If $f,g$ are bounded and Riemann integrable, then $\max{f,g}$ and $\min{f,g}$ are Riemann integrable I know this might be a duplicate, but I just wanted to see if my proof is correct.
I want to prove that if $f,g: [a,b]\rightarrow \mathbb{R}$ is bounded and Riemann integrable then $\max\{f,g\}$ and $\min\{f,g\}$ will also be Riemann integrable on $[a,b]$.
In order to show this, I've decided to prove that if $f$ is RI, then $|f|$ is RI. And then by arithmetic of limits, $$\max(f,g)(x) = \frac{f(x)+g(x) + |f(x)-g(x)|}{2}$$ and $$\min(f,g)(x) = \frac{f(x)+g(x) - |f(x)-g(x)|}{2}$$ will also be RI. My proof is as follows:

Since $f,g$ are RI on $[a,b]$, then by arithmetic of limits we know that $f+g$ is RI on $[a,b]$ and similarly $f-g$ is RI on $[a,b]$** (do I need to prove this or can this be safely assumed from the fact that $f-g = f+(-g)$?. And so, all there is left to show is that if $f$ is RI, then $|f|$ is RI.
By triangle inequality, we have that:
$\begin{align}\sup_{x \in [x_{j-1},x_j]}|f|(x) - \inf_{x \in [x_{j-1},x_j]}|f|(x) \\&=\sup_{x,y \in [x_{j-1},x_j]}| |f(x)|-|f(y)|| \\ &\leqslant \sup_{x,y \in [x_{j-1},x_j]}| f(x)-f(y)| \\ & \leq \sup_{x \in [x_{j-1},x_j]}f(x) - \inf_{x \in [x_{j-1},x_j]}f(x).\end{align}$
And thus, there exists a partition $P$ such that:
$U(P,|f|) - L(P,|f|) = \sum_{j=1}^n \left(\sup_{x \in [x_{j-1},x_j]}|f|(x) - \inf_{x \in [x_{j-1},x_j]}|f|(x)\right)(x_j - x_{j-1})\\ \leqslant  \sum_{j=1}^n \left(\sup_{x \in [x_{j-1},x_j]}f(x) - \inf_{x \in [x_{j-1},x_j]}f(x)\right)(x_j - x_{j-1})\\= U(P,f) - L(P,f)$
Since $f$ is RI, we have that:
$U(P, |f|) - L(P, |f|) \leqslant U(P,f)-L(P,f) < \epsilon.$
Therefore, by the arithmetic of RI functions and the fact that if $f$ is RI then $|f|$ is RI, we have that $\max\{f,g\}$ and $\min\{f,g\}$ are RI.

Is this a sufficient proof? Or am I oversimplifying things/do I have to do a direct $\epsilon$ proof?
 A: Why not prove this directly with the Riemann criterion. 
For example, in the first case, $h = \max\{f,g\}$, consider a partition $P = (x_0,x_1, \ldots, x_n)$ of $[a,b].$  With $I_k = [x_{k-1},x_k]$ define $M_k(h) = \sup_{x \in I_k}h(x)$ and $m_k(h) = \inf_{x \in I_k}h(x),$ with similar notational definitions for $M_k(f)$, $M_k(g)$, etc.
We have for all $x \in I_k$,
$$h(x) = \max[f(x),g(x)] \leqslant \max[M_k(f),M_k(g)] \\ \implies M_k(h) = \sup_{x \in I_k}h(x) \leqslant \max[M_k(f),M_k(g)], $$
and
$$h(x) = \max[f(x),g(x)] \geqslant f(x), \, g(x) \implies \\ m_k(h) = \inf_{x \in I_k}h(x) \geqslant \inf_{x \in I_k}f(x) = m_k(f) \\  m_k(h) = \inf_{x \in I_k}h(x) \geqslant \inf_{x \in I_k}g(x) = m_k(g) $$
Thus,
$$M_k(h) - m_k(h) \leqslant \max[M_k(f) - m_k(f), M_k(g) - m_k(g)] \\ \leqslant M_k(f) - m_k(f) + M_k(g) - m_k(g). $$
Forming upper and lower sums, it follows that 
$$U(P,h) - L(P,h) \leqslant U(P,f) - L(P,f) + U(P,g) - L(P,g),$$
and applying the Riemann criterion given that $f$ and $g$ are Riemann integrable , we can find a partition $P_\epsilon$ for any $\epsilon > 0$ such that $U(P_\epsilon,h) - L(P_\epsilon,h) < \epsilon.$
A: I should leave this in the comment but I do not have enough reputation point, so sorry for bothering.
I agree with RRL's answer but I am confused at this step:

Thus,(ℎ)−(ℎ)⩽max[()−(),()−()]⩽()−()+()−().


Why don't we write down $M_k(h) - m_k(h)\le max[M_k(f) - m_k(f),M_k(f) - m_k(g),M_k(g) - m_k(f),M_k(g) - m_k(g)]$
And all others remain the same. The following steps work out exactly in the same way, we will have
$M_k(h) - m_k(h)\le2*(M_kf-m_kf+M_kg-m_kg)$
then.
