# 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.

• The Cantor set has measure zero, but is uncountable. See en.wikipedia.org/wiki/Cantor_set – user254433 Jul 9 '17 at 2:22
• A measure zero set can have any cardinality less than or equal to $2^{\aleph_0},$ the cardinality of $\mathbb R.$ A measurable set of positive measure can only have cardinality $2^{\aleph_0};$ cardinality $\aleph_1$ is possible only if $\aleph_1=2^{\aleph_0}.$ – bof Jul 9 '17 at 2:29

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.

1. $$A=\{1\}$$
2. $$B = \mathbb{N}$$
3. $$C =$$ 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 $$2^{\aleph_0}$$.