I have the following proof, but I don't think it's right. Could somebody please tell me what's wrong with it, and how to fix it? Thanks 🙂
Let $B$ be the set of rational numbers in the interval $[0,1]$ Let $I_{k}$, $k=1,2,\cdots, n$ be a finite collection of open intervals that covers $B$. Prove that $\sum_{k=1}^{n}l(I_{k})\geq 1$.
($l(I)$ denotes the length of the interval $I$).
$\underline{\mathbf{\text{Proof}}}$: Suppose $a,b \in [0,1]$ such that $a < b$. By the density of the rational numbers in $\mathbb{R}$, $\exists q_{i} \in \mathbb{Q}$ such that $\forall a_{i} < b_{i}$, $q_{i}\in(a_{i},b_{i})$.
Since the rationals are dense in $\mathbb{R}$ and hence in $[0,1]$, we can cover $[0,1]$ as follows:
$[0,1] \subseteq \cup_{k=1}^{\infty}I_{k}=\cup_{k=1}^{\infty} (a_{k},b_{k}) $.
(In the case where the $(a_{k},b_{k})$ do not completely cover $[0,1]$ completely, $\exists$ a point between $(a_{i},b_{i})$ and $(a_{i+1}, b_{i+1})$ that remains uncovered. However, because there are countably many intervals, there will be only countably many such points. Thus, the set of points, call it $S$, being a countable set will have $m^{*}(S)=0$. So, these points will be negligible in terms of their effect on the sum of the lengths of the intervals.)
By Heine-Borel, since $[0,1]$ is closed and bounded, every such open cover has a finite subcover. So, $[0,1] \subset \cup_{k=1}^{n}(a_{k},b_{k}) = \cup_{k=1}^{n}I_{k}$.
Now, the outer measure of an interval is its length. So $m^{*}([0,1])=l([0,1]) = 1$, and by the definition of the outer measure, $1 = m^{*}([0,1]) = \inf_{k=1}^{n}l(I_{k}) \leq \sum_{k=1}^{n}l(I_{k})$
Q.E.D.
I also realize this question has been asked before on here, but I am terribly dissatisfied with the answers that have thus far been given.
Best Answer
Let $B$ be the set of rational numbers in the interval $[0,1]$, and let $\{I_k\}_{k=1}^n$ be a finite collection of open intervals that covers $B$. Prove that $\sum_{k=1}^n m^*(I_k) \geq 1$.
Proof:
First, we have $B = \mathbb{Q}\cap [0,1] \subset [0,1]$ and $B \subset \bigcup_{k=1}^n I_k$, where each $I_k$ is an open interval. It follows immediately that since $B$ is countable has outer measure zero and $$B \subset [0,1] \Rightarrow m^*(B) \leq m^*([0,1]) = 1,$$ $$B \subset \bigcup_{k=1}^n I_k \Rightarrow m^*(B) \leq m^*(\bigcup_{k=1}^n I_k) \leq \sum_{k=1}^{\infty} m^*(I_k).$$ Now, since $0 \in B$, there exist one of the $I_k$'s that contains $0$, denote such interval by $(a_1,b_1)$. By a similar reasoning, there exist one of the $I_k$'s that contains $1$ (Note that $1 \in B$), denote such interval by $(a_N,b_N)$. We thus have: $$a_1 < 0 < b_1$$ $$a_N < 1 < b_N$$ Now, lets assume the contrary, that is $\sum_{k=1}^n m^*(I_k) < 1$. In particular, we are assuming the following: $$\sum_{k=1}^n m^*(I_k) < m^*([0,1])$$ We claim that if $b_1 \geq 1$, we obtain a contradiction, since: $$a_1 < 0 < 1 \leq b_1$$ and $$ l((a_1,b_1)) = b_1 - a_1 \leq \sum_{k=1}^n m^*(I_k) < 1,$$ but $b_1 - a_1 > 1$. Otherwise, $b_1 \in [0,1)$ and since $b_1 \notin (a_1,b_1)$, there exist an interval in the collection $\{I_k\}_{k=1}^{n}$ which we label as $(a_2,b_2) \ni b_1$ and $$a_2 < b_1 < b_2.$$ We claim that if $b_2 \geq 1$, we obtain a contradiction, since: $$b_2 - a_1 < (b_2 -a_2)+(b_1 -a_1) \leq \sum_{k=1}^n m^*(I_k) < 1$$ but $b_1 -a_0 > 1$ because $$a_0 < 0 < 1 \leq b_1.$$ We can continue this selection process until it terminates, as it must since there are only $n$ intervals in the collection $\{I_k\}_{k=1}^{n}$. We thus obtain a sub-collection $\{(a_k,b_k)\}_{k=1}^N$ of $\{I_k\}_{k=1}^{n}$ for which $a_1 < 0,$ while $a_{k+1} < b_k$ for $1 \leq k \leq N-1$ and $b_N>1$. We thus have: $$\begin{align*} b_N-a_1 &< b_N - (a_N-b_{N-1})- \cdots-(a_2-b_1) -a_1 \\ &\leq (b_N-a_N)+(b_{N-1}-a_{N-1})+\cdots +(b_1-a_1) \\ &= \sum_{k=1}^N l(I_k) \leq \sum_{k=1}^n m^*(I_k)<1 \end{align*}$$ a contradiction, since $$l((0,1)) = 1 < b_N-a_1$$ and $$a_1 < 0 < 1 < b_N.$$ Thus, we conclude that: $$\sum_{k=1}^n m^*(I_k) \geq 1.$$