[proofplan]
Fix an interval $I\subset\mathbb R/\mathbb Z$ and compare its indicator with Beurling-Selberg trigonometric polynomials of degree $H$ from above and below. Averaging these polynomial inequalities over the points $x_1,\dots,x_N$ expresses the discrepancy on $I$ in terms of the nonzero Fourier modes. The Beurling-Selberg coefficient bounds convert those modes into the exponential sums in the statement, and taking the supremum over $I$ gives the desired inequality.
[/proofplan]
[step:Approximate the interval indicator by degree $H$ trigonometric polynomials]
Let $I\subset\mathbb R/\mathbb Z$ be an interval, and let $\mathbb{1}_I:\mathbb R/\mathbb Z\to\{0,1\}$ be its indicator function. Since $I$ is a Borel interval in the compact abelian group $\mathbb R/\mathbb Z$ and $H\in\mathbb N$ is a positive integer, the hypotheses of the Beurling-Selberg approximation theorem for intervals on the circle are satisfied. Applying that theorem with degree parameter $H$, there exist real-valued trigonometric polynomials
\begin{align*}
P^+:\mathbb R/\mathbb Z\to\mathbb R
\end{align*}
and
\begin{align*}
P^-:\mathbb R/\mathbb Z\to\mathbb R
\end{align*}
of the form
\begin{align*}
P^\pm(t)=\sum_{|h|\le H} c_h^\pm e(ht)
\end{align*}
such that
\begin{align*}
P^-(t)\le \mathbb{1}_I(t)\le P^+(t)
\end{align*}
for every $t\in\mathbb R/\mathbb Z$,
\begin{align*}
c_0^+\le \mu(I)+\frac{3}{H+1}
\end{align*}
and
\begin{align*}
c_0^-\ge \mu(I)-\frac{3}{H+1},
\end{align*}
while for every integer $h$ with $1\le |h|\le H$ one has
\begin{align*}
|c_h^\pm|\le \frac{3}{2|h|}.
\end{align*}
Here $c_0^\pm=\int_{\mathbb R/\mathbb Z}P^\pm(t)\,d\mu(t)$ because normalized Haar measure annihilates every nonzero character $t\mapsto e(ht)$.
[guided]
The purpose of this step is to replace the discontinuous indicator $\mathbb{1}_I$ by finite [Fourier series](/page/Fourier%20Series) whose coefficients are controlled. Fix an interval $I\subset\mathbb R/\mathbb Z$. The Beurling-Selberg approximation theorem for intervals on the circle applies because $I$ is a Borel interval in the compact abelian group $\mathbb R/\mathbb Z$ and the degree parameter $H$ is a positive integer. It gives two real-valued trigonometric polynomials
\begin{align*}
P^+:\mathbb R/\mathbb Z\to\mathbb R
\end{align*}
and
\begin{align*}
P^-:\mathbb R/\mathbb Z\to\mathbb R
\end{align*}
with degree at most $H$, so each has an expansion
\begin{align*}
P^\pm(t)=\sum_{|h|\le H}c_h^\pm e(ht).
\end{align*}
They are chosen so that $P^-$ lies below the interval indicator and $P^+$ lies above it:
\begin{align*}
P^-(t)\le \mathbb{1}_I(t)\le P^+(t)
\end{align*}
for every $t\in\mathbb R/\mathbb Z$. The zeroth Fourier coefficients approximate the measure of the interval with error at most
\begin{align*}
\frac{3}{H+1}.
\end{align*}
For the upper approximant this gives
\begin{align*}
c_0^+\le \mu(I)+\frac{3}{H+1}
\end{align*}
and
\begin{align*}
c_0^-\ge \mu(I)-\frac{3}{H+1}.
\end{align*}
The remaining coefficients satisfy, for every integer $h$ with $1\le |h|\le H$,
\begin{align*}
|c_h^\pm|\le \frac{3}{2|h|}.
\end{align*}
This coefficient estimate is the quantitative input that will produce the weighted sum with reciprocal frequency weights in the final inequality.
Finally, because $\mu$ is normalized Haar measure on $\mathbb R/\mathbb Z$, the nonzero characters have integral zero:
\begin{align*}
\int_{\mathbb R/\mathbb Z}e(ht)\,d\mu(t)=0
\end{align*}
for every nonzero integer $h$. Therefore
\begin{align*}
\int_{\mathbb R/\mathbb Z}P^\pm(t)\,d\mu(t)=c_0^\pm.
\end{align*}
[/guided]
[/step]
[step:Bound the interval count from above]
Define the normalized counting average over the given points by
\begin{align*}
A_I:=\frac{1}{N}\#\{1\le n\le N:\ x_n\bmod 1\in I\}.
\end{align*}
Since $\mathbb{1}_I\le P^+$, we have
\begin{align*}
A_I\le \frac{1}{N}\sum_{n=1}^{N}P^+(x_n\bmod 1).
\end{align*}
Using the Fourier expansion of $P^+$ and the identity $e(h(x_n\bmod 1))=e(hx_n)$ for integers $h$, this becomes
\begin{align*}
A_I\le c_0^+ + \frac{1}{N}\sum_{\substack{|h|\le H, h\ne 0}}c_h^+\sum_{n=1}^{N}e(hx_n).
\end{align*}
Taking absolute values in the nonzero-frequency contribution and using the coefficient bound gives
\begin{align*}
A_I-\mu(I)\le \frac{3}{H+1}+\frac{1}{N}\sum_{\substack{|h|\le H, h\ne 0}}\frac{3}{2|h|}\left|\sum_{n=1}^{N}e(hx_n)\right|.
\end{align*}
For $h\ge 1$,
\begin{align*}
\left|\sum_{n=1}^{N}e(-hx_n)\right|=\left|\overline{\sum_{n=1}^{N}e(hx_n)}\right|=\left|\sum_{n=1}^{N}e(hx_n)\right|.
\end{align*}
Pairing the $h$ and $-h$ terms yields
\begin{align*}
A_I-\mu(I)\le \frac{3}{H+1}+\frac{3}{N}\sum_{h=1}^{H}\frac{1}{h}\left|\sum_{n=1}^{N}e(hx_n)\right|.
\end{align*}
[/step]
[step:Bound the interval count from below]
Since $P^-\le \mathbb{1}_I$, the same averaging argument gives
\begin{align*}
A_I\ge c_0^- + \frac{1}{N}\sum_{\substack{|h|\le H, h\ne 0}}c_h^-\sum_{n=1}^{N}e(hx_n).
\end{align*}
Using
\begin{align*}
c_0^-\ge \mu(I)-\frac{3}{H+1}
\end{align*}
and applying the triangle inequality to the nonzero-frequency terms, we obtain
\begin{align*}
\mu(I)-A_I\le \frac{3}{H+1}+\frac{1}{N}\sum_{\substack{|h|\le H, h\ne 0}}\frac{3}{2|h|}\left|\sum_{n=1}^{N}e(hx_n)\right|.
\end{align*}
Pairing the frequencies $h$ and $-h$ as before gives
\begin{align*}
\mu(I)-A_I\le \frac{3}{H+1}+\frac{3}{N}\sum_{h=1}^{H}\frac{1}{h}\left|\sum_{n=1}^{N}e(hx_n)\right|.
\end{align*}
[/step]
[step:Take the supremum over intervals]
Combining the upper and lower bounds, for every interval $I\subset\mathbb R/\mathbb Z$ we have
\begin{align*}
\left|A_I-\mu(I)\right|\le \frac{3}{H+1}+\frac{3}{N}\sum_{h=1}^{H}\frac{1}{h}\left|\sum_{n=1}^{N}e(hx_n)\right|.
\end{align*}
The right-hand side is independent of $I$. Taking the supremum over all intervals $I\subset\mathbb R/\mathbb Z$ in the defining formula for $D_N(x_1,\dots,x_N)$ gives
\begin{align*}
D_N(x_1,\dots,x_N)
\le \frac{3}{H+1}+\frac{3}{N}\sum_{h=1}^{H}\frac{1}{h}\left|\sum_{n=1}^{N}e(hx_n)\right|.
\end{align*}
This is the desired Erdos-Turan discrepancy inequality.
[/step]
