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.

  • 2
    $\begingroup$ The Cantor set has measure zero, but is uncountable. See en.wikipedia.org/wiki/Cantor_set $\endgroup$ – user254433 Jul 9 '17 at 2:22
  • $\begingroup$ 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 $\aleph_1=2^{\aleph_0}.$ $\endgroup$ – 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 $2^{\aleph_0}$.


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