Let $f$ be a polynomial of degree at most 1 and denote by $I$ the indicator function. For any $c>0$, how can I show that $$\int_0^1 f^2(y)I(\lvert y\rvert\leq c)dy=\lim_{n\to\infty}\sum_{i=1}^n f^2(i/n)I(\lvert i/n\rvert\leq c)?$$
I proved a similar result when the integrand was Lipschitz continuous. In this case, it can be discontinuous (depending on $c$). I struggling to find the right argument.
Can you help me?
Let $g(y)=(f(y))^{2}\chi_{|y|\leq c}$, so $g$ is Riemann integrable, as it is a product to two Riemann integrable functions, which are $f^{2}$ and $\chi_{|y|\leq c}$. Now the right-sided is just a Riemann sum of $g$ with uniform partition of $[0,1]$ with width $1/n$ and taking the right endpoint grids.