The cardinality of Lebesgue sets

Suppose $A=\{S\;|\;S \subset \mathbb R^n, S\text{ is Lebesgue measurable}\}$. What is the cardinality of $A$? Is it the same as the cardinality of all of the real numbers?

If $A$ is the standard Cantor set, then $A\subseteq\mathbb R^1$ is Lebesgue measurable, and has measure zero and the same size as the reals. Any subset of $A$ is also measurable, so there are as many Lebesgue measurable subsets as there are sets of reals (which is strictly larger than the size of the reals).

The same holds in $\mathbb R^n$ with $n>1$, since $[0,1]\times\{0\}^{n-1}$ is Lebesgue measurable of measure $0$ and has the same size as $\mathbb R$.

The key here is that Lebesgue measure is complete, so any subset of a measure zero set is measurable. In contrast, there are only as many Borel sets as there are real numbers.

Hint:

Note that the Cantor set, or some copy of it in $\Bbb R^n$ is Lebesgue measurable with measure zero.

1. Every subset of a null set is measurable.
2. The cardinality of the Cantor set is $2^{\aleph_0}$.