Can the space of Riemann integrable function can be defined as the closure of step functions? 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. 
 A: 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 $|a_1| > 3/4$. 
A: First, just saying that $$\int_a^b f_n\to\int_a^b f$$does not imply that $f_n\to f$ in any interesting or reasonable sense. For example, if $f$ is Riemann integrable you  could just let each $f_n$ be the constant function
$$f_n=\frac1{b-a}\int_a^b f.$$It seems more reasonable to ask this: If $f$ is Riemann integrable does there exist a sequence of step functions $f_n$ with $$\int_a^b|f-f_n|\to0?$$The answer to that is yes; this follows from the characterization of the Riemann integral in terms of Darboux sums. Hint: If $m_j\le f\le M_j$ on the interval $I_j$ then $|f-m_j|\le M_j-m_j$ on $I_j$.
Can this be used to give an alternate definition of the Riemann integral? I don't see how - if we don't know what the integral is then we don't have a definition for $\int_a^b|f-f_n|$.
No, a Riemann  integrable function need not be regulated. Because for example a regulated function has one-sided limits at every point.
A: For Q1), the fact that simple function are dense in Lebesgue integrable functions is a consequence, not a definition ! Normally, $f$ is integrable in lebesgue sense if $$\sup\left\{\int_a^b \varphi\mid \varphi\leq f\text{ simple}\right\}=\inf\left\{\int_a^b \varphi\mid  f\leq \varphi\text{ simple}\right\},$$
and we write $\int_a^b f$ this number. For Riemann, we say that $f$ is Riemann integrable if 
$$\sup\left\{\int_a^b \varphi\mid \varphi\leq f\text{ step function}\right\}=\inf\left\{\int_a^b \varphi\mid  f\leq \varphi\text{ step function}\right\}.$$
But to define $\int_a^b f$ as "there is a sequence of step function $(f_n)_n$ s.t. $$\int_a^b f=\lim_{n\to \infty }\int_a^bf_n$$
has unfortunately no sense. May be you mean that : "if there is a sequence $(f_n)_n$ s.t. $f_n(x)\to f(x)$ for all $x\in [a,b]$, then we define $\int_a^b f$ as $$\int_a^b f=\lim_{n\to \infty }\int_a^b f_n,$$
but unfortunately, you can have an other sequence of step function $(g_n)$ s.t. $g_n(x)\to f(x)$ for all $x\in [a,b]$" but unfortunately it can happen that$$\lim_{n\to \infty }\int_a^b f_n\neq \lim_{n\to \infty }\int_a^b g_n,$$
and thus $\int_a^bf$ wouldn't be well defined (since depending on the sequence). 
