We know that almost all numbers are normal and that a set of non normal numbers has Lebesgue Measure zero.
This mean that, given a basis $b$, the set of all normal numbers in basis $b$ has a Lebesgue Measure different from zero, right?
Well, how can we demonstrate that the set of all normal numbers in basis b is Lebesgue Measurable?
This is my first time here, hope I did everything ok
Thank you all
The Lebesgue measure is complete: every subset of a Lebesgue null set is Lebesgue measurable (and has measure $0$, of course). Since you know that the set of non-normal numbers has measure zero, it follows that the set of numbers that are not normal to base $b$ is measurable, and has measure zero. Therefore, its complement is measurable (and is a set of full measure).
Ultimately, the source of measurability is the following argument: one shows that the set of non-normal numbers (on $[0,1]$ for simplicity) is contained in some set $E_n$ with $m(E_n)<1/n$, for all $n$. The intersection $\bigcap E_n$ is measurable and has measure $0$. By completeness of the Lebesgue measure, the set of non-normal numbers is measurable too.
