The easiest way to solve this is to use the theorem: If $g$ be a continuous function in $[M, m] \subset [j, k]$ and $f$ be bounded and Riemann integrable, then $g \circ f$ is Riemann integrable too. Using $g(x) = x^{1/2}$ solves the problem, but I want to prove it more directly.
Here is my approach:
From the necessary and sufficient condition of integrability, we have
$\forall \epsilon>0$, $\exists$ a $\delta >0$ such that, $0 \leq W(P,f) = \sum_{r=1}^{n} (M_r-m_r)\delta_r=U(P,f)-L(P,f) < \epsilon $ $\ \forall \ ||P||< \delta$, $P$ being any partition of $[a,b]$.
We now create a set ( more precisely, a subset of the index set $I$ ), $Z$, defined as $Z=\{r \in I: \sup_{[x_{r-1},x_r]}f=\sup_{[x_{r-1},x_r]}\sqrt{f}=0\}$. Since the function is non negative over the entire interval, therefore, in every interval indexed in $Z$ must have $0$ as the infimum.
By the property of non negative real numbers, $m_r \leq f \leq M_r \implies \sqrt{m_r} \leq \sqrt{f} \leq \sqrt{M_r}$
Now taking $W(P^*,\sqrt{f})= \sum_{r\in I\Z}(\sqrt{M_r}-\sqrt{m_r})\delta_r =\sum_{r\in I\Z}\frac{({M_r}-{m_r})}{(\sqrt{M_r}+\sqrt{m_r})}\delta_r \leq \mu\sum_{r\in I\Z}({M_r}-{m_r})\delta_r $
choosing $\min \{\sqrt{M_r}+\sqrt{m_r}: r \in I\Z\}= K = 1/\mu >0 (\neq0$, by our choice of $r$) .
We further have $\sqrt{M_r}-\sqrt{m_r} = 0 =M_r -m_r \ \forall r\in Z.$ Hence $W(P^*,\sqrt{f})= W(P,\sqrt{f}) $
Now, $\forall \epsilon >0$, $\exists$ a $\delta_\mu >0$ such that, $0 \leq W(P,f) = \sum_{r=1}^{n} (M_r-m_r)\delta_r< \epsilon/\mu $ [$\mu$ is a fixed positive real number] $\ \forall \ ||P||< \delta_\mu$, $P$ being any partition of $[a,b]$.
Thus, we finally have the following:
$\forall \epsilon>0$, $\exists$ a $\delta_\mu >0$ such that, $0 \leq W(P,\sqrt{f}) \leq \mu \ W(P,f) < \epsilon $, $\forall \ ||P||< \delta_\mu.$
P.S : I am aware of the duplicates. I just want my solution verified.
Best Answer
Following your notation, note that we have $$\sqrt{M_i} - \sqrt{m_i} \leq \sqrt {M_i-m_i}$$ since $\sqrt{(M_i-m_i) + m_i} \leq \sqrt {M_i-m_i} + \sqrt{m_i}$.
To bound the Riemann sum, $$\sum_i (\sqrt{M_i} - \sqrt{m_i}) \delta_i \leq \sum_i \sqrt {(M_i-m_i)\delta_i} \sqrt \delta_i \leq \left( \sum_i (M_i-m_i)\delta_i\right)^{1/2} \left( \sum_i \delta_i\right)^{1/2} $$ because $f$ is Riemann integrable, $\sum_i (M_i-m_i)\delta_i$ can be made arbritrily small, and $\sum_i \delta_i = b-a$ is bounded, we are done.