The problem is related to this question (https://math.stackexchange.com/posts/3073378/edit).
Is there any general way to construct a non-zero continous function $r(x)$ on $[0,1]$, independent of parameters $a$ and $b$ such that $$\lim_{n\to\infty}\sum_{k=0}^{n}\frac{1}{n}g\left(\frac kn,a,b\right)r\left(\frac kn\right)=0$$ where $g(x,a,b)$ is GIVEN continuous (hence bounded) function on $[0,1]$ with parameters $a,b,$ with $0<a<1$ and $b\in\mathbb{R}$.
As an example, we could possibly work with $g(x,a,b)=e^{bx+a}$ though my original $g$ is something else.
Since $g$ is bounded, my intution tells me that there should exist some non-zero $f$ independent of parameters $a,b$ killing that limit.
Since we have $$ \lim_{n\to\infty}\sum_{k=0}^{n}\frac{1}{n}g\left(\frac kn,a,b\right)r\left(\frac kn\right)=\int_0^1g(x,a,b)r(x)\, dx, $$ we want the function $r(x)$ to satisfy $$ \int_0^1g(x,a,b)r(x)\, dx=0 $$ for all $(a,b)\in(0,1)\times\Bbb R$. In general, it is not true that there would exists a nonzero $r(x)$ with the above property.
Indeed, consider the family of bounded continuous functions $$ \mathcal F=\left\{ \frac1{1+b^2}+ \sin(x)^{1/a} : (a,b)\in(0,1)\times\Bbb R \right\}, $$ it is not hard to see that $D=\{1,\sin(x),\sin(x)^2,\dots\}$ is a subset of $\mathcal F$. According to the Stone-Weierstrass theorem, the span of $D$ is dense in $C[0,1]$. Since $$ \int_0^1 g(x)r(x)\, dx=0 $$ for all $g\in D$, by passing to the limit (using the density result above) we conclude that $$ \int_0^1 f(x)r(x)\, dx=0 $$ for all $f\in C[0,1]$. It follows that the only $r(x)$ that works is $r(x)\equiv 0$.