\begin{equation*}
\frac{1}{2\pi} \int_{-\pi}^\pi D_N = 1 .
\end{equation*}
Write
\begin{equation*}
\begin{split}
s_N(f;x)-f(x) & =
\frac{1}{2\pi} \int_{-\pi}^\pi f(x-t) D_N(t) \, dt
-
f(x)
\frac{1}{2\pi} \int_{-\pi}^\pi D_N(t) \, dt
\\
& =
\frac{1}{2\pi} \int_{-\pi}^\pi \bigl( f(x-t) - f(x) \bigr) D_N(t) \, dt
\\
& =
\frac{1}{2\pi} \int_{-\pi}^\pi \frac{f(x-t) - f(x)}{\sin(\nicefrac{t}{2})} \sin\bigl(
(N+\nicefrac{1}{2})t \bigr) \, dt .
\end{split}
\end{equation*}
By the hypotheses, for small nonzero \(t\text{,}\)
\begin{equation*}
\abs{ \frac{f(x-t) - f(x)}{\sin(\nicefrac{t}{2})} }
\leq
\frac{M\sabs{t}}{\sabs{\sin(\nicefrac{t}{2})}} .
\end{equation*}
As \(\sin(\theta) = \theta + h(\theta)\) where \(\frac{h(\theta)}{\theta} \to
0\) as \(\theta \to 0\text{,}\) we notice that \(\frac{M\sabs{t}}{\sabs{\sin(\nicefrac{t}{2})}}\) is continuous at the origin. Hence, \(\frac{f(x-t) - f(x)}{\sin(\nicefrac{t}{2})}\text{,}\) as a function of \(t\text{,}\) is bounded near the origin. As \(t=0\) is the only place on \([-\pi,\pi]\) where the denominator vanishes, it is the only place where there could be a problem. So, the function is bounded near \(t=0\) and clearly Riemann integrable on any interval not including \(0\text{,}\) and thus it is Riemann integrable on \([-\pi,\pi]\text{.}\) We use the trigonometric identity
\begin{equation*}
\sin\bigl( (N+\nicefrac{1}{2})t \bigr)
=
\cos(\nicefrac{t}{2}) \sin(Nt) +
\sin(\nicefrac{t}{2}) \cos(Nt) ,
\end{equation*}
to compute
\begin{multline*}
\frac{1}{2\pi} \int_{-\pi}^\pi \frac{f(x-t) - f(x)}{\sin(\nicefrac{t}{2})} \sin\bigl(
(N+\nicefrac{1}{2})t \bigr) \, dt
= \\
\frac{1}{2\pi} \int_{-\pi}^\pi
\left( \frac{f(x-t) - f(x)}{\sin(\nicefrac{t}{2})}
\cos (\nicefrac{t}{2}) \right) \sin (Nt) \, dt
+
\frac{1}{2\pi} \int_{-\pi}^\pi \bigl( f(x-t) - f(x) \bigr)
\cos (Nt) \, dt .
\end{multline*}
As functions of
\(t\text{,}\) \(\frac{f(x-t) - f(x)}{\sin(\nicefrac{t}{2})} \cos (\nicefrac{t}{2})\) and
\(\bigl( f(x-t) - f(x) \bigr)\) are bounded Riemann integrable functions and so their Fourier coefficients go to zero by
Theorem 11.8.9. So the two integrals on the right-hand side, which compute the Fourier coefficients for the real version of the Fourier series go to 0 as
\(N\) goes to infinity. This is because
\(\sin(Nt)\) and
\(\cos(Nt)\) are also orthonormal systems with respect to the same inner product. Hence
\(s_N(f;x)-f(x)\) goes to 0, that is,
\(s_N(f;x)\) goes to
\(f(x)\text{.}\)