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I understand there are $|\mathbb{P} (\mathbb{R})|$ many non-measurable sets. However the sets that occur through the Vitali construction are the only sets (that I am aware of), which we can construct in a defined way. Are there any other non-measurable sets we can describe which do not involve Vitali's construction?

Nikolaos Zervos
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    Bernstein sets, an Hamel basis of $\Bbb R$ as a $\Bbb Q$ vector space, squinting a bit a nonprincipal ultrafilter over $\Bbb N$ is a nonmeasurable subset of the Cantor space $2^\Bbb N$ – Alessandro Codenotti Aug 20 '24 at 20:19
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    You can start with any nonmeasurable subset of $[0,1]$ and add to it any subset of $[2,3]$. This creates a new nonmeasurable subset of $\mathbb{R}$ and there are $|Pow([2,3])|=|Pow(\mathbb{R})|$ ways to choose a subset of $[2,3]$. – Michael Aug 20 '24 at 21:33

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Here is an easy way to construct $|Pow(\mathbb{R})|$ new nonmeasurable subsets of $\mathbb{R}$ (starting from just a single one). Let $\mathcal{F}$ be any sigma algebra on $\mathbb{R}$ that satisfies:

i) $[0,1] \in \mathcal{F}$

ii) There is a set $A \subseteq [0,1]$ such that $A \notin \mathcal{F}$.

Both the Lebesgue sigma algebra on $\mathbb{R}$ and the standard Borel sigma algebra on $\mathbb{R}$ satisfy (i) and (ii). Fix $A$ as the set that satisfies point (ii).

Claim 1: For all $B \subseteq [2,3]$ we have $A \cup B \notin \mathcal{F}$.

Proof: Suppose $A \cup B \in \mathcal{F}$. Since $[0,1] \in \mathcal{F}$, we know $(A\cup B) \cap [0,1] \in \mathcal{F}$. But $(A\cup B)\cap [0,1]= A$, contradicting the fact that $A \notin \mathcal{F}$. $\Box$.

Claim 2: There are exactly $|Pow(\mathbb{R})|$ distinct subsets of $\mathbb{R}$ that are not in $\mathcal{F}$.

Proof: There are $|Pow([2,3])|$ distinct ways to choose set $B$, each leading to a distinct set $A \cup B$ that is not in $\mathcal{F}$. Since $|Pow([2,3])|=|Pow(\mathbb{R})|$, there are at least $|Pow(\mathbb{R})|$ subsets of $\mathbb{R}$ that are not in $\mathcal{F}$. On the other hand, the number of subsets of $\mathbb{R}$ that are not in $\mathcal{F}$ is less than or equal to the number of subsets of $\mathbb{R}$, which is $|Pow(\mathbb{R})|$. $\Box$

Michael
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