Riemann integrability and Riemann sums

Question

If we don't use improper integrals, $f(x)=1/\sqrt{x}$ is not Riemann-integrable on $[0,1]$ because only bounded functions are Riemann-integrable. Does that mean that there exists a tagged partition of $[0,1]$ and a corresponding Riemann sum that doesn't converge (to 2) for $f(x)$? (I wasn't able to find one)

Answer 1

If you'd like to see a specific example, here's one. For an arbitrary partition of $[0,1]$, let $[0,x_1]$ be its leftmost subinterval, where $0<x_1<1$ (of course). Then for any $N>1$, the point $x_1^{*}=\frac{x_1^2}{N^2}$ is in $[0,x_1]$, and the first term in the corresponding Riemann sum will be $$f(x_1^{*})\Delta x_1=\frac{1}{\sqrt{x_1^2/N^2}}\cdot x_1=N,$$ which can be arbitrarily large if we let $N\to+\infty$. Since the rest of the terms are nonnegative, we see that Riemann sums are unbounded and don't have a limit.