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