Let $f(x) \geq 0$ be a monotonic decreasing function with $f \rightarrow 0$ and $\int \limits_{0}^{\infty}f(x)dx = \infty$. Split $[0,\infty)$ into intervals $I_n = [x_n,x_{n+1})$, so that $\int \limits_{I_n}f(x)dx = \frac{1}{n}$.
My question is, is it true that $len(I_n) = x_{n+1}-x_n$ is bounded?
(I have the feeling that without further conditions on $f$ (like how fast $f \rightarrow 0$) this might not be true, but I can't really prove my feeling...)
Here is an example of an $f$ that is Riemann integrable, monotonically decreasing and goes to $0$ as $x\to\infty$. Let us partition $[0,\infty)$ via $$ [0,\infty) = \bigcup_{n=0}^{\infty}[x_n,x_{n+1}) $$ where $x_0=0$, and $x_{n+1}-x_n = (n+1)$ (hence, $x_1 = 1, x_2=3 , x_3 = 6,\dots$ and so on). With this, let us construct the function to be constant on the interval $[x_n,x_{n+1})$ with the value of $\left(\frac{1}{n+1}\right)^2$. Clearly, this function has countably many discontinuities (in particular, set of discontinuities of $f$ is precisely $\{x_n : n\in\mathbb{N}\}$); and is bounded, hence it is Riemann integrable. You can actually prove that its Riemann integral is evaluated as $$ \sum_{n=0}^{\infty}\left(\frac{1}{n+1}\right)^2\underbrace{(x_{n+1}-x_n)}_{=n+1}=\sum_{n\geq 1}\frac{1}{n} = \infty $$ while $x_{n+1}-x_n\to\infty$.
Edit : I presume, this construction, with a little bit of spice, works even for strictly decreasing, continuous functions; even though I did not rigorize it. Here is my thought: Just take very small intervals around $x_n$'s and give a slope to function to acquire continuity, and give little offsets in the intervals $[x_n,x_{n+1})$ to adjust for strict decreasing case.