True or False: If $f>0$ is integrable on $[a,b]$ then $\sqrt{f}$ is integrable I'm having trouble proving or finding a counterexample for the following statement:


  If $f&gt;0$ is integrable on $[a,b]$ then $\sqrt{f}$ is integrable


We're using the Riemann integral definition:

If $f$ is integrable on $[a,b]$ then given $\epsilon&gt;0$ there exists a $\delta&gt;0 \ $ s.t. if $P$ is a partition of $[a,b]$ and $\lambda(P)&lt;\delta$ then $|S-I|&lt;\epsilon \ $ (where $S$ is the Riemann sum and $I$ is the integral's value).

I tried using the fact that $\sqrt{x}$ is uniformly continuous on $(0,\infty)$ which means that if $f$'s oscillation gets very small, so does $\sqrt{f}$'s, but I wasn't able to rigorously prove it.

Is this statement actually true? is the uniformly continuous angle of any help?

Much appreciated.
By the Riemann-Lesbesgue criterion, we have that a function $g$ on $[a,b]$ is Riemann-integrable iff it is bounded and it's set of discontinuities has measure zero. So that should get it figured out for ya.
I will write it here, and if its wrong, I will delete it.

Let $g=\sqrt{f}$, we have that $f&gt;0$, hence $g$ is monotonic, using theorem, that if you have a monotonic function on close an bounded interval, then monotonic function is integrable.