Let S([a, b]) denote the real vector space generated by indicator functions of closed intervals i.e. S([a, b]) = $\left\{\sum_{i=0}^{n}{\alpha_i\chi_{I_i}}\right\}$ where n denotes some integer and $I_i \subset [a, b]$ a closed interval which may be just one point. The Riemann integral $\int : S([a, b]) \to \mathbb{R}$ can be defined for the functions in S([a, b]) in an obvious way, and a slight observation shows that $\int$ is uniformly continuous so it extends to S([a, b])’s closure in $\ell^{\infty}([a, b])$ (Note that the latter is a Banach space.)
My question is ‘how large is $\overline{S([a, b])}$?’. It is known that a bounded function defined on a bounded closed interval is Riemann integrable iff the set of its discontinuities is a null set. I doubt that this definition includes such a relatively large ( very small compared to Lebesgue or KH integration though) class of functions. $\overline{S([a, b])}$ obviously includes piecewise continuous functions ( viz. a function with finitely many discontinuities) but how about a function with countable many discontinuities ? $\overline{S([a, b])}$ includes a function like g(x) which returns the denominator of the reduced form for rationals and 0 for irrationals.
Note: Asking about the closure in $\ell^\infty([a,b])$ is the same as asking about the uniform closure, but it's a sufficiently unusual way of saying "uniform closure" to raise some doubt regarding whether that's what you meant. In any case, here I take $\overline{S([a,b])}$ to mean the uniform closure.
Of course since a uniform limit of Riemann integrable functions is Riemann integrable, every function in $\overline{S([a,b])}$ is Riemann integrable. But sad to say no, not every Riemann integrable function is in $\overline{S([a,b])}$.
Take $[a,b]=[0,1]$. Say $x_1>y_1>x_2\dots$ and $x_n\to0$. Let $f=\sum\chi_{I_n}$, where $I_n=[x_n,y_n]$. Then $f$ is Riemann integrable. But if $g\in S([0,1])$ then there exists $\delta>0$ such that $g$ is constant on $(0,\delta)$; hence $||f-g||_\infty\ge1/2$.