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
$$\bigcup\limits_{n = 1}^\infty {{\Delta _{1/n}}} = \Delta $$
Your definition has problems. The most obvious one is that there is nothing in your definition forcing you to "get essentially the same result even when you refine the partition more". But even if you introduce that, you have a more subtle problem. If you specify a fixed partitioning and point evaluation scheme, then certain functions are integrable that shouldn't be. For instance, the indicator function of the rationals in $[0,1]$ is "integrable" in the sense that the left Riemann sums for uniform partitions are all equal to $1$. But it shouldn't be Riemann integrable (it is "too wildly discontinuous").
Being more precise about these problems, we don't want our definition of the Riemann integral to depend on any particular scheme for partitioning the interval. We also don't want it to depend on any scheme for choosing the points in the subintervals to evaluate the integrand. For example, we should not claim that a function is really integrable merely because its left Riemann sums over uniform partitions converge. To enforce this, the definition involves a quantifier over all possible partitions with small mesh and all possible ways to choose evaluation points for such partitions.
One of the basic theorems that makes Riemann integration manageable says that this definition is redundant. This theorem says that Riemann integration and Darboux integration are equivalent. Darboux integration is based on the simpler requirement that for any tolerance, there exists some partition such that the difference between the lower and upper sums for that partition is smaller than the tolerance. Thus, we are free to pick the partition ourselves, but we are forced to check the "worst possible" scheme of evaluation points, rather than all possible ones.
Like your definition, this definition also ostensibly doesn't force you to "get essentially the same result upon refinement", but actually it does. The basic fact, which is key to the proof, is that the upper sum decreases upon refinement, while the lower sum increases upon refinement. Thus the difference goes down upon refinement. So now to prove this result, you need only prove that your definition of integrability is equivalent to the "refinement" definition, which is given after the third paragraph of https://en.wikipedia.org/wiki/Riemann_integral#Riemann_integral
Best Answer
Given $\epsilon>0$, choose a partition $P=\{a=x_{0}<\cdots<x_{n}=b\}$ such that $U(f,P)-L(f,P)<\epsilon$.
Let $c_{i}=\inf_{x\in[x_{i},x_{i+1}]}f(x)$, $i=0,...,n-1$. Set $g=\displaystyle\sum_{i=0}^{n-1}c_{i}\chi_{[x_{i},x_{i+1})}$, then \begin{align*} \int_{a}^{b}|f(x)-g(x)|dx&=\sum_{i=0}^{n-1}\int_{x_{i}}^{x_{i+1}}|f(x)-c_{i}|dx\\ &=\sum_{i=0}^{n-1}\int_{x_{i}}^{x_{i+1}}(f(x)-c_{i})dx\\ &\leq\sum_{i=0}^{n-1}\int_{x_{i}}^{x_{i+1}}\left(\sup_{x\in[x_{i},x_{i+1}]}f(x)-c_{i}\right)\\ &=\sum_{i=0}^{n-1}\left(\sup_{x\in[x_{i},x_{i+1}]}f(x)-\inf_{x\in[x_{i},x_{i+1}]}f(x)\right)(x_{i+1}-x_{i})\\ &=U(f,P)-L(f,P)\\ &<\epsilon. \end{align*}