# 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.