Convergence of Riemann sums

Question

Suppose that $g:[0,1]\rightarrow \mathbb{R}$ is a Riemann integrable function. Consider an equi-spaced partition of $[0,1]$ made of $n$ points

$$ \Pi=\left\{0,\frac{1}{n},\frac{2}{n},\dots,\frac{n-1}{n},1\right\}. $$

For all $n$ let $f^{(n)}$ be a function $f^{(n)}:[0,1]\rightarrow \mathbb{R}$ such that, for all $s\in[0,1]$ we have that

$$ f^{(n)}\left(\frac{1}{n}\cdot\left\lfloor s \cdot n\right\rfloor\right) \to g(s)\quad \tag{1} $$

when $n\to\infty$. Equation $(1)$ is saying that the sequence of functions $f^{(n)}$ when computed in the point $\frac{1}{n}\cdot \left\lfloor s\cdot n\right\rfloor$ which approximates asymptotically $s$, converges to the function $g$ in $s$. My problem is to establish if

$$ \sum_{k=1}^{n}f^{(n)}\left(\frac{k}{n}\right)\,\frac{1}{n}\rightarrow\int_0^1g(s)\,{\rm d}s.\quad \tag{2} $$

The quantity on the left of $(2)$ is exactly the Riemann sum of $f^{(n)}$ in the equi-spaced partition $\Pi$, so intuitively result in $(2)$ should follow from equation $(1)$.

Answer 1

The conclusion is false. Here is a counterexample.

Let $g=0$ and let $f_n (x)= (xn)^2$ if $x<\frac{1}{\sqrt{n}}$ and zero otherwise.

Then for each $s\in [0,1]$, $f_n(n^{-1} \lfloor s n \rfloor)$ is equal to zero for $n$ large enough. But

$$\frac{1}{n}\sum_{k=1}^n f_n ( k/n) \ge \frac{1}{n}\sum_{k=1}^{\lfloor \sqrt{n}\rfloor} k^2 \sim \frac {1}{n} \frac 13 n^{3/2}\to \infty>0=\int_{[0,1]} g(x) d x.$$

If you want to fix it, you can assume that that the functions $s\to f_n (n^{-1}\lfloor ns \rfloor )$ not only converge to $g(s)$ pointwise, but are also uniformly bounded and then you can apply the Dominated Convergence Theorem (Note: boundedness can be relaxed to uniform integrability).