Let's consider the $2\pi$ periodic and Riemann integrable function $f:\mathbb{R}\to\mathbb{R}$ and the function $\varphi:\mathbb{R}\to\mathbb{R}$, with $\varphi(u)=\frac{f(x_0+u)+f(x_0-u)}{2}-f(x_0)$; and $x_0\in\mathbb{R}$ arbitrarily chosen.
We want to show that the function $g:\mathbb{R}\to\mathbb{R}$, where
\begin{align*} &g(u)=\begin{cases} g(\pi),&\text{ if } u=-\pi\\ \frac{\varphi(u)}{2\sin(\frac{u}{2})},&\text{ if } u\in (-\pi,\pi)\setminus\{0,-\pi\}\\ 0,&\text{ if } u=0. \end{cases} \end{align*} is Riemann-integrable on $(-\pi,\pi)$. We can assume that the two limits \begin{align*} &\lim\limits_{\underset{u>0}{u\to 0}} g(u):=a \text{ and }\lim\limits_{\underset{u<0}{u\to 0}} g(u):=b \end{align*} exist.
My approach:
We know that $h:(-\pi,\pi)\to\mathbb{R}$ with \begin{align*} &h(u)=\begin{cases} \frac{\varphi(u)}{2\sin(\frac{u}{2})},&\text{ if } u\in (-\pi,\pi)\setminus\{0\}\\ 0,&\text{ if } u=0 \end{cases} \end{align*} is continuous at all points $u\neq0$. If we restrict $h$ to $(0,\pi)$, then we can make $h_{\mid (0,\pi) }$ continuous at point $0$ by setting $h_{\mid 0,\pi) }(0):=a$. We do the same with $h_{\mid (-\pi,0) }$, so $h_{\mid (-\pi,0) }(0):=b$. Hence, $h_{\mid (0,\pi)}$ and $h_{\mid (-\pi,0) }$ are continuous in each point and therefore Riemann-integrable. If I amend the following points again by setting \begin{align*} &h_{\mid (-\pi,0) }(-\pi):=g(\pi),~h_{\mid (-\pi,0) }(0):=0 \text{ and } h_{\mid (0,\pi) }(0):=0, \end{align*} the restrictions $h_{\mid (-\pi,0)}$ and $h_{\mid (0,\pi) }$ remain Riemann-integrable. Now merging both restrictions $h_{\mid (-\pi,0)}$ and $h_{\mid (0,\pi) }$ gives us a new function, which we denote by $\hat{h}$, which is also Riemann integrable on $(-\pi,0)\cup (0,\pi)$. Finally, we see that $\hat{h}=g$ for all $u\in(-\pi,\pi)$, so $g$ is Riemann integrable on $(-\pi,\pi)$.
Is this correct?
My professor gave me the hint that I had assumed that $f$ is continuous. However, this might not be the case in general; we only know that $f$ is Riemann-integrable. Without $f$ being continuous I can't say that $\varphi$ and $h$ are continuous. So at this point my proof breaks.