# Lebesgue outer measure of irrationals by definition

Is it possible to prove that lebesgue outer measure for irrationals in $[0,1]$ is 1 by definition? (not using the properties of lebesgue outer measure)

If $A \subset \mathbb{R}$ then the lebesgue outer measure is $\lambda^*(A)=\text {inf}\{ \sum_{i=1}^{\infty} |I_i| : A \subset \bigcup_{i=1}^{\infty} I_i \}$ (where $I_i$ is a bounded interval)

My idea is to prove this by definition: first we need to prove that given $\mathbb {I} \cap [0,1] \subset \bigcup_{i=1}^{\infty} I_i$ then $1\le \sum_{i=1}^{\infty} |I_i|$. I tried to prove that if $\mathbb {I} \cap [0,1] \subset \bigcup_{i=1}^{\infty} I_i$ then $[0,1] \subset \bigcup_{i=1}^{\infty} I_i$ but this is not true if we consider $[0,1/2) \cup (1/2,1]$

I´ve been thinking about this problem but unfortunately I haven´t come up with something.

I would really appreciate any hints or ideas

## 1 Answer

You cannot avoid the countability of the rationals . And the fact that $[0,1]$ \ $\mathbb Q$ is uncountable and dense is also insufficient as there are dense measurable subsets of $[0,1]$ with positive measure $<1.$

(1). If $S\subset \mathbb R$ and $S$ is countable then $\lambda^*(S)=0.$ For if $S\subset \{x_n:n\in \mathbb N\}$ and $\epsilon >0$ let $I_n=[-2^{-n-1}\epsilon +x_n,2^{-n-1}\epsilon +x_n].$ Then $\cup_{n\in \mathbb N}I_n\supset S$ and $\lambda^* (S)\leq \sum_{n\in \mathbb N}|I_n|=\epsilon.$

(2). Let $i(S)$ be the set of every countable family F of bounded intervals such that $\cup F\supset S.$

(2i). It should be obvious that $A\subset B\implies \lambda^*(A)\leq \lambda^*(B).$ Because $A\subset B\implies i(B)\subset i(A)\implies \{\sum_{I\in F}|I|:F\in i(B)\}\subset \{\sum_{I\in F}|I|:F\in i(A)\}.$

(2ii). Let $j(S,T)=\{F\cup G:f\in i(S)\land g\in i(t)\}.$

Then $j(S,T)\subset i(S\cup T).$ Also, if $F\in i(S)$ and $G\in i(T)$ then $\sum_{I\in F\cup G}|I|\leq \sum_{I'\in F}|I'|+\sum_{I''\in G}|I''|.$

Therefore $$\lambda^*(S\cup T)=\inf \left\{\sum_{I\in H}|I|:h\in i(S\cup T)\right\}\leq \inf \left\{\sum_{I\in H}|I|:H\in j(S,T)\right\} \leq$$ $$\leq \inf \left\{\sum_{I'\in F}|I'|+\sum_{I''\in G}|I''|:F\in i(S)\land g\in i(T)\right\}=$$ $$=\inf \left\{\sum_{I'\in F}|I'|:F\in i(S)\right\}+\inf \left\{\sum_{I''\in G}|I''|:G\in i(T)\right\}=$$ $$=\lambda^*(S)+\lambda^*(T).$$

(3). For $S=[0,1]\cap \mathbb Q$ and $T=[0,1]$ \ $S$ we have $\lambda^*(S)=0$ and $\lambda^* (T)\leq \lambda^* ([0,1])\leq \lambda^* (S)+\lambda^* (T)=\lambda (T).$

Therefore $\lambda^*(T)=\lambda^* ([0,1]).$ It remains to show that $\lambda^* ([0,1])=1.$

(3i). We have $F_0=\{[0,1]\}\in i([0,1])$ so $\lambda^*([01,1])\leq \sum_{I\in F_0}|I|=1.$

(3ii). To prove the reverse inequality $\lambda ([0,1])\geq 1$ it is more convenient to consider the set $i^o([0,1])$ of countable families $F$ of bounded OPEN intervals such that $\cup F\supset [0,1] .$

For brevity let $V=\inf \{\sum_{I\in F}|I|: F\in i^o([0,1])\}.$

Since $i^o([0,1])\subset i([0,1])$ we have $\lambda^*([0,1])\leq V.$

But we also have $V\leq \lambda^*([0,1])$, which we obtain by showing that $V\leq \lambda^*([0,1])+\epsilon$ for any $\epsilon >0:$

For $\{f_n:n\in \mathbb N\}= F\in i([0,1])$ with $(a_n,b_n)\subset f_n\subset [a_n,b_n]$ let $$F'=\{(a_n-\epsilon 2^{-n-1},b_n+\epsilon 2^{-n-1}):n\in \mathbb N\}.$$ Now $\{F':F\in i([0,1])\} \subset i^o([0,1])$ so $$V\leq \inf\{\sum_{I'\in F'}|I'|: F\in i([0,1])\} =\inf \{\epsilon+\sum_{I\in F}|I|:F\in i([0,1])\}=\lambda ([0,1]).$$

So $V=\lambda ^*([0,1]).$

(3iii). Finally to show that $V=1:$ For any $F\in i^o([0,1])$ there is a finite $G\subset F$ with $\cup G\supset [0,1]$ because $[0,1]$ is COMPACT. We show that $\sum_{I\in G}|I|\}>1$ as follows:

Let $0\in I_0\in G.$ For $n\geq 0,$ if $\sup I_n\leq 1$ then let $\sup I_n\in I_{n+1}\in G.$ (If $\sup I_n >1$ then $I_{n+1}$ will not be defined.)

By induction on $n,$ if $I_n$ exists then (a)... $\sup I_n>\sup \{\sup I_j:j<n\}$ and (b)...$\cup_{j\leq n}I_j\supset [0,\sup I_n).$ Now by (a), the sequence $(I_0,...)$ is a non-repeating sequence of members of the finite set $G ,$ so it is a finite sequence with a last member $I_{n^*}$, which must satisfy $\sup I_{n^*}>1,$ and by (b) we have $\cup_{0\leq j\leq n^*}I_j\supset [0,\sup I_{n^*})\supset [0,1].$

Now if $n^*=0$ we have $[0,1]\subset I_0\in G$ and therefore $\sum_{I\in G}|I|\geq|I_0|>1.$

And if $n^*>0$ note that $\inf I_j<\sup I_{j-1}$ when $1\leq j\leq n^*$ , so we have $$\sum_{I\in G}|I|\geq \sum_{j=1}^{n^*}|I_j|=\sum_{j=0}^{n^*}(\sup I_j-\inf I_j)>$$ $$>(\sup I_0-\inf I_0)+\sum_{j=1}^{n^*}(\sup I_j-\sup I_{j-1}) =$$ $$=\sup I_{n^*}-\inf I_0>1.$$

• This is a much better and more thorough answer than mine! Thank-you, I'll delete mine now :) – postmortes Jun 26 '17 at 11:34
• @postmortes. I was originally going to give a one-word answer : "No." – DanielWainfleet Jun 26 '17 at 18:59