Suppose that $f:[0,\infty)\to[0,\infty)$ is a non-increasing function such that $\int_0^\infty f(x)dx<\infty$. Suppose that $\{a_n\}_{n\ge1}$ and $\{b_n\}_{n\ge1}$ are two sequences of positive integers such that $a_n\to\infty$, $b_n\to\infty$, but $b_n/a_n\to\infty$ as $n\to\infty$.
Is it true that \begin{equation}\label{eq}\tag{*} \frac1{a_n}\sum_{j=1}^{b_n}f(j/a_n)\to\int_0^\infty f(x)dx \end{equation} as $n\to\infty$? How can I show this?
For a positive integer $m$, we have that $$ \frac1{a_n}\sum_{j=1}^{ma_n}f(j/a_n)\to\int_0^m f(x)dx\quad\text{as}\quad n\to\infty. $$ Intuitively, it seems that if $b_n/a_n\to\infty$ as $n\to\infty$, \eqref{eq} could be true. However, I am not sure if it is true or not.
If $f$ is not necessarily non-increasing function, then \eqref{eq} is false (see this question).
Any help is much appreciated! References are also welcome.
For monotonic functions, all works out nicely. Since $f$ is non-increasing, we have
$$\int_{\frac{j}{a_n}}^{\frac{j+1}{a_n}} f(x)\,dx \leqslant \frac{1}{a_n}f(j/a_n) \leqslant \int_{\frac{j-1}{a_n}}^{\frac{j}{a_n}} f(x)\,dx$$
for every $j \geqslant 1$, and therefore
$$\int_{1/a_n}^{(b_n+1)/a_n} f(x)\,dx \leqslant \frac{1}{a_n}\sum_{j = 1}^{b_n} f(j/a_n) \leqslant \int_0^{b_n/a_n} f(x)\,dx\,.$$
Both of the outer terms converge to
$$\int_0^{\infty} f(x)\,dx\,,$$
so the squeeze theorem finishes the argument.