A Riemann-integrable map that is not regulated

For a Banach space \(X\), a function \(f : [a,b] \to X\) is said to be regulated if there exists a sequence of step functions \(\varphi_n : [a,b] \to X\) converging uniformly to \(f\).

One can prove that a regulated function \(f : [a,b] \to X\) is Riemann-integrable. Is the converse true? The answer is negative and we provide below an example of a Riemann-integrable real function that is not regulated. Let’s first prove following theorem.

THEOREM A bounded function \(f : [a,b] \to \mathbb R\) that is (Riemann) integrable on all intervals \([c, b]\) with \(a < c < b\) is integrable on \([a,b]\).

PROOF Take \(M > 0\) such that for all \(x \in [a,b]\) we have \(\vert f(x) \vert < M\). For \(\epsilon > 0\), denote \(c = \inf(a + \frac{\epsilon}{4M},b + \frac{b-a}{2})\). As \(f\) is supposed to be integrable on \([c,b]\), one can find a partition \(P\): \(c=x_1 < x_2 < \dots < x_n =b\) such that \(0 \le U(f,P) - L(f,P) < \frac{\epsilon}{2}\) where \(L(f,P),U(f,P)\) are the lower and upper Darboux sums. For the partition \(P^\prime\): \(a= x_0 < c=x_1 < x_2 < \dots < x_n =b\), we have \[ \begin{aligned} 0 \le U(f,P^\prime) - L(f,P^\prime) &\le 2M(c-a) + \left(U(f,P) - L(f,P)\right)\\ &< 2M \frac{\epsilon}{4M} + \frac{\epsilon}{2} = \epsilon \end{aligned}\] We now prove that the function \(f : [0,1] \to [0,1]\) defined by \[ f(x)=\begin{cases} 1 &\text{ if } x \in \{2^{-k} \ ; \ k \in \mathbb N\}\\ 0 &\text{otherwise} \end{cases}\] is Riemann-integrable (that follows from above theorem) and not regulated. Let's prove it. If \(f\) was regulated, there would exist a step function \(g\) such that \(\vert f(x)-g(x) \vert < \frac{1}{3}\) for all \(x \in [0,1]\). If \(0=x_0 < x_1 < \dots < x_n=1\) is a partition associated to \(g\) and \(c_1\) the value of \(g\) on the interval \((0,x_1)\), we must have \(\vert 1-c_1 \vert < \frac{1}{3}\) as \(f\) takes (an infinite number of times) the value \(1\) on \((0,x_1)\). But \(f\) also takes (an infinite number of times) the value \(0\) on \((0,x_1)\). Hence we must have \(\vert c_1 \vert < \frac{1}{3}\). We get a contradiction as those two inequalities are not compatible.

A discontinuous midpoint convex function

Let’s recall that a real function \(f: \mathbb R \to \mathbb R\) is called convex if for all \(x, y \in \mathbb R\) and \(\lambda \in [0,1]\) we have \[
f((1- \lambda) x + \lambda y) \le (1- \lambda) f(x) + \lambda f(y)\] \(f\) is called midpoint convex if for all \(x, y \in \mathbb R\) \[
f \left(\frac{x+y}{2}\right) \le \frac{f(x)+f(y)}{2}\] One can prove that a continuous midpoint convex function is convex. Sierpinski proved the stronger theorem, that a real-valued Lebesgue measurable function that is midpoint convex will be convex.

Can one find a discontinuous midpoint convex function? The answer is positive but requires the Axiom of Choice. Why? Because Robert M. Solovay constructed a model of Zermelo-Fraenkel set theory (ZF), exclusive of the axiom of choice where all functions are Lebesgue measurable. Hence convex according to Sierpinski theorem. And one knows that convex functions defined on \(\mathbb R\) are continuous.

Referring to my previous article on the existence of discontinuous additive map, let’s use a Hamel basis \(\mathcal B = (b_i)_{i \in I}\) of \(\mathbb R\) considered as a vector space on \(\mathbb Q\). Take \(i_1 \in I\), define \(f(i_1)=1\) and \(f(i)=0\) for \(i \in I\setminus \{i_1\}\) and extend \(f\) linearly on \(\mathbb R\). \(f\) is midpoint convex as it is linear. As the image of \(\mathbb R\) under \(f\) is \(\mathbb Q\), \(f\) is discontinuous as explained in the discontinuous additive map counterexample.

Moreover, \(f\) is unbounded on all open real subsets. By linearity, it is sufficient to prove that \(f\) is unbounded around \(0\). Let’s consider \(i_1 \neq i_2 \in I\). \(G= b_{i_1} \mathbb Z + b_{i_2} \mathbb Z\) is a proper subgroup of the additive \(\mathbb R\) group. Hence \(G\) is either dense of discrete. It cannot be discrete as the set of vectors \(\{b_1,b_2\}\) is linearly independent. Hence \(G\) is dense in \(\mathbb R\). Therefore, one can find a non vanishing sequence \((x_n)_{n \in \mathbb N}=(q_n^1 b_{i_1} + q_n^2 b_{i_2})_{n \in \mathbb N}\) (with \((q_n^1,q_n^2) \in \mathbb Q^2\) for all \(n \in \mathbb N\)) converging to \(0\). As \(\{b_1,b_2\}\) is linearly independent, this implies \(\vert q_n^1 \vert, \vert q_n^2 \vert \underset{n\to+\infty}{\longrightarrow} \infty\) and therefore \[
\lim\limits_{n \to \infty} \vert f(x_n) \vert = \lim\limits_{n \to \infty} \vert f(q_n^1 b_{i_1} + q_n^2 b_{i_2}) \vert = \lim\limits_{n \to \infty} \vert q_n^1 \vert = \infty.\]

A discontinuous additive map

A function \(f\) defined on \(\mathbb R\) into \(\mathbb R\) is said to be additive if and only if for all \(x, y \in \mathbb R\)
\[f(x+y) = f(x) + f(y).\] If \(f\) is supposed to be continuous at zero, \(f\) must have the form \(f(x)=cx\) where \(c=f(1)\). This can be shown using following steps:

  • \(f(0) = 0\) as \(f(0) = f(0+0)= f(0)+f(0)\).
  • For \(q \in \mathbb N\) \(f(1)=f(q \cdot \frac{1}{q})=q f(\frac{1}{q})\). Hence \(f(\frac{1}{q}) = \frac{f(1)}{q}\). Then for \(p,q \in \mathbb N\), \(f(\frac{p}{q}) = p f(\frac{1}{q})= f(1) \frac{p}{q}\).
  • As \(f(-x) = -f(x)\) for all \(x \in\mathbb R\), we get that for all rational number \(\frac{p}{q} \in \mathbb Q\), \(f(\frac{p}{q})=f(1)\frac{p}{q}\).
  • The equality \(f(x+y) = f(x) + f(y)\) implies that \(f\) is continuous on \(\mathbb R\) if it is continuous at \(0\).
  • We can finally conclude to \(f(x)=cx\) for all real \(x \in \mathbb R\) as the rational numbers are dense in \(\mathbb R\).

We’ll use a Hamel basis to construct a discontinuous linear function. The set \(\mathbb R\) can be endowed with a vector space structure over \(\mathbb Q\) using the standard addition and the multiplication by a rational for the scalar multiplication.

Using the axiom of choice, one can find a (Hamel) basis \(\mathcal B = (b_i)_{i \in I}\) of \(\mathbb R\) over \(\mathbb Q\). That means that every real number \(x\) is a unique linear combination of elements of \(\mathcal B\): \[
x= q_1 b_{i_1} + \dots + q_n b_{i_n}\] with rational coefficients \(q_1, \dots, q_n\). The function \(f\) is then defined as \[
f(x) = q_1 + \dots + q_n.\] The linearity of \(f\) follows from its definition. \(f\) is not continuous as it only takes rational values which are not all equal. And one knows that the image of \(\mathbb R\) under a continuous map is an interval.