# An example of Lebesgue measurable set but not Borel measurable besides the “subset of Cantor set” example. [duplicate]

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!

## marked as duplicate by Jonas Meyer real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jul 24 '15 at 4:56

• 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. – Tony Jul 24 '15 at 4:48
• 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 $2^{2^{\aleph_0}}>2^{\aleph_0}$ – Jonas Meyer Jul 24 '15 at 4:51
• 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. – Jonas Meyer Jul 24 '15 at 5:05