Let us consider real-valued functions on the bounded interval $[0,1]$. A "step function" means an element of the vector space spanned by indicator functions of (points and) intervals in $[0,1]$ (the integral of step functions is, of course, unproblematic). The following definitions/properties are standard, but are recalled to put my question in context:

A

**regulated function**$f\colon [0,1]\to\mathbb{R}$ is one such that for every $\varepsilon>0$ there exists a step function $h$ such that for all $x$ we have $|f(x)-h(x)|\leq\varepsilon$. (Equivalently, this means that $f$ has a left limit and a right limit at each point.)A

**Riemann-integrable function**$f\colon [0,1]\to\mathbb{R}$ is one such that for every $\varepsilon>0$ there exist step functions $h,\psi$ such that for all $x$ we have $|f(x)-h(x)|\leq\psi(x)$ and $\int_0^1\psi\leq\varepsilon$. (Equivalently, this means that it is a bounded function whose set of points of discontinuity is of (Lebesgue-)measure zero.) In this case, the Riemann integral of $f$ can be defined as real number whose distance to $\int_0^1 h$ is $\leq\varepsilon$ for every such $h,\psi$.One possible characterization of a

**Lebesgue-integrable function**$f\colon [0,1]\to\mathbb{R}$ is that there exists a sequence $(h_n)$ of step functions such that $\sum_{n=0}^{+\infty}\int_0^1|h_n|$ converges and such that $f(x) = \sum_{n=0}^{+\infty} h_n(x)$ wherever the RHS converges absolutely. (In which case, the Lebesgue integral of $f$ can be defined as $\sum_{n=0}^{+\infty}\int_0^1 h_n$, which necessarily converges.)

Now the $\varepsilon$-definitions above are a bit tedious. We can reformulate the first and third as follows:

The space of regulated functions is the closure of the space of step functions in the topology given by uniform convergence.

The space of Lebesgue-integrable functions is the completion of the space of step functions for the $L^1$-norm (of course, this glosses over how we identify them with functions).

In either case, the integral is defined as the continuous linear function extending the trivially-defined integral of step functions.

So this suggests the following:

**Question:** Can we define the set of Riemann-integrable functions $[0,1]\to\mathbb{R}$ as the closure or completion of the space of step functions for some topology / uniform structure / norm (and so that the Riemann integral itself will then follow as the unique continuous extension of the integral of step functions)? Or is there some reason to think this is impossible?

Alternatively, is there a more sophisticated and more topological way to rephrase the elementary definition given in terms of $h,\psi$ above (and which seems to be a kind of mix between "uniform" and "$L^1$" notions)?

7more comments