \int_0^1f\left(t\right)\,dt $$
The idea is to define the sequence of functions
$$ g_n(t)=f(t)\,(1-p^{\left\lfloor t*n\right\rfloor}). $$
so that $g_n(t)\rightarrow f(t)$ point-wise on $[0,1]$. However I cannot Prove that the convergence is uniform since
$$ \sup_{t\in[0,1]}\left|g_n(t)-f(t)\right| = \sup_{t\in[0,1]}p^{\left\lfloor t*n\right\rfloor}=1. $$
We have
$$\frac{1}{n}\,\sum_{j=0}^nf\left(\frac{j}{n}\right)\left(1-p^{j}\right) = \frac{1}{n}\,\sum_{j=0}^nf\left(\frac{j}{n}\right)- \frac{1}{n}\,\sum_{j=0}^nf\left(\frac{j}{n}\right)p^{j},$$
and it is enough to show the second term converges to $0$.
Since $f$ is Riemann integrable and, hence, bounded,
$$\left|\frac{1}{n}\,\sum_{j=0}^nf\left(\frac{j}{n}\right)p^{j} \right| \leqslant \frac{1}{n}\sum_{j=0}^n\left|f\left(\frac{j}{n}\right)\right|p^{j} \\ \leqslant \frac{\sup_{x \in [0,1]} |f(x)|}{n}\frac{1- p^{n+1}}{1-p} \\ \leqslant \frac{\sup_{x \in [0,1]} |f(x)|}{n(1-p)}\\ \longrightarrow_{n \to \infty} 0$$