Covering the Rationals with Nothing

Cover the rationals one by one with

\varepsilon/2^k

covers — the total length drops below

\varepsilon

. Riemann can't move.

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".