Question
How the boundedness of $g(t)\cos\left(\frac{t}{2}\right)$ and $g(t)\sin\left(\frac{t}{2}\right)$ on $[-\pi,\pi]$ shows their Riemann-integrability in Rudin's proof? This is needed in the end of the proof, as assumed condition for Theorem 8.12, for applying $\eqref{74}$ to these functions as $f$ and $\left\{\frac{\sin(nx)}{\sqrt{\pi}}\right\}$, $\left\{\frac{\cos(nx)}{\sqrt{\pi}}\right\}$ as $\{\phi_{n}\}$, respectively.
Prerequisites
If $\{\phi_{n}\}$ is orthonormal on $[a,b]$, and if $$c_{n}=\int_{a}^{b}f(t)\overline{\phi_{n}(t)}dt\quad(n=1,2,3,…),$$ we call $c_{n}$ the $n$th Fourier coefficient of $f$ relative to $\{\phi_{n}\}$. We write $$f(x)\sim\sum_{1}^{\infty}c_{n}\phi_{n}(x)$$ and call this series the Fourier series of $f$(relative to $\{\phi_{n}\}$).
Theorem 8.12: Assumed that $f$ is Riemann-integrable on $[a,b]$. If $\{\phi_{n}\}$ is orthonormal on $[a,b]$, and if $$f(x)\sim\sum_{n=1}^{\infty}c_{n}\phi_{n}(x),$$ then $$\sum_{n=1}^{\infty}|c_{n}|^{2}\le\int_{a}^{b}|f(x)|^{2}dx.$$ In particular, $$\lim_{n\to\infty}c_{n}=0\tag{74}\label{74}.$$
Dirichlet kernel: $$D_{N}(x)=\sum_{n=-N}^{N}e^{inx}=\frac{\sin\left(\left(N+\frac{1}{2}\right)x\right)}{\sin(x/2)}\tag{77}\label{77}.$$
$f:\mathbb{R}\to\mathbb{C}$, Riemann-integrable on $[-\pi,\pi]$ and has period $2\pi$.
The Fourier series of $f$ is the series $$\sum_{-\infty}^{\infty}c_{n}e^{inx}\quad(x\space\mathrm{real})$$ whose coefficients $c_{n}$ are given by the integrals $$c_{n}=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-inx}dx,$$ and $$s_{N}(x)=s_{N}(f;x)=\sum_{-N}^{N}c_{n}e^{inx}$$ is the $N$th partial sum of the Fourier series of $f$.
$$s_{N}(f;x)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)D_{N}(x-t)dt=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x-t)D_{N}(t)dt\tag{78}\label{78}.$$
Theorem 8.14
Theorem 8.14: If, for some $x$, there are constants $\delta\gt0$ and $M\lt\infty$ such that: $$|f(x+t)-f(x)|\leq M|t|\tag{79}\label{79}$$ for all $t\in(-\delta,\delta)$, then $$\lim_{N\to\infty}s_{N}(f;x)=f(x)\tag{80}\label{80}.$$
Proof: Define $$g(t)=\frac{f(x-t)-f(x)}{\sin(t/2)}\tag{81}\label{81}$$ for $0\lt|t|\leq\pi$, and put $g(0)=0$. By the definition $\eqref{77}$,$$\frac{1}{2\pi}\int_{-\pi}^{\pi}D_{N}(x)dx=1.$$ Hence $\eqref{78}$ shows that: $$s_{N}(f;x)-f(x)=\frac{1}{2\pi}\int_{-\pi}^{\pi}g(t)\sin\left(\left(N+\frac{1}{2}\right)t\right)=\frac{1}{2\pi}\int_{-\pi}^{\pi}\left[g(t)\cos\left(\frac{t}{2}\right)\right]\sin(Nt)dt+\frac{1}{2\pi}\int_{-\pi}^{\pi}\left[g(t)\sin\left(\frac{t}{2}\right)\right]\cos(Nt)dt.$$ By $\eqref{79}$ and $\eqref{81}$, $g(t)\cos\left(\frac{t}{2}\right)$ and $g(t)\sin\left(\frac{t}{2}\right)$ are bounded. The last two integrals thus tend to $0$ as $N\to\infty$, by $\eqref{74}$. This proves $\eqref{80}$.
Best Answer
It's easily seen from \eqref{81} that both functions are Riemann-integrable on any $[a,b]\subset[-\pi,\pi]$ if $0\notin[a,b]$. Let's denote one of them by $f=(f_{1},f_{2})$. Boundedness of $f$ implies that there exists such $M>0$ that for any $t\in[-\pi,\pi]$ $|f_{1}(t)|\leq M$. For any $\epsilon>0$ such that $\epsilon<12\pi M$ let's take $\phi=\frac{\epsilon}{12M}$ so that $0<\phi<\pi$. Since $f_{1}$ is Riemann-integrable on $[-\pi,-\phi]$, there exists such partition of $[-\pi,-\phi]$ $P^{*}$ that $$U(P^{*},f_{1})-L(P^{*},f_{1})<\frac{\epsilon}{3}.$$ Analogously, there exists such partition of $[\phi,\pi]$ $P^{**}$ that $$U(P^{**},f_{1})-L(P^{**},f_{1})<\frac{\epsilon}{3}.$$ So that if we take partition of $[-\pi,\pi]$ $P=P^{*}\cup P^{**}$, $$U(P,f_{1})-L(P,f_{1})=U(P^{*},f_{1})-L(P^{*},f_{1})+U(P^{**},f_{1})-L(P^{**},f_{1})+2\phi M^{*}-2\phi m^{*}.$$ where $$M^{*}=\sup f(x)\quad (x\in[-\phi,\phi]),$$ $$m^{*}=\inf f(x)\quad (x\in[-\phi,\phi]).$$ Since $M^{*}-m^{*}\leq2M$, $$U(P,f_{1})-L(P,f_{1})<\epsilon,$$ which proves Riemann-integrability of $f_{1}$ on $[-\pi,\pi]$. Analogously, $f_{2}$ is Riemann-integrable on $[-\pi,\pi]$, hence $f$ is as well. Analogous argument holds for second function in question.