We show that the Dirichlet function f defined by \[ f(x) = \begin{cases} 1, & \quad \mbox{if $x$ is rational}, \\ 0, & \quad \mbox{if $x$ is irrational}. \end{cases} \] is not of bounded variation on any interval.

Let n ∈ ℤ and n > 0. Let [𝑎, b] be a closed interval in ℝ. We construct a partition P = {x₀, x₁, … , x_n+2} of [𝑎, b] such that V(f, [𝑎, b]) ≥ ∑ⁿ⁺²_i=1 |f(x_i) − f(x_i−1)| > n as follows.

Recall that by definition x₀ = 𝑎. It is known that between any two real numbers there is a rational number and an irrational number. Take x₁ to be an irrational number between 𝑎 and b. Then take x₂ to be a rational number between x₁ and b. Continue like this, taking x_2i+1 to be an irrational number between x_2i and b, and x_2i to be a rational number between x_2i−1 and b. Finally x_n+2 = b. Thus, we have created a partition that begins with 𝑎 and then alternates between rational and irrational numbers, until it finally ends with b. Now consider the sum ∑ⁿ⁺²_i=1 |f(x_i) − f(x_i−1)|, which we know is at most the variation of f on [𝑎, b]. Hence, \begin{align*} V(f, [a,b]) &\ge \sum_{i=1}^{n+2} \left\vert f(x_i ) - f(x_{i-1}) \right\vert \\ &\ge \sum_{i=2}^{n+1} \left\vert f(x_i ) - f(x_{i-1}) \right\vert \\ &= \left\vert f(x_2 ) - f(x_1 ) \right\vert + \cdots + \left\vert f(x_{n+1}) - f(x_n ) \right\vert \\ &= \left\vert 1-0 \right\vert + \left\vert 0-1 \right\vert + \cdots + \left\vert 1-0 \right\vert \\ &= 1 + 1 + 1 + \cdots + 1 \\ &= n . \end{align*} Thus, V(f, [𝑎, b]) is arbitrarily large, and so V(f, [𝑎, b]) = ∞.    ■

We begin by making a partition of [0, 1]. We assume, without loss of generality, that the number of partition points is even, so n is even. We make our partition as follows:     xₙ = 1, x_n−(2k+1) = 1/(k+3), x_{n−2k = 2/(2k+3), for k = 1, 2, … and x₀ = 0.}

Then \begin{align*} V(f, [0,1]) &\ge \sum_{i=1}^n \left\vert f(x_i ) - f(x_{i-1}) \right\vert \\ &\ge \sum_{i=1}^{n/2} \left\vert f(x_i ) - f(x_{i-1}) \right\vert . \end{align*} That is to say that we remove every other interval from the sum. Over the remaining intervals, our function has some convenient properties. The intervals we are now considering have the form [1/(k + 3), 2/(2k + 3)]. Also notice that \( \displaystyle \quad |f (2/(2k + 3))| = \sqrt[3]{2/(2k + 3)} \quad \) and    f(1/(k + 3)) = 0. Because \( \displaystyle \ \sqrt[3]{x} \ \) is an increasing function, sin(π/x) is the only component affecting when f is increasing or decreasing. Knowing how the sine function behaves we can see that f is monotone on each interval of the form [1/(k + 3), 2/((2k + 3)]. Thus, \begin{align*} V(f, [0,1]) &\ge \sum_{i=1}^{n/2} \left\vert f(x_{2i}) - f(x_{2i-1}) \right\vert \\ &= \sum_{i=1}^n \left\vert f(1/(k+3)) - f(2/(2k+3)) \right\vert \\ &= \sum_{i=1}^n \left\vert \sqrt[3]{\frac{2}{2k+3}} \right\vert = \sum_{i=1}^n \frac{\sqrt[3]{2}}{\sqrt[3]{2k+3}} . \end{align*} Notice that each term is smaller than the last, since the denominator is getting larger as k increases. Hence, \[ \sum_{i=1}^n \frac{\sqrt[3]{2}}{\sqrt[3]{2k+3}} \ge \sum_{i=1}^n \frac{1}{\sqrt[3]{k+3}} = \sum_{i=2}^{n+1} \frac{1}{\sqrt[3]{k}} . \] This is a p-series, with p = 1/3, so we know that it diverges. This means that by choosing n large enough we can make the sum \[ \sum_{i=1}^n \frac{\sqrt[3]{2}}{\sqrt[3]{2k+3}} \] as large as we like and thus V(f, [0, 1]) = ∞. Thus, though the function is continuous it is not of bounded variation on [0, 1].    ■