Jtdylktuy

Studying the Lebesgue measure on the line I've found the following argument which concludes that $m(mathbb{R}) < +infty$ (where $m$ denotes the Lebesgue measure on $mathbb{R}$). Obviously it must be flawed, but I haven't been able to find the flaw so far.

Recall that for a Lebesgue measurable set $A$ we have by definition that
$$
m(A)=infleft{sum_{n=1}^infty(b_n-a_n):cup_{n=1}^infty(a_n,b_n]supset Aright}.
$$
Pick your favorite summable sequence of positive terms, ${a_n=1/n^2}_{n=1}^infty$ for instance. We know the rational numbers are countable, so we can index them in a sequence ${q_n}_{n=1}^infty$. Now consider the intervals
$$
I_n=left(q_n-frac{a_n}{2},q_n+frac{a_n}{2}right].
$$
As the rationals are dense in $mathbb{R}$ we must have $mathbb{R}subsetcup_{n=1}^infty I_n$ but then, having into account the definition of Lebesgue measure we have
$$
m(mathbb{R})leqsum_{n=1}^inftyleft(big(q_n+frac{a_n}{2}big)-big(q_n-frac{a_n}{2}big)right)=sum_{n=1}^infty a_n<+infty.
$$
For instance with $a_n=1/n^2$ we get $m(mathbb{R})leqpi^2/6$.

As I said, I know this is flawed but I've spent almost two hours trying to find the flaw so any help would be appreciated!

"As the rationals are dense in $mathbb{R}$ we must have $mathbb{R}subset bigcup_{n=1}^infty I_n$."

This is false. Pick your favorite irrational number $x$. For every $n$, there exist infinitely many rationals $q_m$ such that $q_m$ is within $a_n/2$ of $x$. But the least such $m$ may be much larger than $n$. In particular, it does not follow that there exists some $n$ such that $x$ is within $a_n/2$ of $q_n$, which is what is necessary to have $xin bigcup_{n=1}^infty I_n$.

One can explicitly demonstrate the failure of the argument for $mathbb R_+$. Generalize to $mathbb R$ as needed.

Consider the following maps $f:mathbb N_+^2 rightarrow mathbb N_+$ and $g: mathbb N_+^2 rightarrow mathbb Q_+$
begin{eqnarray}
f({k,m}) &=& (k + m - 2 )(k + m - 1) + m \
g({k,m}) &=& k/m.
end{eqnarray}
You can verify that $f$ is a bijection and $g$ is onto, and thus $h(n) = g(f^{-1}(n))$ maps from $mathbb N_+$ onto a dense subset of $mathbb R_+$. Now consider the set
$$
U = bigcup_{n=1}^infty left(h(n) - frac{a_n}{2},h(n) + frac{a_n}{2}right).
$$
Using the sequence ${a_n = n^{-2}}_{n=1}^infty$, I can prove $phi = (1+sqrt{5})/2notin U$.

First, note that $h$ orders the ratios $k/m$ first by increasing $k+m$, then by increasing $m$. Because of this, if $h(n)$ is a continued fraction convergent of $phi$, then $|h(n') - phi| > |h(n) - phi|$ for all $n' < n$. The continued fraction convergents of $phi$ are of the form $F_{i+1}/F_i$, where $F_i$ are Fibonnaci numbers, and occur when $n_i = (F_{i+1} + F_{i} - 2)(F_{i+1} + F_{i} - 1)/2 + F_{i}ge F_{i}F_{i+1}$. One can prove from the Fibonacci formula that the error on these convergents is always larger than $0.4F_{i}^2$.

Now consider $n_i le n < n_{i+1}$. Then we have $a_n = n^{-2} le n_i^{-2}le F_{i}^{-2}F_{i+1}^{-2}$ and $|h(n) - phi| > |h(n_{i+1})-phi| > 0.4 F_{i+1}^{-2}$. If $phiin U$, there must be an $n$ with $|h(n) - phi| < a_n/2$ for some $n$. This would then imply $0.4 < F_{i}^{-2}/2$, which can only be satisfied for $F_i = 1$. Thus such an $n$ would have to satisfy $n < n_3 = 8$. Since it can be checked that none of these works, $phi notin U$.

(Extended from a comment) Recall that a subset $mathcal{D}$ of $mathbb{R}$ is called dense if its closure $bar{mathcal{D}}$ is all of $mathbb{R}$. From this definition, you can readily check that the followings are equivalent for dense subsets $mathcal{D} subset mathbb{R}$:

$$ text{$mathcal{D}$ is closed in $mathbb{R}$} qquad Leftrightarrow qquad text{$mathcal{D} = mathbb{R}$}. $$

In particular, you can't tell much how porous your set $mathcal{D}$ will be only by knowing the density of $mathcal{D}$. There are lots of dense subsets of $mathbb{R}$ which is not closed (and hence not all of $mathbb{R})$.

Example. If we pick a sequence $(r_n)_{n=1}^{infty}$ of positive numbers satisfying $sum_{n=1}^{infty} r_n < infty$, then

$$ U = bigcup_{n=1}^{infty} (q_n - r_n, q_n + r_n)$$

is an open set such that $mathbb{Q} subset U$ but $operatorname{Leb}(U) leq sum_{n=1}^{infty} 2r_n < infty$. So there are open dense subsets of $mathbb{R}$ with finite length. Notice also that $operatorname{Leb}(U)$ can be made arbitrarily small by choosing suitable $(r_n)$. Another interesting feature of this example is that the boundary $partial U$ has infinite length! In this way, you may think of $U$ as tiny pores of a highly porous material, such as a sponge.

The rationals are all contained in $bigcup I_n$ but not all irrationals
are.