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{{Short description|Indicator function of rational numbers}}
{{for|the other function sometimes incorrectly called the Dirichlet function|Dirichlet kernel}}
In [[mathematics]], the '''Dirichlet function'''<ref>{{springer|title=Dirichlet-function|id=p/d032860}}</ref><ref>[http://mathworld.wolfram.com/DirichletFunction.html Dirichlet Function — from MathWorld]</ref> is the [[indicator function]] <math>\mathbf{1}_\Q</math> of the set of [[rational number|rational numbers]] <math>\Q</math> over the set of [[real number]]s <math>\R</math>, i.e. <math>\mathbf{1}_\Q(x) = 1</math> for a real number {{mvar|x}} if {{mvar|x}} is a rational number and <math>\mathbf{1}_\Q(x) = 0</math> if {{mvar|x}} is not a rational number (i.e. is an [[irrational number]]).
<math display="block">\mathbf 1_\Q(x) = \begin{cases}
1 & x \in \Q \\
0 & x \notin \Q
\end{cases}</math>

It is named after the mathematician [[Peter Gustav Lejeune Dirichlet]].<ref>{{cite journal| first = Peter Gustav | last = Lejeune Dirichlet | title = Sur la convergence des séries trigonométriques qui servent à représenter une fonction arbitraire entre des limites données| journal = Journal für die reine und angewandte Mathematik |volume = 4 | year = 1829 | url = https://eudml.org/doc/183134 | pages = 157–169 }} The function is defined on page 169</ref> It is an example of a [[Pathological (mathematics)|pathological function]] which provides counterexamples to many situations.

== Topological properties ==
{{unordered list
| The Dirichlet function is [[nowhere continuous function|nowhere continuous]]. We can prove this by reference to the definition of a [[continuous function]] to show that it violates the continuity properties at both rational and irrational arguments:
{{Math proof| drop=hidden|proof=*If {{mvar|y}} is rational, then {{nowrap|{{math|{{var|f}}({{var|y}}) {{=}} 1}}}}. To show the function is not continuous at {{mvar|y}}, we need to find an {{mvar|ε}} such that no matter how small we choose {{mvar|δ}}, there will be points {{mvar|z}} within {{mvar|δ}} of {{mvar|y}} such that {{math|{{var|f}}({{var|z}})}} is not within {{mvar|ε}} of {{nowrap|{{math|{{var|f}}({{var|y}}) {{=}} 1}}}}.  In fact, {{frac|1|2}} is such an {{mvar|ε}}. Because the [[irrational number]]s are [[dense set|dense]] in the reals, no matter what {{mvar|δ}} we choose we can always find an irrational {{mvar|z}} within {{mvar|δ}} of {{mvar|y}}, and {{nowrap|{{math|{{var|f}}({{var|z}}) {{=}} 0}}}} is at least {{frac|1|2}} away from 1. 
*If {{mvar|y}} is irrational, then {{nowrap|{{math|{{var|f}}({{var|y}}) {{=}} 0}}}}. Again, we can take {{nowrap|{{math|{{var|ε}} {{=}} {{frac|1|2}}}}}}, and this time, because the rational numbers are dense in the reals, we can pick {{mvar|z}} to be a rational number as close to {{mvar|y}} as is required. Again, {{nowrap|{{math|{{var|f}}({{var|z}}) {{=}} 1}}}} is more than {{frac|1|2}} away from {{nowrap|{{math|{{var|f}}({{var|y}}) {{=}} 0}}}}.}}
Its restrictions to the set of rational numbers and to the set of irrational numbers are [[constant function|constants]] and therefore continuous. The Dirichlet function is an archetypal example of the [[Blumberg theorem]].

| The Dirichlet function can be constructed as the double pointwise limit of a sequence of continuous functions, as follows:
<math display="block">\forall x \in \R, \quad \mathbf{1}_{\Q}(x) = \lim_{k \to \infty} \left(\lim_{j\to\infty}\left(\cos(k!\pi x)\right)^{2j}\right)</math>
for integer {{mvar|j}} and {{mvar|k}}. This shows that the Dirichlet function is a [[Baire function|Baire class]] 2 function. It cannot be a Baire class 1 function because a Baire class 1 function can only be discontinuous on a [[meagre set]].<ref>{{cite book
  | last = Dunham
  | first = William
  | title = The Calculus Gallery
  | publisher = [[Princeton University Press]]
  | date = 2005
  | pages = 197
  | isbn = 0-691-09565-5 }}</ref>
}}

== Periodicity == 

For any real number {{mvar|x}} and any positive rational number {{mvar|T}}, <math>\mathbf{1}_\Q(x + T) = \mathbf{1}_\Q(x)</math>. The Dirichlet function is therefore an example of a real [[periodic function]] which is not [[constant function|constant]] but whose set of periods, the set of rational numbers, is a [[Dense set|dense subset]] of <math>\R</math>.

== Integration properties ==
{{unordered list
| The Dirichlet function is not [[Riemann integral|Riemann-integrable]] on any segment of <math>\R</math> despite being bounded because the set of its discontinuity points is not [[negligible set|negligible]] (for the [[Lebesgue measure]]).
| The Dirichlet function has both an upper [[Darboux integral]] (namely, <math>b-a</math>) and a lower Darboux integral (0) over any bounded interval <math>[a,b]</math> — but they are not equal if <math> a < b</math>, so the Dirichlet function is not Darboux-integrable (and therefore not Riemann-integrable) over any nondegenerate interval.
| The Dirichlet function provides a counterexample showing that the [[monotone convergence theorem]] is not true in the context of the Riemann integral.
{{Math proof|drop=hidden|proof=Using an [[enumeration]] of the rational numbers between 0 and 1, we define the function {{math|{{var|f}}{{sub|{{var|n}}}}}} (for all nonnegative integer {{mvar|n}}) as the indicator function of the set of the first {{mvar|n}} terms of this sequence of rational numbers. The increasing sequence of functions {{math|{{var|f}}{{sub|{{var|n}}}}}} (which are nonnegative, Riemann-integrable with a vanishing integral) pointwise converges to the Dirichlet function which is not Riemann-integrable.}}
| The Dirichlet function is [[Lebesgue integral|Lebesgue-integrable]] on <math>\R</math> and its integral over <math>\R</math> is zero because it is zero except on the set of rational numbers which is negligible (for the Lebesgue measure).
}}

==See also==
* [[Thomae's function]], a variation that is discontinuous only at the rational numbers

== References ==
{{Reflist}}

[[Category:Elementary special functions|Dirichlet]]
[[Category:Real analysis]]