Pointwise maximum of integrables is Darboux integrable $\newcommand{\set}[1]{\left\{#1\right\}}$
Let $f,g$ be integrable in $[a,b]$. Let $h(x) = \max\set{f(x),g(x)}$. I need to prove that $h$ is integrable using the Riemann criterion of integrability (and not using linearity of integrals), which states


A function $f:[a,b]\to\mathbb{R}$ is Darboux integrable if and only if, for every $\epsilon > 0$, there exists a partition $P$ of $[a,b]$ such that $U(f,P)-L(f,P)\leq\epsilon$


So the strategy is to take a partition $P$, look at an arbitrary subinterval, and look at $M_k(h),m_k(h)$, the supremum and infimum of $h$ in $[x_{k-1},x_k]\subset [a,b]$, and try to bound them from above using the fact that $f,g$ are integrable.
After some thinking, I concluded that
$$
M_k(h) = \max\set{M_k(f),M_k(g)},\ \ m_k(h) \geq \max\set{m_k(f),m_k(g)}
$$
Which leads to the fact that
$$
M_k(h) - m_k(h) \leq \max\set{M_k(f),M_k(g)} - \max\set{m_k(f),m_k(g)} 
$$
$$
\leq \max\set{M_k(f),M_k(g)} - \min\set{m_k(f),m_k(g)}
$$
But this is where I am stuck. For any given $a,b,c,d$, I have nothing to say about $\max\set{a,b}-\min\set{c,d}$, even if I know that $a>c, b>d$
I have seen another question where in one of the answers the writer concluded that the above is less then or equal to
$$
\max\set{M_k(f)-m_k(f),M_k(g)-m_k(g)}
$$
which, as I figured out, is not true in general.
 A: The referenced proof in Is the pointwise maximum of two Riemann integrable functions Riemann integrable? that $h = \max(f,g)$ is integrable, uses the intermediate result
$$M_k(h) - m_k(h) \leqslant \max[M_k(f) - m_k(f), M_k(g) - m_k(g)].$$
To prove this, note that on $I_k = [x_{k-1},x_k]$ we have
$$h(x) = \max[f(x),g(x)] \leqslant \max[M_k(f),M_k(g)].$$
Hence,
$$M_k(h) = \sup_{x \in I_k}h(x) \leqslant \max[M_k(f),M_k(g)].$$
We also have for $x \in I_k$,
$$h(x) = \max[f(x),g(x)] \geqslant f(x) \geqslant m_k(f), \\ m_k(h) = \inf_{x \in I_k}h(x) \geqslant m_k(f), $$
and
$$h(x) = \max[f(x),g(x)] \geqslant g(x) \geqslant m_k(g), \\ m_k(h) = \inf_{x \in I_k}h(x) \geqslant m_k(g). $$
Either $M_k(h) \leqslant M_k(f)$ or $M_k(h) \leqslant M_k(g).$  
In the first case,
$$m_k(f) \leqslant m_k(h) \leqslant M_k(h) \leqslant M_k(f), \\ M_k(h) - m_k(h) \leqslant M_k(f) - m_k(f).$$
In the second case,
$$m_k(g) \leqslant m_k(h) \leqslant M_k(h) \leqslant M_k(g), \\ M_k(h) - m_k(h) \leqslant M_k(g) - m_k(g).$$
Since the two cases are exhaustive, it follows that
$$M_k(h) - m_k(h) \leqslant \max[M_k(f) - m_k(f), M_k(g) - m_k(g)].$$
Also, since $\max[a,b] \leqslant a + b$ for $a,b \geqslant 0$ it follows that
$$M_k(h) - m_k(h) \leqslant M_k(f) - m_k(f) + M_k(g) - m_k(g).$$
This second inequality is the essential result needed to complete the proof that $h$ is integrable.
