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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.

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

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  • $\begingroup$ +1...this is a nice example...unfortunately i do not know ordinal theory to understand the construction...but still good $\endgroup$ Jan 3 '18 at 17:42
  • $\begingroup$ @MariosGretsas: You can treat ordinal as generalisation of natural number. In other word, you can imagine constructing Bernstein set using natural numbers instead. $\endgroup$
    – Idonknow
    Jan 3 '18 at 17:46
  • $\begingroup$ I will try that..thanks for the answer. $\endgroup$ Jan 3 '18 at 17:46

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