0
$\begingroup$

This question already has an answer here:

The question is to give and example of Lebesgue measurable set but not Borel measurable.

I know there exists subset of Cantor set that is not Borel measurable, since the cardinality of all Borel sets in $[0,1]$ is $\aleph_1$ while the cardinality of the subsets of Cantor set (defined on $[0,1]$) is $\aleph_2$, so it must contains subsets that are non-Borel measurable.

The problem is, the "cardinality of all Borel sets in $[0,1]$ is $\aleph_1$" requires something that is beyond the scope of first-year graduate level real analysis course. I cannot prove it using what I learned.

Can anyone give another example? Thank you!

$\endgroup$

marked as duplicate by Jonas Meyer real-analysis Jul 24 '15 at 4:56

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ The Continuum Hypothesis doesn't seem relevant. $\endgroup$ – Jonas Meyer Jul 24 '15 at 4:46
  • $\begingroup$ The Cantor set is assumed to be defined on $[0,1]$, its subsets has cardinality $\aleph_2$, more than Borel measurable sets in $[0,1]$, which is $\aleph_1$, so there must be a subset of Cantor set that is not Borel measurable. $\endgroup$ – Tony Jul 24 '15 at 4:48
  • $\begingroup$ It is $\aleph_2$ if you assume the Continuum Hypothesis, but you do not need CH for Cantor's theorem that $2^n>n$ even when $n$ is an infinite cardinal number. So $2^{2^{\aleph_0}}>2^{\aleph_0}$ $\endgroup$ – Jonas Meyer Jul 24 '15 at 4:51
  • $\begingroup$ Thank you for your reply. But I am not saying that because of Continuum Hypothesis so I cannot prove it. I am saying this example requires knowledge beyond the real analysis stuff, involving the structure of Borel sets, etc. So I am asking if there exists other examples within the scope of general real analysis level. $\endgroup$ – Tony Jul 24 '15 at 5:02
  • $\begingroup$ I understand what you were asking, but you were stating stuff about $\aleph$s that depend on the (generalized) CH without relevance to the argument. Sorry I was unclear. $\endgroup$ – Jonas Meyer Jul 24 '15 at 5:05