Let $f(x)$ be a $2\pi$-periodic function on $[0,2\pi]$ (so we extend $f$ to the reals by $f(x)=f(x+2\pi)$. ) $f$ is also absolutely improperly integrable, i.e., $$ \int_0^{2 \pi } |f| \, dx := \lim_{x \rightarrow 0} \int_x^c |f| \, dx + \lim_{y \rightarrow 2 \pi} \int_c^y |f| \, dx .$$ exists, where the terms in limits are Riemann Integrals exists. (So $|f|$ is Riemann Integrable in any interval $(\alpha, \beta) \subset [0, 2\pi] $.)
Is the function $G(x):= \int_0^x |f(t)| \, dt $ uniformly continuous on $\mathbb{R}$?
Thoughts: We show $G$ is continuous on $x \in [0, 2\pi ]$. Suppose $x,y \in [0, 2\pi] $. Since $f$ is absolutely improperly integrable, exists $\delta_1 > 0$ such that $|G(x)-G(0)| < \varepsilon/2$ and $|G(y)- G(0)| < \varepsilon/2$. when $\max \{ |x|, |y-2\pi| \} < \delta_1$. So we have continuity at end points.
If $a \in ( \delta, 2\pi - \delta) \subseteq [0, 2\pi ] $. Then as $|f|$ is Riemann Integrable on $[\delta, 2 \pi - \delta]$, it is bounded on this interval. Hence, $|G(x) - G(a)| \rightarrow 0 $ as $x \rightarrow 0$.
So $G(x)$ is continuous on $[0,2\pi]$. As $[0,2 \pi]$ is compact, it is uniformly continuous. There exists $\delta_2>0$, such that $|x-y|<\delta_2 \Rightarrow |G(x)-G(y)|<\varepsilon$ for $x,y \in [0, 2\pi]$.
Thus, for uniform convergence of real line, choose $\delta:= \min \{ \delta_1, \delta_2 \}$.