Dirichlet Function

Definition

1.The Dirichlet function $D : \mathbb{R} \to \{0, 1\}$ is the indicator function $\mathbf{1}_{\mathbb{Q}}$ of the rational numbers. It assigns the value $1$ to rational numbers and $0$ to irrational numbers:

$$ D(x) = \mathbf{1}_{\mathbb{Q}}(x) = \begin{cases} 1 & \text{if } x \in \mathbb{Q} \\[6pt] 0 & \text{if } x \in \mathbb{R} \setminus \mathbb{Q} \end{cases} $$

Properties

3.For a rational number $r,$ the number $x + r$ is rational exactly when $x$ is rational. This gives $D(x + r) = D(x)$ for every $x \in \mathbb{R},$ so every nonzero rational number is a period. Conversely, if $r$ is a period, setting $x = 0$ gives $D(r) = D(0) = 1,$ which forces $r \in \mathbb{Q}.$ The positive rational periods have no least element, so $D$ has no fundamental period.

4.$\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ are dense in the real line. Fix $x_0 \in \mathbb{R}$ and take $\varepsilon = 1/2.$ If $x_0$ is rational, every $\delta$-neighborhood of $x_0$ contains an irrational number $x$ satisfying $0 < |x - x_0| < \delta.$ The function values satisfy:

$$|D(x) - D(x_0)| = 1 > \varepsilon$$

5.If $x_0$ is irrational, the argument uses a rational number $x$ in the $\delta$-neighborhood. Thus the $\varepsilon$-$\delta$ condition for continuity fails at every $x_0,$ and $D$ is discontinuous everywhere.

6.The function $D$ is an iterated pointwise limit of continuous functions:

$$D(x) = \lim_{k \to \infty} \lim_{j \to \infty} \cos^{2j}(k!\pi x)$$

7.If $x = p/q$ is rational, with $p \in \mathbb{Z}$ and $q \in \mathbb{N}_{>0},$ then the factorial $k!$ is divisible by $q$ for every $k \geq q.$ For these values of $k,$ the number $k!x$ is an integer, $\cos(k!\pi x) = \pm 1,$ and the inner limit is $1.$ If $x$ is irrational, then $|\cos(k!\pi x)| < 1$ for every positive integer $k,$ and the inner limit is $0.$ The outer limit agrees with $D(x).$

8.For fixed $k,$ denote the inner limit by $g_k.$ Each $g_k$ is of Baire class one because it is a pointwise limit of continuous functions. The identity $D = \lim_{k \to \infty} g_k$ shows that $D$ is of Baire class at most two. A function of Baire class one is continuous on a dense set, whereas $D$ is discontinuous everywhere. This excludes class one, so the Baire class of $D$ is two.

9.Differentiability at a point implies continuity at that point. Because $D$ is discontinuous everywhere, it is nowhere differentiable.

Riemann and Lebesgue integrability

10.Let $[a, b]$ be a compact interval with $a < b.$ Every subinterval of a partition $P$ contains rational and irrational points. The supremum of $D$ on that subinterval is $1,$ and the infimum is $0.$ For each partition, the upper and lower Darboux sums are given by:

$$ \begin{align} U(D, P) &= b - a \\[6pt] L(D, P) &= 0 \end{align} $$

11.For $a < b,$ these sums are unequal, which violates the Darboux criterion. The function $D$ is not Riemann integrable on any non-degenerate compact interval.

12.The countable set $\mathbb{Q}$ is Borel and has Lebesgue measure zero. Its indicator function $D$ is measurable and equals zero almost everywhere. For every Lebesgue-measurable set $E \subseteq \mathbb{R},$ the integral of $D$ over $E$ is equal to:

$$\int_E D(x) \ d\lambda = \lambda(E \cap \mathbb{Q}) = 0$$

13.In particular, taking $E = \mathbb{R}$ shows that $D$ is Lebesgue integrable on $\mathbb{R},$ with integral zero.