As Martin mentioned, $$\text{Riemann Integrable}\implies \text{Lebesgue Integrable}$$
THEOREM Let $f:[a,b]\to\Bbb R$ be a monotone function. We show that $f$ is Riemann integrable.
PROOF Let $P=\{t_0,\dots,t_n\}$ be a partition of $[a,b]$. We set the upper and lower sums:
$$U(f,P)=\sum_{k=1}^n M_k (t_k-t_{k-1})$$
$$L(f,P)=\sum_{k=1}^n m_k (t_k-t_{k-1})$$
where $$m_k=\inf\limits_{[t_{k-1},t_k]}f(x)$$
$$M_k=\sup\limits_{[t_{k-1},t_k]}f(x)$$
Assume $f$ increasing. Then $f$ is automatically bounded $f(a)\leq f(x)\leq f(b)$.
$$m_k=\inf\limits_{[t_{k-1},t_k]}f(x)=f(t_{k-1})$$
$$M_k=\sup\limits_{[t_{k-1},t_k]}f(x)=f(t_k)$$
This means that, for any partition $P$, we will have
$$U(f,P)-L(f,p)=\sum_{k=1}^n (f(t_k)-f(t_{k-1}))(t_k-t_{k-1})$$
Now, choose the partition $P$ such that $t_k-t_{k-1}<\delta$. Then $$\displaylines{
U(f,P) - L(f,p) = \sum\limits_{k = 1}^n {(f(} {t_k}) - f({t_{k - 1}}))({t_k} - {t_{k - 1}}) \cr
< \delta \sum\limits_{k = 1}^n {(f(} {t_k}) - f({t_{k - 1}})) \cr
< \delta \left( {f\left( b \right) - f\left( a \right)} \right) \cr} $$
Thus, it suffices to take $$\delta = \frac{\varepsilon }{{f\left( b \right) - f\left( a \right)}}$$ and we will have $$U(f,P) - L(f,p) < \varepsilon $$ whence $f$ will be (Riemann) integrable. For $f$ nonincreasing, apply the result to $-f$; which is non decreasing.
Another interesting fact is
THEOREM Let $f:[a,b]\to\Bbb R$ be monotone. Then the set $$\Delta=\{x\in[a,b]:f \text{ is discontinuous at } x\}$$ is at most countable.
PROOF
Define the function $$s\left( x \right) = \mathop {\lim }\limits_{y \to {x^ + }} f\left( y \right) - \mathop {\lim }\limits_{y \to {x^ - }} f\left( y \right)$$
since $f$ is monotone the left and right handed limits will always exist. It is readily seen that $f$ is discontinuous at $x=a$ if and only if $s(a)>0$. Note that, for any $x_1,x_2,\dots,x_n\in[a,b]$, we have $$\tag 1 0\leq \sum_{k=1}^n s(x_k)\leq f(b)-f(a)$$ (the sum of the gaps can't be greater than the whole $f(b)-f(a)$ gap). Let $L>0$ be given, and consider the set $$\Delta_L=\{x\in[a,b]:s(x)>L\}$$ We show this set is finite for each $L$. Indeed, suppose there were infinitely many points in $\Delta_L$. Then, we'd have $$\sum\limits_{k = 1}^n {s\left( {{x_k}} \right)} > \sum\limits_{k = 1}^n L = nL$$ and we could make this greater than $f(b)-f(a)$ by choosing $n$ large enough, which we can for we have infintely many points to take. But this contradicts $(1)$. Now, consider $\Delta_{1/n}$, for $n\in \Bbb N$. We know that this set is finite for each $n$, so $$\bigcup\limits_{n = 1}^\infty {{\Delta _{1/n}}} $$ is at most countable. But
$$\bigcup\limits_{n = 1}^\infty {{\Delta _{1/n}}} = \Delta $$
Best Answer
Because the mesh gets finer and finer, you might as well start with a mesh so fine that each point of difference is in its own interval. You can do this because with a finite set of points there is a minimum distance between any pair. Now let there be $n$ points where $f,g$ disagree. If the mesh spacing is $h$, the difference in Riemann sums is bounded by $h$ times the sum of the absolute value of the differences. As $h \to 0$.....