The Dirichlet Function | Real Analysis

\mathbf{1}_{\mathbb{Q}}

is 1 on the rationals and 0 on the irrationals of

[0, 1]

— every subinterval holds both, so the upper sum is always 1 and the lower sum always 0. The canonical non-Riemann-integrable function: no refinement closes the gap.

Examples & Exercises

Let $f = \mathbf{1}_{\mathbb{Q}}$ be the Dirichlet function on $[0, 1]$ . Using that every subinterval of $[0, 1]$ contains a rational, prove completely that $U(f, P) = 1$ for every partition $P$ .

U(f, P) = \sum_{i=1}^{n} M_i\,\Delta x_i = 1 \quad \forall P

n = 8

: all 8 subintervals carry an upper rectangle (light red, height 1) —

U(f, P_8) = 8 \cdot \frac18 = 1

Show proof

First, density of $\mathbb{Q}$ : if $u < v$ then $(u, v)$ contains a rational. Pick $N \in \mathbb{N}$ with $\frac1N < v - u$ (Archimedean property) and set $m = \lfloor Nu \rfloor + 1$ ; then $\frac{m}{N} - u = \frac{\lfloor Nu \rfloor + 1 - Nu}{N} \le \frac1N < v - u$ , so $u < \frac{m}{N} < v$ .
Take a partition $P = \{x_0 < x_1 < \cdots < x_n\}$ . Each subinterval $[x_{i-1}, x_i]$ contains a rational $r_i$ by the above, so $M_i = \sup_{[x_{i-1}, x_i]} f \ge f(r_i) = 1$ . But $f \le 1$ always, so $M_i = 1$ .
Hence $U(f, P) = \sum_{i=1}^{n} M_i\,\Delta x_i = \sum_{i=1}^{n} 1 \cdot (x_i - x_{i-1}) = x_n - x_0 = 1$ . However the partition is chosen, the upper sum is always $1$ — the red dots (rationals) sit in every interval.

Using that every subinterval of $[0, 1]$ contains an irrational, prove completely that $L(f, P) = 0$ for every partition $P$ .

L(f, P) = \sum_{i=1}^{n} m_i\,\Delta x_i = 0 \quad \forall P

n = 8

: the lower bars (blue) draw as thin zero-height lines —

L(f, P_8) = 8 \cdot 0 = 0

Show proof

Irrationals are dense too: if $u < v$ then $(u, v)$ contains an irrational. Density of $\mathbb{Q}$ gives a rational $q$ in $(u - \sqrt{2},\, v - \sqrt{2})$ , and then $q + \sqrt{2}$ is an irrational in $(u, v)$ (since $\sqrt{2}$ is irrational, rational $+$ irrational is irrational).
For a partition $P = \{x_0 < \cdots < x_n\}$ , each subinterval $[x_{i-1}, x_i]$ contains an irrational $s_i$ , so $m_i = \inf_{[x_{i-1}, x_i]} f \le f(s_i) = 0$ . Since $f \ge 0$ , $m_i = 0$ .
Hence $L(f, P) = \sum_{i=1}^{n} 0 \cdot \Delta x_i = 0$ . Together with the previous problem, $U(f, P) - L(f, P) = 1$ for every partition — the gap never closes, so $f$ is not Riemann integrable.

Prove completely, using sequences, that the Dirichlet function $f = \mathbf{1}_{\mathbb{Q}}$ is discontinuous at every point of $[0, 1]$ .

r_k \to x,\ f(r_k) = 1 \qquad s_k \to x,\ f(s_k) = 0

n = 8

: red dots (rationals,

f = 1

) alternate with the gaps between them (irrationals,

f = 0

) in every interval — at no point do the nearby function values settle on one number.

Show proof

Fix any $x \in [0, 1]$ . By density of rationals and irrationals, for each $k \ge 1$ we can pick a rational $r_k$ and an irrational $s_k$ in $(x - \tfrac1k,\, x + \tfrac1k) \cap [0, 1]$ . Since $|r_k - x| < \frac1k$ and $|s_k - x| < \frac1k$ , we have $r_k \to x$ and $s_k \to x$ .
But $f(r_k) = 1$ , so $f(r_k) \to 1$ , and $f(s_k) = 0$ , so $f(s_k) \to 0$ . If $f$ were continuous at $x$ , sequential continuity would force both limits to equal $f(x)$ — impossible since $1 \ne 0$ .
Hence $f$ is discontinuous at every point of $[0, 1]$ — "too discontinuous" for the Riemann integral. The Lebesgue integral instead uses that the rationals have measure $0$ to get $\int f = 0$ (see the Lebesgue scene).