I'm trying to show that assuming $f:[a,b] \rightarrow \mathbb{R}$ is nonnegative, bounded function that is Riemann integrable on [a,b], then $\sqrt{f}$ is Riemann integrable on [a,b].

I know we have to construct an $\epsilon$ proof using the fact that f is Riemann integrable. We know that $\exists P$, a partition, such that $U(f,p) - L(f,p) < \epsilon$.

And we want to show that $| \sqrt{f(x)} - \sqrt{f(y)}|$ is somehow less than $\sum (\sup_{[x_i-1, x_i]} f - \inf_{[x_i-1, x_i]} f) \Delta x_i$, which in turn will be less than $\epsilon$. I can't seem to figure this bit out, so a hint or help would be much appreciated! Here's what I've tried (but I'm not sure if it's correct):

$$| \sqrt{f(x)} - \sqrt{f(y)}| \leq |f(x) - f(y)| \leq \sup_{[x_{i-1}, x_i]} f - \inf_{[x_{i-1}, x_i]} f \text{ } (\forall 1 \leq i \leq N) $$

And so $\exists$ partition P such that:

$$U(\sqrt{f}, p) - L (\sqrt{f}, p) \leq U(f, p) - L (f,p) < \epsilon. $$

  • $\begingroup$ Why would be $|\sqrt{f(x)} - \sqrt{f(y)}|\ \le\ |f(x)-f(y)|$? $\endgroup$
    – Berci
    Dec 20, 2016 at 22:45
  • $\begingroup$ Indeed, that inequality is true only when $\left|\sqrt{f(x)}+\sqrt{f(y)}\right|\geq 1$ $\endgroup$ Dec 20, 2016 at 22:53
  • $\begingroup$ @Berci This is the part where I'm desperately grasping at straws. I'm assuming since f:[a,b] is nonnegative, then $\sqrt{f(x} \leq f(x)$, although now I'm realizing this isn't true because for example, $\frac{1}{/sqrt{4}} > \frac{1}{4}$ $\endgroup$
    – Nikitau
    Dec 20, 2016 at 22:54
  • $\begingroup$ Do you have to construct your own $\varepsilon$-type proof? You can simply refer to the Lebesgue criterion instead. $\endgroup$
    – zipirovich
    Dec 20, 2016 at 23:35
  • $\begingroup$ @zipirovich Unfortunately, my analysis class did not cover Lebesgue criterion, so we have to to use $\epsilon$ proofs to prove Riemann integrability for our final. $\endgroup$
    – Nikitau
    Dec 21, 2016 at 0:22

2 Answers 2


Note that $|\sqrt{f(x)}-\sqrt{f(y)}|$ is not necessarily smaller than $|\,f(x)-f(y)|$, if $\,f$ is less that $1$.

But, $$ |\sqrt{x}-\sqrt{y}|=\frac{|x-y|}{|\sqrt{x}+\sqrt{y}|}\le \frac{|x-y|}{\max\{\sqrt{x},\sqrt{y}\}}. $$ Let $\varepsilon>0$. As $f$ is Riemann integrable over $[a,b]$ we may choose a partition $P=\{t_0<\cdots<t_n\}$ of $[a,b]$, with $\,m_i,\,M_i$, the infima and suprema of $f$ in the subinterval $[t_{i-1},t_i]$, such that $U(\,f,P)-L(\,f,P)<\varepsilon^2$, then $$ U(\sqrt{f},P)-L(\sqrt{f},P)=\sum_{i=1}^n(\sqrt{M_i}-\sqrt{m_i})(t_i-t_{i-1})\\=\sum_{M_i\ge\varepsilon^2}(\sqrt{M_i}-\sqrt{m_i})(t_i-t_{i-1})+ \sum_{M_i<\varepsilon^2}(\sqrt{M_i}-\sqrt{m_i})(t_i-t_{i-1}) \\ \le \frac{1}{\varepsilon}\sum_{i=1}^n(M_i-m_i)(t_i-t_{i-1})+\varepsilon\sum_{i=1}^n(t_i-t_{i-1})<\varepsilon+\varepsilon(b-a). $$ Since $$ \sum_{M_i\ge\varepsilon^2}(\sqrt{M_i}-\sqrt{m_i})(t_i-t_{i-1})\le \sum_{M_i\ge\varepsilon^2}\frac{1}{\sqrt{M_i}}(M_i-m_i)(t_i-t_{i-1})\le\frac{1}{{\varepsilon}}\big(U(\,f,P)-L(\,f,P)\big)<\varepsilon, $$ while $$ \sum_{M_i<\varepsilon^2}(\sqrt{M_i}-\sqrt{m_i})(t_i-t_{i-1})\le \sum_{M_i<\varepsilon^2}\varepsilon(t_i-t_{i-1})\le \varepsilon(b-a). $$

  • $\begingroup$ I don't think I quite follow after the first line. Won't $U(f,P) - L(f,P) < \epsilon$ and not $epsilon^2$? I don't think I quite understand the third line either. Where did $(b-a)$ come from? $\endgroup$
    – Nikitau
    Dec 21, 2016 at 0:56
  • $\begingroup$ @Nikitau I made some corrections, and I tried to explain things a bit better. $\endgroup$ Dec 21, 2016 at 8:37

In fact, if $f$ is Riemann integrable on $[a,b],$ $f([a,b]) \subset [c,d],$ and $g$ is continuous on $[c,d],$ then $g\circ f$ is Riemann integrable on $[a,b].$ Your problem is a special case of this.

  • 1
    $\begingroup$ But this is a highly nontrivial theorem. I don't see how you could prove this without using the characterization of Riemann-integrable functions as functions that are almost everywhere continuous. $\endgroup$
    – Dominik
    Dec 21, 2016 at 9:17

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