Cardinality of the class of $G_\delta$ subset of $\mathbb{R}$ of Lebesgue measure zero Let $\mathcal{N}$ be the class of all subsets of $\mathbb{R}$ of Lebesgue measure zero
and let $\mathcal{G}_\delta$ be the class of all $G_\delta$ subsets of $\mathbb{R}$.
How do I show that $|\mathcal{N}\cap \mathcal{G}_\delta| = 2^{\omega}$ ?
 A: To show that there are at most $2^\omega$ such sets, it suffices to show that there are at most $2^\omega$ $G_\delta$-sets. $\Bbb R$ has a countable base $\mathscr{B}$ (e.g., the family of open intervals with rational endpoints), and every open set is the union of some subfamily of $\mathscr{B}$, so there are at most $|\wp(\mathscr{B})|=2^\omega$ open sets. Each $G_\delta$ is the intersection of a countable family of open sets, and there are only $(2^\omega)^\omega=2^{\omega\cdot\omega}=2^\omega$ such families, so there are at most $2^\omega$ $G_\delta$-sets in $\Bbb R$.
A: You can do this in two easy steps:

*

*Show that there are only $2^{\aleph_0}$ sets which are $G_\delta$ at all. To do so first show that there are only $2^{\aleph_0}$ open sets, which is true because every open set in the real numbers is a countable union of disjoint intervals. Since every interval has the form $(a,b)$ where $a,b\in\Bbb R\cup\{\infty,-\infty\}$ we have that there are only $2^{\aleph_0}$ intervals; and therefore there are only $(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}$ countable families of intervals, and so we conclude that there are only $2^{\aleph_0}$ open sets.
By the same argument as before there are exactly $2^{\aleph_0}$ countable families of open sets, therefore there are at most $2^{\aleph_0}$ sets which are $G_\delta$ sets.


*Show that every singleton is a $G_\delta$ set, and in particular it is a $G_\delta$ set which has measure zero.
Therefore there are $2^{\aleph_0}$ measure zero $G_\delta$ sets, but there are only $2^{\aleph_0}$ $G_\delta$ sets to begin with, so this is the exact cardinality that you are looking for.
