Let $f:[a,b]\to \mathbb{R}$ be a monotone function (say strictly increasing). 

Then, do for every $\epsilon>0$ exist two step functions $h,g$ so that $g\le f\le h$ and $0\le h-g\le \epsilon$?

Does there exist some closed form of these functions like the one below?

I encountered the problem when trying to prove the Riemann Integrability of monotone functions via the traditional definition of the Riemann Integral (not the one with Darboux sums). The book I am reading proves that a continuous function is integrable via the process below:

Let $\mathcal{P}=\left\{ a=x\_0<...<x\_n=b \right\}$ be a partition of $[a,b]$. Define $m\_i=\min\_{x\in [x\_{i-1},x\_i]}f(x)$ and $M\_i=\max\_{x\in [x\_{i-1},x\_i]}f(x)$. By the Extreme Value theorem $M\_i,m\_i$ are well defined. We approximate $f$ with step functions:
\begin{gather}g=m\_1\chi\_{[x\_0,x\_1]}+\sum\_{i=2}^nm\_i\chi\_{(x\_{i-1},x\_i]}\\
h=M\_1\chi\_{[x\_0,x\_1]}+\sum\_{i=2}^nM\_i\chi\_{(x\_{i-1},x\_i]}\end{gather}
It is easily seen that they satisfy the "step function approximation" and as $0\le \int\_a^bh-g\le \epsilon$ (uniform continuity) by a previous theorem $f$ is integrable.

The book then goes on to generalise by discussing regulated functions. I would like however to see a self contained proof similar to the previous one if $f$ is monotone. This question is reduced to the two questions asked in the beggining of the post