TITLE: What is the cardinality of the set of all non-measurable sets in $\Bbb R^n$?
QUESTION [5 upvotes]: The cardinality of the set of all measurable sets in $\Bbb{R}^n$ can be shown to be the same as the power set of $\Bbb{R}$ by looking into Cantor set.

Denote $M=$$\{$$Ω⊆\Bbb{R}^n:Ω$ is measurable$\}$, then $card(M)≤card(2^\Bbb{R})$. The Cantor set is measurable, has measure zero and has the same cardinality of $\Bbb{R} ⇒$ every subset of Cantor set is measurable $⇒ card(M)≥card(2^\Bbb{R} )⇒card(M)=card(2^\Bbb{R} )$. 

I am wondering what is the cardinality of the set of all non-measurable sets in $\Bbb{R}^n$? I know a theorem guarantees every non-empty open sets in $\Bbb{R}$ has a non-measurable subsets, is this useful to show the cardinality of all non-measurable sets in $\Bbb{R}^n$? Thank you!

REPLY [11 votes]: Given that $(0,1)$ has a non-measurable subset, adding points in $(1,2)$ can't make it measurable.  There are $2^{\mathfrak c}$ subsets of $(1,2)$, so take your non-measurable subset of $(0,1)$ union each subset of $(1,2)$.

REPLY [8 votes]: Take your favorite nonmeasurable set $E \subset (1,2).$ Let $K\subset [0,1]$ be the Cantor set. Then $\{E \cup F: F\subset K\}$ is a collection of non measurable sets (because throwing in a set of measure $0$ can't turn a nonmeasurable set into a measurable one). Since the cardinality of $K$ is $c,$ we see there are $2^c$ sets $F$ that can be used in the above union, and distinct $F$'s give distinct unions. It follows that there are $2^c$ nonmeasurable subsets of $\mathbb R .$