Pointwise convergence of Fourier series from Rudin 
This theorem about pointwise convergence of Fourier series. And some moments from here little bit confuses me.
1) How rigorously show that the last two integrals tends to zero? How to use that $|f(x+t)-f(x)|\le M|t|$ for $t\in (-\delta,\delta)$?
I would be very grateful for your help!
 A: A) The assumption that $|f(x+t)-f(x)|\le M|t|$ is used to infer that $g(t)$ is bounded. 
In more detail, assuming $\delta$ sufficiently small, for $0<|t|<\delta$ we have $|\sin(t/2)|\geq |t|/4$ by the Taylor series.  Thus, if $|f(x+t)-f(x)|\le M|t|$ for $|t|\le\delta$ (so that $|f(x-t)-f(x)|\le M|t|$ also), we get $$|g(t)|=\left|{f(x-t)-f(x)\over \sin(t/2)}\right|=\left|{f(x-t)-f(x)\over t}\cdot {t\over \sin(t/2)}\right|\leq 4M|t|\leq 4M\delta$$
For $\delta < |t| <\pi$, we have that $|1/\!\sin (t/2)|$ is bounded because it is $\leq 1/\!\sin(\delta/2)$, a fixed number.  $f$ is assumed Riemann integrable on $[-\pi,\pi]$ and periodic (Rudin 8.13); it is therefore bounded (Rudin 6.1, intuitively because there has to be an upper sum for each partition).  This lets us bound $g(t)$ for $\delta<|t|<\pi$.
B) The fact that the integrals go to zero is deduced from $(74)$ which says that the Fourier coefficients of a Riemann-integrable function go to zero.  To apply it we need to verify that $g(t)\cos t$ and $g(t)\sin t$ are Riemann-integrable.  To do this easily I jump ahead to Rudin 11.33, which just requires me to show that these functions are bounded and continuous almost everywhere, and that in turn follows from $g$ being bounded and from $f$ and $\cos$ and $\sin$ being continuous almost everywhere. (Apologies for jumping ahead; my Riemann integrability skills are rusty.)
