Equidistributed sequence and Riemann integrable function Let $f$ be a function of period 1, Riemann integrable on [0,1]. Let $\xi_n$ be a sequence
which is equidistributed in $[0,1)$. 
(a) Is it true that $$\frac{1}{N}\sum\limits_{n=1}^N f(x+\xi_n)$$ converges to the constant $\int_0^1 f(y) dy$
for each $x\in \mathbb{R}$ as $N\to \infty$ ?
(b) If so, is the convergence uniform over all $x$?
 A: For question (a), see this (the section "Properties"); for a proof, see this.
A: The answer to both questions (a) and (b) is yes, and is a consequence of the fact that Riemann integrable functions can be approximated from above and below by step functions. The proof follows in a fairly straightforward manner once you realize that this allows you to reduce the problem to that of indicator functions of intervals. To be precise, if $f\colon[0,1]\to\mathbb{R}$ is Riemann integrable then, for each fixed $\epsilon > 0$, there are step functions $g\le f\le h$ with $\int_0^1(h(y)-g(y))\,dy < \epsilon$.
$$
\begin{align}
\frac1N\sum_{n=1}^Nf(\xi_n)-\int_0^1 f(y)\,dy&\le\frac1N\sum_{n=1}^Nh(\xi_n)-\int_0^1 h(y)\,dy+\int_0^1(h(y)-f(y))\,dy\\
&\le\frac1N\sum_{n=1}^Nh(\xi_n)-\int_0^1 h(y)\,dy+\epsilon.
\end{align}
$$
Similarly, the reverse inequality holds with $g$ replacing $h$ and $-\epsilon$ replacing $\epsilon$ on the right hand side. So, convergence for step functions implies convergence for Riemann integrable functions. Also, replacing $\xi_n$ by $x+\xi_n$ (mod 1) above, the problem of uniform convergence over all $x$ is also reduced to that of step functions. Linearity further reduces the problem to that of indicator functions of intervals in $(0,1]$.
However, if $f=1_A$ is an indicator function of an interval $A\subseteq(0,1]$ then the limit $\frac1N\sum_{n=1}^Nf(x+\xi_n)=\int_0^1f(y)\,dy$ follows by definition of equidistribution.
We can also obtain uniform convergence (in $x$) as follows: note that $y\mapsto f(x+y)=1_{\{y\in A-x{\rm\ (mod\ 1)}\}}$ is again an indicator function of an interval (taken mod 1). Convergence is uniform simultaneously on all intervals. For any $M > 0$ consider the finite collection of intervals $B_i=((i-1)/M,i/M]$. Then, any interval $A\subseteq(0,1]$ can be sandwiched between unions of the $B_i$. That is, $\cup_{i\in I}B_i\subseteq A\subseteq\cup_{j\in J}B_j$ for $I,J\subseteq\{1,2,\ldots,M\}$ with $I\subseteq J$ and $J\setminus I$ containing no more than two elements.
$$
\frac1N\sum_{n=1}^N 1_A(\xi_n)-\int_0^1 1_A(y)\,dy\le\sum_{i\in J}\left(\frac1N\sum_{n=1}^N 1_{B_i}(\xi_n)-\int_0^1 1_{B_i}(y)\,dy\right) + 2/M.
$$
The reverse inequality also holds if we make the sum run over $i\in I$ and replace $2/M$ by $-2/M$ on the right hand side. So,
$$
\left\vert\frac1N\sum_{n=1}^N 1_A(\xi_n)-\int_0^1 1_A(y)\,dy\right\vert\le\sum_{i=1}^M\left\vert\frac1N\sum_{n=1}^N 1_{B_i}(\xi_n)-\int_0^1 1_{B_i}(y)\,dy\right\vert+2/M.
$$
For any $\epsilon > 0$ we can choose $2/M < \epsilon/2$ and, using equidistribution, $N$ can be chosen large enough that the first term on the right hand side is less than $\epsilon/2$. So, the left hand side can be made less than $\epsilon$ by choosing $N$ large enough. This choice was independent of $A$ so, in fact, convergence is uniform on the set of all intervals $A$.
