If $f : [a,b] \to \mathbb{R} $ is integrable, then so is $\sqrt{f}$? Let $f:[a,b] \to \mathbb{R}$ is integrable on $[a,b]$ withn $f \ge 0$, show that $g = \sqrt{f}$ is also integrable.
I have seen some hints says using $\sqrt{x}-\sqrt{y}= \frac{x - y}{\sqrt{x} + \sqrt{y}}$, but I get stuck here since I dont know how to make $\frac{1}{\sqrt{x}+ \sqrt{y}}$ smaller than some fixed number.
Can someone help me?
 A: Because $f$ is (proper) Riemann integrable, it follows that $f$ is bounded and continuous almost everywhere, and so is $\sqrt f$. Now use the fact that bounded +  continuous almost everywhere implies Riemann integrable (see this).
A: If you are supposed to work directly with upper and lower sums (i.e., the Darboux definition of the integral), the way to deal with the problematic $\sqrt x+\sqrt y$ in the denominator, goes roughly as follows:
Pick any $\varepsilon>0$. When comparing upper and lower sums for the integral, consider an interval where the upper step function takes value $y$, and the lower one takes value $x$. So $0\le x\le y$. If $y<\varepsilon$, then $\sqrt y-\sqrt x<\sqrt\varepsilon$ by direct comparison, so this is no problem. In other intervals, you have $\sqrt x+\sqrt y\ge\varepsilon$, so $\sqrt y-\sqrt x\le (y-x)/\epsilon$. Now keep that $\epsilon$ and refine your partition so that the difference between upper and lower sum is less than $\varepsilon^2$.
I'll leave the somewhat messy details for you to clean up, if you don't mind.
PS. Actually, it is probably easier just to use the uniform continuity of the square root function. If you look at my outline above, you may agree that this is what it does, albeit in a much too explicit form.
