I made this post because i would like to see some examples of non-measurable sets with respect to the Lebesgue measure on $\Bbb{R}^d$
A first example is the classical Vitaly set.
Another example which involves dynamical systems is this:
Let $([0,1),\mathcal{B},m,T)$ where $Tx=(x+a)mod1=\{x+a\}=x+a-[x+a]$ where $a \in \Bbb{R}\setminus \Bbb{Q}$
$T$ is an irrational rotation.
Every orbit $\{T^nx|n \in \Bbb{Z}\}$ is infinite for all $x \in [0,1)$
We use the axiom of choice and we construct the set $A \subseteq [0,1)$ which contains exactly $\text{one}$ element of every orbit of $T$
In other words $\forall x \in [0,1)$ exists unique $m \in \Bbb{Z}$ such that $T^mx \in A$
It is clear that $A$ is uncountable.
Not let $B_i=T^i(A)$.
It is not difficult to see that $B_i \cap B_j=\emptyset,\forall i \neq j$ in $\Bbb{Z}$
Now if $A$ is Lebesgue measurable then:
$m(\bigcup_{i \in \Bbb{Z}}B_i)=\sum_{i \in \Bbb{Z}}m(B_i)=\sum_{i \in \Bbb{Z}}m(A)$ because the Lebesgue measure is translation-invariant(or $T-$invariant).
Also $1=m(\bigcup_{i \in \Bbb{Z}}B_i)$ because $\bigcup_{i \in \Bbb{Z}}B_i=[0,1)$
Thus if we assume that $m(A)>0$ or $m(A)=0$ we easily derive a contradiction.
Thus $A$ is not Lebesgue measurable.
Can someone provide me other interesting example of non-Lebesgue measureable set in $\Bbb{R}$ or $\Bbb{R}^d$ in general?
Thank you in advance.
Best Answer
In Measure and Category by Oxtoby, Chapter $5,$ the author stated that Bernstein set is non-measurable (Theorem $5.4$).