Pointwise convergence of Fourier series - Dirichlet theorem2 I want to prove the following theorem:

I have corrected the definition of the nth partial sum of the Fourier series of $f$ to $$S_{n}(f)(t_{0}) = \sum_{N=-n}^{n}c_{N}(f)e^{iNt_{0}}$$,
And I proved that:
$$ S_{n}(f)(t_{0}) = \frac{1}{2\pi} \int_{-\pi}^{\pi}f(x) (\sum_{N=-n}^{n} e^{iN(t_{0} - x)})dx = (f*D_{n})(t_{0}),$$
where $D_{n}$ is the Dirichlet kernel,but now if I put $t_{0} - x = u$ to calculate letter(b) I will find that it will be $f(t_{0} - u)$ not $f(t_{0} + u)$ as required inside the integration am I right?also is question(2) so trivial that it`s answer is $$S_{n}(f)(t_{0}) =  \frac{1}{2\pi} \int_{-\pi}^{0}f(x) (\sum_{N=-n}^{n} e^{iN(t_{0} - x)})dx +  \frac{1}{2\pi} \int_{0}^{\pi}f(x) (\sum_{N=-n}^{n} e^{iN(t_{0} - x)})dx $$,
Am I right?now how can I prove letter(c)? I am stucked,could anyone help me?   
 A: They should have said that $f$ is extended to $\mathbb{R}$ in such a way that it is periodic with period $2\pi$. Otherwise the integrals stated in the theorem are not necessarily defined. When you do that, the Dirichlet kernel and the function are both periodic with period $2\pi$, and you can shift the limits of integration to be any interval $[t_0-\pi,t_0+\pi]$:
\begin{align}
     S_N(f)(t_0)& = \int_{t_0-\pi}^{t_0+\pi}f(t)D_N(t-t_0)dt \\
    &= \int_{-\pi}^{0}f(t+t_0)D_N(t)dt+\int_{0}^{\pi}f(t+t_0)D_N(t)dt \\
    &= \int_{-\pi}^{0}\frac{f(t+t_0)-f(t_0^-)}{t}\{ tD_N(t)\}dt +\frac{f(t_0^-)}{2}\\
    & +\int_{0}^{\pi}\frac{f(t+t_0)-f(t_0^+)}{t}\{ tD_N(t)\}dt+\frac{f(t_0^+)}{2}
\end{align}
The Riemann-Lebesgue lemma now applies because
$$
    \frac{f(t+t_0)-f(t_0^-)}{t},\; \frac{f(t+t_0)-f(t_0^+)}{t}
$$
are integrable for $t\in[-\pi,0]$ and $t\in[0,\pi]$, respectively, and because the following is a bounded function times $\sin((n+1/2)\pi t)$:
$$
       tD_N(t) = \frac{t}{\sin(t/2)}\sin((n+1/2)\pi t).
$$
That is,
$$
     S_N(f)(t_0)-\frac{f(t_0^-)+f(t_0^+)}{2}= \int_{-\pi}^{\pi}g(t)\sin((n+1/2)t)dt,
$$
where $g(t)$ is integrable on $[-\pi,\pi]$; hence the integral expression on the right tends to $0$ as $n\rightarrow\infty$, which gives the desired limit for the Fourier series.
