TITLE: Can the space of Riemann integrable function can be defined as the closure of step functions?
QUESTION [5 upvotes]: Q1) We often defined the Riemann integral of a function with Darboux sum, but could we define the space of Riemann integrable function as the closure of step functions ? (but in "$L^1$"-sense) (as Lebesgue Integrable function are defined as the closure of simple function). I mean, $f$ is a Riemann integrable function on $[a,b]$ $\iff$ there is a sequence a sequence of step functions s.t. $$\lim_{n\to \infty }\int_a^b f_n=\int_a^b f$$where $f$ are step functions. 
Q2) I have an other small question (but related to the previous one). Regulated function are Riemann integrable, but is the converse also true ? I.e. all Riemann integrable function are Regulated. I recall that a regulated function is a uniform limit of step function.

REPLY [4 votes]: For Q2, not every Riemann integrable function is regulated.
A simple counterexample is $f:[0,1] \to [0,1]$ where 
$$f(x) = \begin{cases}1, \quad x = 1/2^k, \,\, k \in \mathbb{N}\\ 0, \quad \text{otherwise} \end{cases}$$
Clearly, $f$ is Riemann integrable since the set of discontinuities has measure zero.
Consider a partition $(x_0,x_1,\ldots, x_n)$ of $[0,1]$ and a step function $\phi(x) = \sum_{j=1}^n a_j \mathbf{1}_{(x_{j-1},x_j]}.$
If $f$ were regulated there would exist a step function $\phi$ such that $|f(x) -\phi(x)| < 1/4$ for all $x \in [0,1]$. Thus $|f(x) - a_1| < 1/4$ for all $x \in [0,x_1]$.  
This is impossible since $f$ takes both the values $0$ and $1$ at infinitely many points in $[0,x_1].$ We cannot have both $|0 - a_1| = |a_1| < 1/4$ and $|1 - a_1| < 1/4$, since $1 - |a_1| < |1 - a_1| < 1/4 $, which implies $|a_1| > 3/4$.