It was posted in an answer to my previous question that
If a function $f$ is Riemann-integrable on $[0,\infty)$, then,
$$\int_0^\infty f(x)\,dx = \lim_{h\rightarrow 0}\,h\sum_{n=0}^\infty f(nh)$$
How can we prove this proposition? I only know of the standard definition of Riemann Sums involving lim sup and lim inf.
This is false.
Strictly speaking it's meaningless, because there's no such thing as "Riemann integrable on $[0,\infty)$"; the Riemann integral is defined for functions that have a compact interval for their domain. Presumably here $\int_0^\infty$ is supposed to be an improper Riemann integral, which is to say the reasonable interpretation of the proposition in question is
If a function $f$ is Riemann integrable on $[0,A]$ for every $A>0$ and $\int_0^\infty f=\lim_{A\to\infty}\int_0^A f$ exists then
$$\int_0^\infty f(x)\,dx = \lim_{h\rightarrow 0}\,h\sum_{n=0}^\infty f(nh).$$
That is false.
Choose $A_n$ increasing to infinity such that $A_1=0$ and $A_{n+1}-A_n$ decreases to $0$ (for example, $A_n=\log(n)$). Let $f(t)=(-1)^n$ for $A_n\le t<A_{n+1}$. Then $\int_0^\infty f$ exists, more or less by the proof of the alternating series test for sums, but $\sum_nf(nh)$ does not exist, since the terms do not even tend to $0$.
Details: Regarding existence of $\int_0^\infty f$: First, the alternating series test for sums shows that $$\lim_{n\to\infty}\int_0^{A_n}f$$exists. And it's clear that $$\left|\int_0^Af-\int_0^{A_n}f\right|\le A_{n+1}-A_n\quad(A_n\le A\le A_{n+1}).$$Since $A_{n+1}-A_n\to0$ this shows that $\lim_{A\to\infty}\int_0^A$ exists.
