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 $(1)$.


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).

  • $\begingroup$ Hello! How would you prove convergence provided the $f^{(n)}$ are uniformly bounded? I am not sure how the Dominated Convergence Theorem works here. $\endgroup$
    – acrendic
    Jun 19 at 17:39
  • $\begingroup$ If there exists some $M$ such that $|f_n (t)|\le M$ for all $t \in [0,1]$ and $n=1,2,3,\dots$, then you can apply the dominate convergence theorem noting that i) $\int_0^1 f_n (t) dt$ is in fact the Riemann sum in question; and (ii) $f_n (t)\to g(t)$ pointwise. $\endgroup$
    – Fnacool
    Jun 20 at 20:18

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