# Examples of non-measurable sets(Lebesgue)

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?

In Measure and Category by Oxtoby, Chapter $5,$ the author stated that Bernstein set is non-measurable (Theorem $5.4$).