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
$$
m(A)=\inf\left\{\sum_{n=1}^\infty(b_n-a_n):\cup_{n=1}^\infty(a_n,b_n]\supset A\right\}.
$$
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 $\mathbb{R}\subset\cup_{n=1}^\infty I_n$ but then, having into account the definition of Lebesgue measure we have
$$
m(\mathbb{R})\leq\sum_{n=1}^\infty\left(\big(q_n+\frac{a_n}{2}\big)-\big(q_n-\frac{a_n}{2}\big)\right)=\sum_{n=1}^\infty a_n<+\infty.
$$
For instance with $a_n=1/n^2$ we get $m(\mathbb{R})\leq\pi^2/6$.
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!
Best Answer
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 $x\in \bigcup_{n=1}^\infty I_n$.