Originally Posted by
Plato In fact that is not a counterexample. $\displaystyle f(x)=\frac{1}{\sqrt{x}}$ is not Riemann integrable on $\displaystyle [0,1]$.
Note that the function is not bounded on $\displaystyle [0,1]$.
This happens to be a standard theorem/problem in the study Riemann integrals.
Here is the standard hint: $\displaystyle f(x)^2 - f(y)^2 = \left( {f(x) + f(y)} \right)\left( {f(x) - f(y)} \right)$.
Use that to show that $\displaystyle U\left[ {f^2 ;P} \right] - L\left[ {f^2 ;P} \right] \leqslant 2 B\left( {U\left[ {f;P} \right] - L\left[ {f;P} \right]} \right)$ where $\displaystyle B$ is the bound of $\displaystyle f$ on $\displaystyle [a,b]$