3

I'm having trouble proving or finding a counterexample for the following statement:

If $f>0$ 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 $\epsilon>0$ there exists a $\delta>0 \ $ s.t. if $P$ is a partition of $[a,b]$ and $\lambda(P)<\delta$ then $|S-I|<\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.

Paz
  • 789

2 Answers2

1

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.

AlexM
  • 931
  • 1
  • 5
  • 6
  • Can you add some more details? Is that a tip as to what I'm looking for when thinking of a counterexample? – Paz Apr 14 '13 at 12:39
  • No, I thought it was clear enough as an answer. Try to prove (since this is homework) that $\sqrt(f)$ is still bounded, and the set of discontinuities as measure zero. – AlexM Apr 14 '13 at 17:41
  • Sorry, I haven't learned that yet. We were advised to use the Cauchy-Schwarz inequality... – Paz Apr 16 '13 at 16:55
-1

I will write it here, and if its wrong, I will delete it.

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

17SI.34SA
  • 2,063
  • 2
  • 19
  • 27