# Infinite Riemann integral implies infinite lower Riemann sum?

Suppose that we have a function $f$ defined on $[0,\infty)$ that is Riemann integrable on every bounded set $[0,a]$, but whose improper integral diverges: $$\int_{0}^{\infty} f(x)\,dx := \lim_{a\rightarrow\infty} \int_{0}^{a} f(x) \,dx = \infty.$$ Then clearly the upper (infinite) Riemann sum must be infinite, however is it necessarily true that the lower (infinite) Riemann sum be infinite. If $f$ is smooth, it's intuitively obvious that it should be infinite. However, if $f$ is not smooth is it still necessarily true? Proof or counterexample would be appreciated!

Actually this is a dumb question. Of course both the infinite upper and lower Riemann sums are infinite. Since $f$ is Riemann integrable on $[0,a]$ the upper and lower Riemann sums are equal and finite on $[0,a]$. As $a$ increases we can make $\int_{0}^{a} f(x)\,dx$ arbitrarily large, hence their upper and lower Riemann sums (in particular it's lower Riemann sum) arbitrarily large.