real analysis - Lebesgue measure paradox

See Lebesgue outer measure of $[0,1]\cap\mathbb{Q}$

Lebesgue measure:

$$m(A) = \inf \left\{ \sum |I_n| : A \subset \bigcup I_n \right\}$$

We know that $m(\mathbb{Q} \cap [0,1]) = 0$.

Proof:

Enumerate the rationals $\mathbb{Q} \cap [0,1] = \{ q_n \}_{n \in \mathbb{N}}$.

Let $A_n$ be the covering of rationals $\{ q_n \}_{n \in \mathbb{N}}$ in $[0,1]$ by intervals $A_n = (q_n - \varepsilon/2^n , q_n + \varepsilon/2^n)$. Then,

$$m(\mathbb{Q} \cap [0,1]) \le m( \bigcup A_n ) \le \sum_n |A_n| = \sum_n \frac{\varepsilon}{2^n} = \varepsilon$$
and since it is true for every $\epsilon > 0$ as small as we want, we have $m(\mathbb{Q} \cap [0,1]) = 0$. $\square$

Paradox?

Every number in $[0,1]$ is contained in an open neighborhood of a rational number. We reach a paradox if

$$\bigcup_{n \in \mathbb{N}} A_n = [0,1]$$
since then
$$m([0,1]) \le m( \bigcup A_n ) \le \sum m(A_n) = \varepsilon$$
and then $m([0,1])=0$ and not $1$ as it should be.

To contradict this we need to show

$$\bigcup A_n \ne [0,1] \ .$$

We need to build an irrational number $r$ such that $\forall n : |r - q_n| > \varepsilon/2^n$. Can you construct such a number (or prove it exists without using Lebesgue measure argument)?

Moreover, I think we need to show the set $[0,1]-\bigcup A_n$ has $\aleph = 2^{\aleph_0}$ elements. Otherwise, $m([0,1]-\bigcup A_n) = 0$ and we reach a contradiction.

Answer

If I may choose the enumeration of the rationals:

Enumerate the rationals in increasing order of the denominator in the reduced form, rationals with the same denominator in increasing order, so

$$q_1 = \frac01,\, q_2 = \frac11,\, q_3 = \frac12,\, q_4 = \frac13,\, q_5 = \frac23,\, q_6 = \frac14,\,q_7 = \frac34,\,\ldots$$

Then for $q_n = \frac{r_n}{s_n}$, we have $s_n \leqslant n$.

For $x = \frac{1}{\sqrt{2}}$, and a rational $0 \leqslant \frac{r}{s} \leqslant 1$, we have

$$\left\lvert\frac{r}{s} -\frac{1}{\sqrt{2}}\right\rvert = \frac{\left\lvert \frac{r^2}{s^2} - \frac12\right\rvert}{\left\lvert\frac{r}{s} +\frac{1}{\sqrt{2}}\right\rvert} \geqslant \frac{\lvert 2r^2-s^2\rvert}{4s^2} \geqslant \frac{1}{4s^2}.$$

For $0 < \varepsilon < \frac29$, we have $\frac{1}{4n^2} > \frac{\varepsilon}{2^n}$ for all $n \geqslant 1$, hence $1/\sqrt{2}$ is not contained in any $(q_n - \varepsilon 2^{-n},\,q_n + \varepsilon 2^{-n})$.

By a similar reasoning, we find $2^{\aleph_0}$ irrationals that aren't covered:

For an irrational $x \in (0,\,1)$ with continued fraction expansion

$$x = [a_0,\, a_1,\, a_2,\, \dotsc],$$

convergents $\frac{r_n}{s_n}$, and the complete quotients $\alpha_n$ - that means

$$x = \frac{\alpha_{n+1}r_n + r_{n-1}}{\alpha_{n+1}s_n + s_{n-1}},$$

and thus the continued fraction expansion $\alpha_n = [a_n,\, a_{n+1},\, \dotsc]$, whence $a_n < \alpha_n < a_n + 1$ - we have

$$\begin{align} \left\lvert x - \frac{r_n}{s_n}\right\rvert &= \left\lvert \frac{\alpha_{n+1}r_n + r_{n-1}}{\alpha_{n+1}s_n + s_{n-1}} - \frac{r_n}{s_n}\right\rvert\\ &= \left\lvert \frac{\alpha_{n+1}(r_n s_n - s_nr_n) + (r_{n-1}s_n - r_ns_{n-1})}{(\alpha_{n+1}s_n+s_{n-1})s_n}\right\rvert\\ &= \left\lvert\frac{(-1)^{n}}{(\alpha_{n+1}s_n+s_{n-1})s_n}\right\rvert\\ &> \frac{1}{\bigl((a_{n+1}+1)s_n + s_{n-1}\bigr)s_n}\\ &\geqslant \frac{1}{(a_{n+1}+2)s_n^2}.\tag{1} \end{align}$$

For a rational number $\frac{r}{s}$ with $s \leqslant s_n$, $n \geqslant 1$, we have

$$\lvert sx - r\rvert \geqslant \lvert s_nx - r_n\rvert$$

with equality if and only if $s = s_n$ and $r = r_n$.

This immediately leads to

$$\left\lvert x - \frac{r}{s}\right\rvert > \frac{1}{(a_{n+1}+2)s\cdot s_n}.\tag{2}$$

Thus, for any fixed $M \in \mathbb{Z}^+$, let $A_M$ be the set of all irrational numbers in $(0,\,1)$ whose continued fraction expansion has all partial quotients $\leqslant M$.

For $M \geqslant 2$, the set $A_M$ has cardinality $2^{\aleph_0}$.

For $x \in A_M$, the denominators of the convergents satisfy

$$s_{n+1} = a_{n+1}s_n + s_{n-1} \leqslant M\cdot s_n + s_{n-1} \leqslant (M+1)\cdot s_n.\tag{3}$$

Choosing the minimal $n$ with $s_n \geqslant s$ in $(2)$, we deduce

$$\left\lvert x - \frac{r}{s}\right\rvert > \frac{1}{(M+2)(M+1)s^2}$$

from $(3)$ for $x \in A_M$.

If $\varepsilon$ is small enough, so that

$$\frac{1}{(M+2)^2n^2} > \frac{\varepsilon}{2^n}$$

for all $n$, the set

$$U_\varepsilon = \bigcup_{n=1}^\infty \left(q_n - \frac{\varepsilon}{2^n},\,q_n + \frac{\varepsilon}{2^n}\right)$$

doesn't intersect $A_M$.