Covering the Rationals with Nothing

Cover the rationals one by one and the total length drops below any bound. The Riemann integral cannot move here.

The rationals are countably many. So they can be numbered and given shrinking covers. The covers' total can be made as small as you wish.

For example, cover 🍬 candies scattered on the floor, one by one, with smaller and smaller cups, and the cups' total size shrinks as low as you like.

Examples & Exercises

Prove by covering that a single point $\{x_0\}$ has measure 0.

m(\{x_0\}) \le \varepsilon \quad \forall\,\varepsilon > 0

K = 1

: the first rational

0

gets a cover of length

0.05

Show proof

For any $\varepsilon > 0$ , the interval $(x_0 - \varepsilon/2,\, x_0 + \varepsilon/2)$ covers $\{x_0\}$ and has length $\varepsilon$ .
So the outer measure is at most $\varepsilon$ , and since $\varepsilon$ is arbitrary, the measure is $0$ .

Prove $m(\mathbb{Q}\cap[0,1]) = 0$ with the $\varepsilon/2^k$ covering.

\sum_{k=1}^{\infty}\frac{\varepsilon}{2^k} = \varepsilon

\varepsilon = 0.05

K = 32

: total length

\approx 0.05

covers 32 rationals.

Show proof

The rationals are countable, so enumerate those in $[0,1]$ as $r_1, r_2, \dots$ .
Covering $r_k$ with an interval of length $\varepsilon/2^k$ gives total cover length $\sum_{k=1}^{\infty}\varepsilon/2^k = \varepsilon$ .
Since $\varepsilon > 0$ is arbitrary, the outer measure is $0$ , hence so is the measure.

Prove (briefly) that the Lebesgue integral of the Dirichlet function is 0, and contrast with Riemann.

\int_{[0,1]} D = 1\cdot m(\mathbb{Q}\cap[0,1]) + 0 = 0

The set where

D = 1

(the rationals) has measure 0 — elsewhere

D = 0

Show proof

The set where $D = 1$ is $\mathbb{Q}\cap[0,1]$ , of measure $0$ ; elsewhere $D = 0$ .
So the Lebesgue integral is $1 \cdot 0 + 0 \cdot 1 = 0$ .
Riemann, by contrast, is stuck at $U = 1$ , $L = 0$ for every partition — Lebesgue counts "how often each value occurs".