By using Fourier Series Prove that $\frac{1}{n+1}\sum_{k=0}^n f(x+k\alpha)\to\int_0^1 f(y)dy$. Here we assume that $f$ is an $1$-periodic continuous function, $x\in[0,\,1]$ and $\alpha$ is an irrational number in $(0,\,1)$.
I couldn't recall which theorem in the Fourier series has a similar form to the limit I want to prove.
 A: The OP is basically Weyl's theorem on equi-distribution.
This result can be proved by methods of Calculus (Fourier series as required by the OP) with a little help from the Stone-Weierstrass theorem about density of trigonometric polynomials on the space of continuous functions on the circle (or continuous periodic functions).
Consider the unit circle $\mathbb{S}^1=\{z\in\mathbb{C}:|z|=1\}$ equipped with the $\sigma$-algebra inherited as a subspace of $\mathbb{R}^2$ and the measure $\lambda_{\mathbb{S}^1}:=\frac{1}{2\pi}\lambda_1$, where $\lambda_1$ is the the arch-length measure. (Equivalently, this is the real line  $\mod 1$ with the Borel $\sigma$-algebra and Lebesgue measure restricted to $[0,1]$.)
For any $\theta\in(0,1)$ define the rotation map
$$ 
\begin{align}
R_\theta&:\mathbb{S}^1\rightarrow\mathbb{S}^1\\
z&\mapsto z e^{2\pi i\theta}
\end{align}
$$
Or equivalently, if $z=e^{i2\pi x}$, $R_\theta(x)=x+\theta\mod 1$
Define $R^0_\theta(z)=z$, and or $k\geq1$, $R^k_\theta(z)=R_\theta(R^{k-1}_\theta(z))$. That is,
$$R^k_\theta(z)=ze^{2\pi i k\theta}, \qquad z\in\mathbb{S}^1$$
or equivalently, in terms of sum $\mod 1$, if $z=e^{2\pi i x}$,
$$ R^k_\theta(x)=x+k\theta\mod 1$$

Theorem (Weyl): Suppose $R_\theta$ is an irrational rotation. Then, for every bounded Riemann integrable function $f:\mathbb{S}^1:\rightarrow\mathbb{C}^1$, and any $z\in\mathbb{S}^1$, (or equivalently, any $1$-periodic continuous function on the real line Riemann integrable in $[0,1]$)
$$
\lim_\limits{n\rightarrow\infty}\frac{1}{n}\sum^{n-1}_{k=0}f\circ R^k_\theta(z)=\int_{\mathbb{S}^1} f(w)\,\lambda_{\mathbb{S}^1}(dw)
$$
or in terms of the real line $\mod 1$, if $z=e^{2\pi ix}$
$$
\lim_\limits{n\rightarrow\infty}\frac{1}{n}\sum^{n-1}_{k=0}f(x+k\theta)=\int^1_0 f(t)\,dt
$$

Here is a sketch of the proof:
For any function $f$ on $\mathbb{S}^1$, let $S_nf(z)=\frac{1}{n}\sum^{n-1}_{j=0}f(R^j_\theta(z))$. Consider polynomials $f_k(z)=z^k$, $k\in\mathbb{\mathbb{Z}}$.
For $k=0$, $f_0\equiv 1$ and so,
$$\begin{align}S_nf_0(x)\equiv1&=\int_{\mathbb{S}^1} f_0(z)\,dw\\
&=\frac{1}{2\pi}\int^{2\pi}_0f_0(e^{ikx})\,dx=\int^1_0f_0(e^{2\pi it})\,dt
\end{align}
$$
For $|k|\geq1$,
$$\begin{align}
S_nf_k(z)&=\frac{z^k}{n}\sum^{n-1}_{j=0}e^{i2\pi\theta k j}=
\frac{z^k}{n}\frac{1-e^{in 2\pi k\theta}}{1-e^{i2\pi k\theta}}\xrightarrow{n\rightarrow\infty}0\\
&=\int_{\mathbb{S}^1}f_k\,\lambda_{\mathbb{S}^1}=\frac{1}{2\pi}\int^{2\pi}_0 e^{ikx}\,dx=\int^1_0 e^{2\pi ikt}\,dt
\end{align}$$
since $\theta$ is irrational and so, $e^{i2\pi k\theta}\neq1$ for all $k\in\mathbb{Z}$.
This means that the statement holds for all trigonometric polynomials. By the (complex) Stone-Weierstrass theorem, the result then holds for any $f\in\mathcal{C}(\mathbb{S}^1)$.

To extend the result to any  Riemann integrable function $f$, it is enough to assume that $f$ is real valued. Given $\varepsilon>0$, we can choose continuous functions $g_\varepsilon$ and $h_\varepsilon$ such that
$ g_\varepsilon < f\leq h_\varepsilon$,  and
$$ \int_{\mathbb{S}^1}f\,\lambda_{\mathbb{S}^1} -\varepsilon <\int_{\mathbb{S}^1} g_\varepsilon\, \lambda_{\mathbb{S}^1}\leq \int_{\mathbb{S}^1}h_\varepsilon \,\lambda_{\mathbb{S}^1}<  \int_{\mathbb{S}^1}f\,\lambda_{\mathbb{S}^1} +\varepsilon
$$
Then
$$
S_ng_\varepsilon-\int g_\varepsilon -\varepsilon \leq S_nf(z)-\int f \leq  S_nh_\varepsilon(z)-\int h_\varepsilon+\varepsilon
$$
whence we conclude that $$-\varepsilon\leq \liminf_{n\rightarrow\infty}S_nf(z)-\int f\leq\limsup_n S_nf-\int f\leq\varepsilon$$
for all $\varepsilon>0$ and $z\in\mathbb{S}^1$. This completes the proof.
A: Otherwise there is a solution using only Parseval's theorem.
For $f$ continuous $1$-periodic let $$F_n(x)= \frac1{n+1}\sum_{k=0}^n f(x+k\alpha)- \int_0^1 f(t)dt$$
Note that $\sup_x |F_n(x+h)-F_n(x)| \le \sup_x |f(x+h)-f(x)|$, ie. the family $(F_n)_{n\ge 1}$ is uniformly continuous.
We find the Fourier coefficients
$$c_0(F_n)=0,\qquad c_m(F_n)=\int_0^1 F_n(x)e^{-2i\pi mx}dx = c_m(f)\frac1{n+1}\frac{1-e^{2i\pi (n+1)m\alpha}}{1-e^{2i\pi m\alpha}}$$
So $|c_m(F_n)| \le |c_m(f)|$ and $\lim_{n\to \infty} c_m(F_n)=0$. This implies that
$$\lim_{n\to \infty}\int_0^1 |F_n(x)|^2dx=\lim_{n\to \infty}\sum_m |c_m(F_n)|^2= 0$$ Which in turn, by uniform continuity of the $F_n$ family, implies that $$\lim_{n\to \infty}\sup_x |F_n(x)|= 0$$
