Cardinality of a measure-zero set On real number line $\mathbb R$, does saying a set is of Lebesgue measure zero equivalent to saying that the set is of $\aleph_0$? 
Does saying a set is of Lebesgue measure $>0$ equivalent to saying that the set is of $\aleph_1$?
I understand that countable sets are always of measure zero, but I am not sure if the inverse is also true.
 A: One of your claims is true and one is not.
Firstly, the real numbers $\mathbb{R}$ have cardinality $\mathfrak{c} = 2^{\aleph_0}$. Therefore every subset of $\mathbb{R}$ has a cardinality less than or equal to $2^{\aleph_0}$.
Claim 1: Every subset of $\mathbb{R}$ with positive measure has cardinality equal to that of the real numbers, $2^{\aleph_0}$.
Proof: Suppose $A \subset \mathbb{R}$ with positive measure. Then the cardinality of $A$ is clearly infinite, so we have $\aleph_0 ≤ |A| ≤ 2^{\aleph_0}$. As $A$ is infinite, we now have $|A| = |A+A|$ (where $A+A = \{a_1+a_2:a_1,a_2\in A\}$). Moreover, $A+A$ is guaranteed to contain an open interval $I$. (See here for a proof.) Thus we get $2^{\aleph_0} = |I| ≤ |A+A| = |A| ≤ |\mathbb{R}| = 2^{\aleph_0}$. QED
Claim 2: If a subset of $\mathbb{R}$ has measure $0$, we can't say anything about its cardinality.
Proof: Consider these three cases.

*

*$A=\{1\}$

*$B = \mathbb{N}$

*$C =$ Cantor ternary set

On a final note, you have suggested that $2^{\aleph_0} = \aleph_1$. This is true only if the continuum hypothesis is true, so it's better to just write $2^{\aleph_0}$.
