Exercise from G. De Barra - *Measure Theory and Integration*

Question

This is my first post in Measure Theory::

Let $\{I_n\}_{{n}}$ be a finite sequence of open intervals which covers the set of all rationals in $[0,1]$.

Show that $\sum_{n} l(I_n)\ge1$.

In order to show this I think that I have to show that if $\{I_n\}$ covers $\Bbb Q\cap [0,1]$ then it covers $[0,1]$. Then my monotonicity of outer measure function we have $\sum_{n}l(I_n)>1$.

Now it is enough to show that $\{I_n\}_{{n}}$ covers $\Bbb Q^c\cap [0,1]$.

Let $i$ be an irrational number such that $I_n$ does not cover $i$ for any $n\in \Bbb N$.Let $r_n$ be an enumeration of rational numbers such that $r_n<i<r_{n+1}$

But I can't arrive at a contradiction from here?

Any help will be great.

Answer 1

$$\mathbb{Q}\cap[0,1] \subseteq \bigcup_{n=1}^k I_n$$

$$\overline{\mathbb{Q}\cap[0,1]} \subseteq \overline{\bigcup_{n=1}^k I_n}= \bigcup_{n=1}^k \overline{I_n}$$

$$1=m[0,1]=m(\overline{\mathbb{Q}\cap[0,1]})\le m(\bigcup_{n=1}^k \overline{I_n})\le \sum_{n=1}^k l(\overline{I_n})=\sum_{n=1}^k l(I_n) $$

Answer 2

Hint: One of the intervals contains $0$. Say that interval is $(a_0,b_0)$. So $a_0<0<b_0$. If $b_0\ge1$ you're done. Assume $b_0<1$.

Choose a sequence of rationals $r_k$ that decrease to $b_0$. Each $r_k$ is in one of the remaining intervals. Since there are only finitely many intervals, at least one of the remaining intervals contains infinitely many $r_k$. Say $(a_1,b_1)$ contains infinitely many $r_k$. Then $a_1\le b_0$ and $b_1>b_0$...