$E \subset \mathbb{R}$ such that $\lambda(E \cap [a,b])=\frac{1}{2}(b-a)$ for all $a,b\in\mathbb{R}$

This is Problem 6.45 from Jones' Lebesgue Integration on Euclidean Space.

Prove that there does not exist a set $E \subset \mathbb{R}$ such that $\lambda(E \cap [a,b])=\frac{1}{2}(b-a)$ for all $-\infty<a<b<\infty$.

It is one of 'some facts about integration', according to the book, but I don't see how this problem can be solved using integration. The statement is similar to $\int_{a}^{b} \chi_E \, d\lambda=\frac{1}{2}(b-a)$, but I'm not even sure if $E$ is measurable or not, so the integral doesn't make sense yet. Can someone give me a hint for this problem?

• $\lambda$ is the lebesgue measure here. – Dilemian Jan 4 '17 at 14:13
• If $E$ isn't measurable, $\lambda(E\cap [a,b])$ doesn't make sense in general. And what do you know about the density of measurable sets? – Daniel Fischer Jan 4 '17 at 14:26

I'm assuming $\lambda$ refers to Lebesgue measure defined on the Lebesgue $\sigma$-algebra (and not the associated outer measure). In that case $E$ belongs to the Lebesgue $\sigma$-algebra.

• Suppose first that $\lambda(E)<\infty$.

Given $\epsilon>0$, there exists $G=\cup_i (a_i,b_i)$ where the union is disjoint and countable such that $E\subset G$ and $\lambda(G\setminus E)\leq \epsilon$.

$\lambda(E)=\lambda(E\cap G)=\sum_1^\infty \lambda(E\cap (a_i,b_i))=\sum_1^\infty \lambda(E\cap [a_i,b_i])=\frac 12 \sum_1^\infty (b_i-a_i)=\frac 12 \lambda(G)$

Thus $\lambda(G) = \lambda(E) + \lambda(G\setminus E)\leq \frac 12 \lambda(G) + \epsilon$

By our finiteness assumption, $\lambda(G)\leq 2\epsilon$ and $\lambda(E)\leq 2\epsilon$ for any $\epsilon >0$, that is $\lambda(E)=0$ which is a contradiction with the initial assumption.

• In the general case, let $E_n=(-n,n)\cap E$ and note that $E_n$ increases to $E$.

Also note that $\lambda(E_n \cap [a,b])=\frac{1}{2}(b-a)$ whenever $[a,b]\subset (-n,n)$.

Let $G$ be an open set such that $E_n\subset G$ and $\lambda(G\setminus E_n)\leq \epsilon$. Let $G'=G\cap (-n,n)$ which is open and satisfies $\lambda (G'\setminus E_n)\leq \lambda (G\setminus E_n)\leq \epsilon$. You may write $G'=\cup_i (a_i,b_i)$ with $(a_i,b_i)\subset (-n,n)$.

The same reasoning as above yields $\lambda(G')\leq 2\epsilon$ and $\lambda(E_n)=0$.

But $\lambda(E)=\lim_n \lambda(E_n)=0$.

I must thank Daniel Fischer for fixing the missing argument in my proof.