Convolution and Riemann sum Let $f,g : \mathbb{R} \to \mathbb{C}$ be continuous, 2$\pi$-periodic function, for any $s \in \mathbb{R}$, define the convolution:  
$g * f(s) := \int_{-\pi}^{\pi}g(x)f(s-x)dx$
Show that for any $\epsilon>0$, there exists a partition $T$ over $[-\pi,\pi]$ such that for any refinement $P := \{-\pi = p_0 < p_1 < \ldots <p_N := \pi\}$ of $T$ with any test values $P^*$, then for any $s \in [-\pi,\pi]$:  
$|g*f(s) - \sum_{k=1}^{N}g(p^*_k)f(s - p^*_k)(p_k - p_{k-1})| < \epsilon$
My attempt: I know that the convolution $g * f$ is continuous on $[-\pi,\pi]$, hence uniformly continuous, I also know that for any specific $s \in \mathbb{R}$ we can find a partition $P_{s}$ (depending on $s$) that satisfy the above condition. But how to choose one partition that suites any $s \in [-\pi,\pi]?$ Thanks for your help.
 A: For any $s\in\mathbb R$, denote by $\varphi_s$ the function $x\mapsto g(x)f(s-x)$. The key point is that the family $(\varphi_s)_{s\in\mathbb R}$ is equi-uniformly-continuous. That is : for any given $\varepsilon >0$, one can find a single $\delta >0$ such that $\vert \varphi_s(y)-\varphi_s(x)\vert\leq\varepsilon$ for all $s\in\mathbb R$ whenever $\vert y-x\vert<\delta$. This holds because $f$ and $g$ are both bounded and uniformly continuous on $\mathbb R$ : indeed, for any $s,x,y$ we have
\begin{align*} \vert \varphi_s(y)-\varphi_s(x)\vert&\leq \vert g(y)-g(x)\vert\,\vert f(s-y)\vert +\vert f(s-y)-f(s-x)\vert\,\vert g(x)\vert\\
&\leq \Vert f\Vert_\infty\, \vert g(y)-g(x)\vert+\Vert f\Vert_\infty\, \vert f(s-y)-f(s-x)\vert.
\end{align*}
Let $\varepsilon >0$ be given, and choose $\delta>0$ as above but associated with $\varepsilon/{2\pi}$. Then, since $f*g(s)=\int_{-\pi}^\pi \varphi_s(x)\, dx$, the usual proof for the convergence of Riemann sums shows that any partition $T$ of $[-\pi,\pi]$ with ``mesh" less than $\delta$ does the job.
