What is the cardinality of the set of all non-measurable sets in $\Bbb R^n$? 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!
 A: 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 $\mathbb R .$
A: 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 $(1,2)$.
A: EDIT: I am trying to do a Schroeder Bernstein: 
Let N be the collection of non-measurable subsets.
1)There are at most $2^{\mathbb c}$ non-measurable subsets, since there are at most $2^{\mathbb c}$ subsets. This is the point I was trying to make in 1), though not in the clearest way. 
2) There are at least $2^{\mathbb c}$ non-measurable subsets, by a translation argument.
1)Well, the total number of  subsets of $\mathbb R^n$ is $|2^{|\mathbb R|}$| and the cardinality of all open sets is $|\mathbb R|$ . Now you have an expression $|A|=|B \cup C|$, where A represents all sets and $B$ represents all measurable sets and $B \cap C = ${}. and you know two infinite cardinalities. Now, choose your set theory and do your cardinal arithmetic. EDIT: this was not a clear way of making 
*So $|N|\leq 2^{\mathbb c} $
2)Let's work in $\mathbb R^1$. I claim there are $2^{\mathbb R}$ non-measurable subsets.
  Let $T$ be non-measurable. Then, for every non-zero Real a, the set $T+a$:={$t+a: t \in T$} is also non-measurable; if it was measurable, then $T=T+a-a$ would be the translation of a measurable set and so measurable. Since there are $\mathbb R$ choices for $a$, this means there are at least $2^{\mathbb R}$ non-measurable subsets of the Reals.
** So $|N| \geq 2^{\mathbb c} $

From *, ** , and Cantor-Shroeder-Bernstein, there are $2^{\mathbb c}$ non-measurable subsets.
EDIT: My apologies, this is unifinished, I don't have time at the moment to
finish it. IfIa m unable to finish it /correct it in a few days, I will just
delete it.
