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: $$\int_{0}^{\infty} \!\!\!f(x)\,\text{d}x \,:=\, \lim_{a\rightarrow\infty} \int_{0}^{a} \!\!\!f(x) \,\text{d}x \,=\, \infty.$$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!
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