How Lebesgue sigma algebra is different from outer regularity.

Question

Give an example of a non-Lebesgue measurable set $A$, such that there exist an open set $O$, $A\subseteq O$ and $m_*(O)\leq m*(A)+\epsilon $ but $m_*(O\backslash A)>\epsilon $?

This question arise because outer regularity doesn't implies Lebesgue measurable , so I want to look at example where it fails. I haven't find any on my own.please help.

Ps. I want to see difference between outer regularity and Lebesgue measure. As it is mentioned in comments, concrete examples are harder to find, it's enough for me to understand difference between regularity and Lebesgue measure intuitively.

Def: A set is $E$ measurable if there exists an open set $O$ such that $E\subseteq O, m_*(O\backslash E)<\epsilon $

Theorem: Outer regularity (without proof) let $E\subset \mathbb R^n$, then $m_*(E)=inf\{m_*(O)|O \text{ is open},E\subset O\}$

Answer 1

Give an example of a non-Lebesgue measurable set A, such that there exist an open set O, A⊆O and m∗(O)≤m∗(A)+ϵ but m∗(O∖A)>ϵ?

Just consider any non-measurable set $V\subset [0,1]$. From the definition of the non-measurable set, there exists $\epsilon>0$ for any open set $O\supset V$, $m^\ast(O\backslash V)\geq \epsilon$. Of course by the definition of the outer measure $m^\ast(O)\leq m^\ast(V)+\epsilon$ is satisfied.

The answer to the question is trivial from the definitions. I think you better lookup the construction (which uses non-constructive axiom in the end) of the non-measurable set. (See https://en.wikipedia.org/wiki/Vitali_set)

I want to see difference between outer regularity and Lebesgue measure. As it is mentioned in comments, concrete examples are harder to find, it's enough for me to understand difference between regularity and Lebesgue measure intuitively.

Loosely speaking, the exterior measure $m^\ast(E)$ assigns to any $E\subset\mathbb{R}^d$ a first notion of size; as the name indicates, attempts to describe the volume of a set $E$ by approximating it from the outside (from Elias M.Stein, Rami Shakarchi; Real Analysis).

On the other hand, Lebesgue measure inherits all the features of the exterior measure, and furthermore have more desirable properties such as if $E_1,E_2,\cdots$ are disjoint measurable sets and $E=\cup_{j=1}^\infty E_j$ then $$m(E)=\sum_jm(E_j)\quad\text{(countable additivity)}.$$This is not generally satisfied in the outer measure (yet, it is hard to construct a counterexample).

Also, some intuitive explanation of the measurable set is, if $E$ is measurable then its outer measure and inner measure coincide : $m^\ast(E)=m_\ast(E)$, i.e. approximating volume of $E$ from the outside and from the inside is same.