# Can the Riemann integral be defined through a closure/completion process?

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)?

• The french school does integration theory with the ''fonctions réglée'', which are uniform limits of step functions.
– user1688
Dec 12, 2017 at 12:08
• Regard the space of step functions as embedded in L^infinity, If we devise a family of bounded functionals on L^infinity such that the intersection of their kernels consists precisely of the Riemann integrable functions, then we can regard that space as the closure of the step functions in the topology. generated by the corresponding seminorms. Mimicing the weak topology, I am trying to cook up two such functionals using one-sided limsup minus liminf (and the sup norm). The idea is that Riemann integrable means continuous 'almost everywhere'. Dec 12, 2017 at 14:41
• @NateEldredge I think the simplest example is $x\mapsto \sin\frac{1}{x}$ (extended arbitrarily at $0$): it is not regulated because it has no right limit at $0$, but it is Riemann-integrable because it is bounded and discontinuous only at $0$. Dec 12, 2017 at 16:16
• I wonder if the following helps. Riemann-integrable functions must be bounded by definition, so let's consider only those bounded by $1$ and supported in the unit interval. The $\ell^1$ norm is the weakest norm under which the integral is continuous, by definition, so any candidate norm that solves this problem must be stronger than the $\ell^1$ norm. By the example of $\sin 1/x$, it must also be weaker than the $\ell^\infty$ norm. Consider the quasimetric between functions $d(f,g) = \inf \{ \mu(K) + \lVert f - g \rVert_{\ell^\infty(K)} \mid K \text{ compact}\}$ /continued below Dec 12, 2017 at 19:49
• This seems to require that the space of Riemann integrable functions admit a norm or uniform structure in which it is complete. Does it? Dec 14, 2017 at 17:09

## 2 Answers

Does $$\|f\|:=\inf \left\{ \int_0^1 g(x) dx : |f|\le g \text{ everywhere}, g \text{ a step function} \right\}$$ work? Then $\|\cdot\|$ is a norm on step functions, Cauchy sequences for $\|\cdot\|$ are Cauchy for $L^1$, and using completeness of $L^1$, pointwise convergence for subsequences of $L^1$ convergent functions, and something like Egorov's theorem, the Riemann integrability of the $L^1$ limit should follow. But I haven't worked out all details :)

• sorry, I meant this to be a "comment", not an answer... Dec 14, 2017 at 16:03
• And no, it doesn't work, as this is just the same as the $L^1$ norm on step functions so the closure is all of $L^1$. The sets you get from Egorov's theorem aren't good enough to create step functions. Dec 14, 2017 at 21:37

Here's a potentially trivial answer.

### Notations

Let $$X = \mathbb{R}^{[0,1]}$$ be the space of all real valued functions. Let $$S\subsetneq X$$ be the subspace of step functions. Elements of $$X$$ may be partially ordered by the product order: $$f \preceq g \iff \forall z\in[0,1] (f(z) \leq g(z)).$$ Given $$f,g\in X$$ denote by $$[f,g]$$ the closed order interval $$[f,g] := \{h\in X | f\preceq h \preceq g\}.$$ Denote by $$I_S$$ the set of all (non-empty) intervals with both end points in $$S$$.

We can define a width functional $$w:I_S \to [0,\infty)$$ by $$w([s_1, s_2]) := \int s_2 - s_1 ~dx$$ This is well-defined as $$s_2 - s_1$$ is a step function.

### Pseudo-metric

Consider the function $$\rho:X^2 \to [0,\infty]$$ defined by $$\rho(f,g) = \begin{cases} 0 & f = g \\ \inf \{ w(i) | i\in I_S; f\in i; g\in i\} & f\neq g\end{cases}$$ We take the convention that the infimum of the empty set is $$+\infty$$.

This function is trivially symmetric and non-negative. It vanishes on the diagonal by definition. And triangle inequality is satisfied by the observation that if $$[s_1, s_2] \cap [t_1,t_2] \neq \emptyset$$, then at every point $$x\in [0,1]$$, $$\max(s_2(x), t_2(x)) - \min(s_1(x), t_1(x)) \leq s_2(x) - s_1(x) + t_2(x) - t_1(x)$$

So $$\rho$$ defines a pseudo-metric on $$X$$.

### Closure of $$S$$ under $$\rho$$

The closure of $$S$$ under $$\rho$$ is simply the set of all points $$f\in X$$ such that $$\inf_{s\in S} \rho(s,f) = 0$$. This is in particular satisfied if for every $$\epsilon > 0$$ there exists $$s_1 \preceq f \preceq s_2$$ such that $$w([s_1,s_2]) < \epsilon$$. This is precisely the statement that $$f$$ is Darboux integrable.

### Extension of Riemann integral

The Riemann integral is uniformly continuous on $$S$$, pretty much by definition. Let $$s_1 \neq s_2\in S$$, the smallest interval in $$I_S$$ that contains both $$s_1$$ and $$s_2$$ is $$[s_1\wedge s_2, s_1\vee s_2]$$. Its width is $$\int |s_2 - s_1| ~dx$$.

In other words, restricted to $$S$$, we have that $$\rho$$ is the $$L^1$$ distance between step functions. And so the Riemann integral is Lipschitz continuous with respect to the pseudometric $$\rho$$, and hence uniformly continuous with respect to it, and hence has unique continuous extension to the closure of $$S$$.

### Final Remarks

A key point is that the pseudometric $$\rho$$ is not translation invariant in $$X$$. Let $$f = 0$$ and $$g$$ be the characteristic function of $$x = 1/2$$, then $$\rho(f,g) = 0$$. But $$\rho(f + D, g+D) = 1$$ where $$D$$ is the Dirichlet function.

Indeed, this is why this answer is sort of "fake". The pseudometric is designed so that every non-Riemann integrable function is an isolated point: if $$f$$ is not Riemann integrable, then there exists some $$\epsilon > 0$$ such that for every pair of step functions $$s_1 \preceq f \preceq s_2$$ we have that $$\int s_2 - s_1 ~dx \geq \epsilon$$. And therefore for any $$g \neq f$$, we have that $$\rho(f,g) \geq \epsilon$$.