ExPreQ1: If $f>0$ is integrable on $[a,b]$, then $\sqrt{f}$ is integrable. I want to prove the following:

If $f>0$ is Riemann integrable on $[a,b]$, then $\sqrt{f}$ is also Riemann integrable.

My attempt:
Since $f$ is integrable on $[a,b]$, $f$ is bounded, thus exist $0<M\in\mathbb{R}$ s.t $0<f(x)<M$. Therefor $\sqrt{f}$ is bounded on that interval, suppose with $\sqrt{M}$. That observation gives us a stimulation to go further. 
For every partition $T$ of $[a,b]$ we will define $\omega_i^f$ and $\omega_i^{\sqrt{f}}$ be the oscillations of $f$ and $\sqrt{f}$, respectively, on $[x_{i-1},x_i]$. Now, for every $x_1,x_2\in[x_{i-1},x_i]$ we will have:
$$|\omega_i^{\sqrt{f}}|=|\sqrt{f(x_1)}-\sqrt{f(x_2)}|\leq \sqrt{f(x_1)-f(x_2)}=\sqrt{\omega_i^f} $$  
Edit:
A progress...
Let us look now at the following sum:
$$\sum^n_{i=1}\omega_i^{\sqrt{f}}\Delta{x_i}$$
We know that,
$$\sum^n_{i=1}\omega_i^{\sqrt{f}}\Delta{x_i}\leq \sum^n_{i=1}\sqrt{\omega_i^{f}}\Delta{x_i}$$
And by Cauchy-Schwarz inequality we have:
$$\sum^n_{i=1}\sqrt{\omega_i^{f}}\Delta{x_i}\leq\sqrt{\sum_{i=1}^n\omega_i^{f}\sum_{i=1}^n\Delta^2{x_i}}$$
Here is the point that I don't know how to proceed...
Edit 2(With a hint of Peter Tamaroff):
For every $x_1,x_2\in [x_{i-1},x_i]$:
$$|\sqrt{f(x_1)}-\sqrt{f(x_2)}|=\frac{|f(x_1)-f(x_2)|}{|\sqrt{f(x_1)}-\sqrt{f(x_2)}|}\leq \frac{\omega^f_i}{|\sqrt{f(x_1)}-\sqrt{f(x_2)}|}\leq\frac{\sqrt{M}\omega^f_i}{2}$$
Therefor:
$$\omega_i^{\sqrt{f}}\leq\frac{\sqrt{M}\omega^f_i}{2}$$
Let us look now on:
$$ 0 \leq \sum \omega_i^{\sqrt{f}} \Delta x_i \leq  \sum\frac{\sqrt{M}\omega^f_i}{2}\Delta x_i$$
Since $f$ is integrable on $[a,b]$ we have that,
$$\lim_{\lambda(T)\to 0}\sum\frac{\sqrt{M}\omega^f_i}{2}\Delta x_i=0$$
From sandwich rule, we have:
$$\lim_{\lambda(T)\to 0}\sum\omega^{\sqrt{f}}_i\Delta x_i=0$$
Therefor by theorem, $\sqrt{f}$ is Riemann integrable on $[a,b]$.
Q.E.D (?)
Please, without Lebesgue name!  
And thank you all!
 A: Hint: Use that $$\sqrt{x}-\sqrt{y}= \frac{x-y}{\sqrt{x}+\sqrt{y}}$$
Alternatively, there is the following theorem:
THM Let $f:[a,b]\to \Bbb R$ be Riemann integrable, and supose $\phi(x):[m,M]\to\Bbb R$ is continuous. Assume $f([a,b])\subset [m,M]$. Then $g=\phi\circ f$ is Riemann integrable.
P Take $\epsilon >0$. Since $\phi$ is  continuous over the compact $[m,M]$, it is uniformly continuous there, so there exists $\delta >0$ such that for each $x,y\in [m,M]$, $|x-y|<\delta\implies |\phi(x)-\phi(y)|<\epsilon$. Since $f$ is Riemann integrable on $[a,b]$ there exists a partition $P_\epsilon$ such that for any refinement $P$ of $P_\epsilon$ we have $$\tag 1 U(f,P)-L(f,P)<\delta^2$$
Assuming $P=\{x_0,x_1,\dots,x_n\}$, let $$M_i=\sup\{f(x):x\in[x_{i-1},x_i]\}$$
$$m_i=\inf\{f(x):x\in[x_{i-1},x_i] \}$$
$$M_i^*=\sup\{g(x):x\in[x_{i-1},x_i]\}$$
$$m_i^*=\inf\{g(x):x\in[x_{i-1},x_i] \}$$
Divide now the numbers $1,\dots,n$ into two classes: $i\in A$ if $M_i-m_i<\delta$, and $i\in B$ if $M_i-m_i\geq \delta$. If $i\in A$, the way we chose $\delta$ gives that $$M_i^*-m_i^*\leq \epsilon$$ For $i\in B$, we have that $$M_i^*-m_i^*\leq 2K$$ where $K=\sup\{|\phi(x)|:x\in[m,M]\}$
We have by $(1)$ that $$\delta\sum_{i\in B}\Delta x_i\leq \sum_{i\in B}(M_i-m_i)\Delta x_i<\delta^2$$
since $B$ is a subset of $\{1,2,\dots,n\}$ and all is positive.
It follows that $$U(g,P)-L(g,P)=\sum_{i\in A}(M_i^*-m_i^*)\Delta x_i+\sum_{i\in B}(M_i^*-m_i^*)\Delta x_i\\ \leq \epsilon(b-a)+2K\delta <\epsilon(b-a+2K)$$
for we may assume $\delta <\epsilon$. Since $\epsilon >0$ was arbitrary, the theorem follows. $\blacktriangle$.
A: another easy solution is available\
$f:[a,b]\rightarrow~R^{+}$ be the given riemann integrable function we consider $g:R^{+}\rightarrow~R^{+}$ defined by $g(x)=\sqrt{x}$. Then $g\circ~f=\sqrt{f}$.\
Sice g is continuous and f is Riemann integrable $\Rightarrow~\sqrt{f}$ is riemann integrable.
A: This question has been asked several times on this site, and outside of some hints that are fairly removed from the actual argument structure, no full proof using upper and lower sums has been offered. Here is one such proof. We will argue the most general case where $f$ is assumed to be non-negative.

Relevant lemmas:
$\forall x,y \in \mathbb R: x-y=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right) \quad$ some trivial manipulations of this equality can be made assuming $x$ and $y$ adhere to appropriate criteria: e.g. $\sqrt{x}+\sqrt{y} \neq 0 \implies \sqrt{x}-\sqrt{y}=\frac{x-y}{\sqrt{x}+\sqrt{y}} \quad (*_1)$
$\forall x\geq y \in \mathbb R: 0 \leq \sqrt{x}-\sqrt{y} \leq \sqrt{x-y} \quad (*_2)$

Consider any arbitrary closed subinterval of $[a,b]$. Call this subinterval $i:=[c,d]\subseteq [a,b]$. We will use the following notation:
$M_i^f:=$ supremum of $f$ on the subinterval $i$
$m_i^f:=$ infimum of $f$ on the subinterval of $i$
$M_i^{\sqrt{f}}:=$ supremum of $\sqrt{f}$ on the subinterval $i$
$m_i^{\sqrt{f}}:=$ infimum of $\sqrt{f}$ on the subinterval $i$
Note that because $\sqrt{\cdot}$ is a strictly increasing function, we have that $\sqrt{M_i^f}=M_i^{\sqrt{f}}$ and $\sqrt{m_i^f}=m_i^{\sqrt{f}} \quad (*_3)$
Further, we know that $M_i^f=0 \implies \sqrt{M_i^f}=0$ and $m_i^f=0 \implies \sqrt{m_i^f}=0$
From $(*_1)$ and $(*_3)$, we know that $M_i^f-m_i^f=(\sqrt{M_i^f}-\sqrt{m_i^f})(\sqrt{M_i^f}+\sqrt{m_i^f})=(M_i^{\sqrt{f}}-m_i^{\sqrt{f}})(M_i^{\sqrt{f}}+m_i^{\sqrt{f}})$. If we assume that $M_i^f\neq m_i^f$, because $f$ is non-negative, we know that $M_i^{\sqrt{f}}+m_i^{\sqrt{f}}\neq 0$. Therefore, we can rewrite the above equation as:
$$\frac{M_i^{\sqrt{f}}-m_i^{\sqrt{f}}}{M_i^f- m_i^f}=\frac{1}{M_i^{\sqrt{f}}+m_i^{\sqrt{f}}} \quad (*_4)$$
$(*_4)$ provides us with two important observations:

*

*If $M_i^{\sqrt{f}}+m_i^{\sqrt{f}} \geq 1$ and $M_i^f\neq m_i^f$, then $\frac{1}{M_i^{\sqrt{f}}+m_i^{\sqrt{f}}}\leq 1$. This means that $M_i^{\sqrt{f}}-m_i^{\sqrt{f}} \leq M_i^f- m_i^f \quad (\dagger_1)$


*If $M_i^{\sqrt{f}}+m_i^{\sqrt{f}} \lt 1$ and $M_i^f\neq m_i^f$, then $\frac{1}{M_i^{\sqrt{f}}+m_i^{\sqrt{f}}}\gt 1$. This means that $M_i^{\sqrt{f}}-m_i^{\sqrt{f}} \gt M_i^f- m_i^f \quad (\dagger_2)$
In the event that $M_i^{\sqrt{f}}= m_i^{\sqrt{f}}$, we also have that $M_i^f= m_i^f$. This provides us with a third observation:


*If  $M_i^{\sqrt{f}}= m_i^{\sqrt{f}}$, then  $M_i^{\sqrt{f}}- m_i^{\sqrt{f}}=M_i^f -m_i^f \quad (\dagger_3)$
These three observations are important for the following reasons:
Consider an arbitrary $\varepsilon \gt 0$. By assumption, we know that $f$ is integrable, which means there exists a partition $P$ of $[a,b]$ defined as $P=\{a,t_1,\cdots,t_{n-1},b\}$ such that: $U(f,P)-L(f,P) =\displaystyle \sum_{i=1}^n(M_i^f-m_i^f)(t_i-t_{i-1}) \lt \varepsilon \quad (*_5)$.
Now, when it comes to determining a partition $Q$ of $[a,b]$ defined as $\{a,q_1,\cdots,q_{k-1},b\}$ that satisfies $U(\sqrt{f},Q)-L(\sqrt{f},Q)=\displaystyle \sum_{j=1}^k(M_j^{\sqrt{f}}-m_j^{\sqrt{f}})(q_{j}-q_{j-1}) \lt \varepsilon \quad (*_6)$, note that we can use our first and third observations $(\dagger_1)$ and $(\dagger_3)$ to carry out the following procedure:

Look at each subinterval $[t_{i-1},t_i]$ that describes $P$. Each subinterval will have a corresponding $M_i^f$ and $m_i^f$. Take the square root of each of these values. This will give us $M_i^{\sqrt{f}}$ and $m_i^{\sqrt{f}}$. If $(\dagger_1)$ or $(\dagger_3)$ is satisfied, we can use these subintervals to populate $Q$ because under these circumstances, $M_i^{\sqrt{f}}-m_i^{\sqrt{f}} \leq M_i^f-m_i^f$, which implies that $(M_i^{\sqrt{f}}-m_i^{\sqrt{f}})(t_i-t_{i-1}) \leq (M_i^f-m_i^f)(t_i-t_{i-1})$. This means that $(M_i^{\sqrt{f}}-m_i^{\sqrt{f}})(t_i-t_{i-1})$'s contribution to $U(\sqrt{f},Q)-L(\sqrt{f},Q)$ behaves nicely. We will call the subintervals defined by $P$ that satisfy $(\dagger_1)$ or $(\dagger_3)$ Good Subintervals, which, for convenience, we will denote as the set $G$ (i.e. the set $G$ contains the Good Subintervals as its elements).

Now, suppose that a given subinterval $[t_{i-1},t_i]$ of $P$ satisfies $(\dagger_2)$. Then we have that  $M_i^{\sqrt{f}}-m_i^{\sqrt{f}} \gt M_i^f- m_i^f$, which implies $(M_i^{\sqrt{f}}-m_i^{\sqrt{f}})(t_i-t_{i-1}) \gt (M_i^f- m_i^f)(t_i-t_{i-1})$. This is a problem.  Why?  For any $i \in \{1,2,\cdots n\}$, we know that $M_i^f \geq m_i^f$. Therefore, we can deduce from $(*_5)$ that $\forall  \in \{1,2,\cdots n\}: 0 \leq (M_i^f -m_i^f)(t_i-t_{i-1}) \lt \varepsilon \quad (*_7)$.
If we were to include the subinterval $[t_{i-1},t_i]$ of $P$ that satisfies $(\dagger_2)$ into $Q$ in the context of $(*_6)$, then we may end up with a partition $Q$ such that  $U(\sqrt{f},Q)-L(\sqrt{f},Q) \geq \varepsilon$ because we may end up with a situation where $M_i^{\sqrt{f}}-m_i^{\sqrt{f}} \gg M_i^f- m_i^f$. $(*_7)$ shows us that if even one subinterval's contribution to $U(\sqrt{f},Q)-L(\sqrt{f},Q)$ exceeds or equals $\varepsilon$, we are in trouble because there is no way of using negative values to counterbalance it.  If a subinterval of $P$ satisfies $(\dagger_2)$, we will call it a Bad Subinterval, which we will describe as the set $B$ (i.e. the set $B$ contains the Bad Subintervals as its elements).
As we go through the subintervals defined by $P$ in the context of $(*_5)$, if we encounter a $[t_{i-1},t_i]\in B$, then we will carry out the following procedure:

Suppose $[t_{i-1},t_i]\in B$. Note that $(*_5)$ provides us with a corresponding $M_i^f$ and $m_i^f$ such that, by $(\dagger_2)$: $M_i^f \neq m_i^f$.   Because $f$ is integrable on $[a,b]$, we know that $f$ is also integrable on $[t_{i-1},t_i]\subseteq [a,b]$. This means that for any $\eta \gt 0$, there exists a partition $H_i$ of $[t_{i-1},t_i]$ defined as $[t_{i-1},h_1,h_2,\cdots,h_{w-2},h_{w-1},t_i]$ such that: $U(f,H_i)-L(f,H_i) = \displaystyle \sum_{j=1}^w (M_j^f-m_j^f)(h_j-h_{j-1}) \lt \eta$. Suppose we let $\eta=\left(M_i^f-m_i^f\right)^2 \gt 0$. A similar argument to $(*_7)$ reveals that $\forall j \in \{1,2,\cdots,w \}: 0 \lt (M_j^f-m_j^f) \lt \left(M_i^f-m_i^f\right)^2 \quad (*_8)$. We can simplify $(*_8)$ by taking the square root of all terms:  $0 \lt \sqrt{M_j^f-m_j^f} \lt M_i^f-m_i^f$. Applying our second lemma $(*_2)$ then gives us: $0\leq \sqrt{M_j^f}-\sqrt{m_j^f}\leq \sqrt{M_j^f-m_j^f} \lt M_i^f-m_i^f$. But $\sqrt{M_j^f}=M_j^{\sqrt{f}}$ and $\sqrt{m_j^f}=m_j^{\sqrt{f}}$, which means we can simplify to: \begin{align}\forall j \in \{1,2,\cdots,w\}:M_j^{\sqrt{f}}-m_j^{\sqrt{f}} \lt M_i^f-m_i^f \quad (*_9) \end{align} Now, if $(*_9)$ is true, then \begin{align}\displaystyle U(\sqrt{f},H_i)-L(\sqrt{f},H_i)&=\sum_{j=1}^w (M_j^{\sqrt{f}}-m_j^{\sqrt{f}})(h_j-h_{j-1})\\&\lt (M_i^f-m_i^f)\cdot \left(\sum_{j=1}^w h_j-h_{j-1}\right)\\&=(M_i^f-m_i^f)(t_{i}-t_{i-1})\end{align} which means that we have: \begin{align} \displaystyle U(\sqrt{f},H_i)-L(\sqrt{f},H_i) \lt (M_i^f-m_i^f)(t_{i}-t_{i-1}) \quad (*_{10})\end{align}

From $(*_{10})$, we see that if $[t_{i-1},t_i] \in B$, then we can further subdivide $[t_{i-1},t_i]$ into the subintervals defined by $H_i$. Therefore, the way we construct $Q$ (in the context of $(*_6)$) is to use $P$ (in the context of $(*_5)$) as a skeleton: if $[t_{i-1},t_i] \in G$, then we keep it as is for $Q$, and if $[t_{i-1},t_i] \in B$, then we further subdivide it in accordance to the partitioning scheme $H_i$ provided by $(*_{10})$. Using this $Q$, we will have that $U(\sqrt{f},Q)-L(\sqrt{f},Q) \lt \varepsilon$, which means that $\sqrt{f}$ is integrable.
