I read a Math Overflow answer recently that claimed the set of numbers in $[0,1]$ whose binary expansion contained a different number of 0 and 1 digits (i.e. doesn't include numbers like $0.\overline{10}_2$) has zero Lebesgue measure, which the user said is implied by the Strong Law of Large Numbers. This is probably my inexperience talking, but I can't even see the connection between those two concepts, much less how one implies the other is true.
It also seems strange to me that this set and not its complement in $[0,1]$ would have zero measure, since there are $\frac {n!}{(\frac n2!)^2}$ binary numbers expressed with half zeroes and ones up to the nth digit after the radix point, which is less than the amount of numbers that do not satisfy this (that being $2^n-\frac {n!}{(\frac n2!)^2}$) for every $n\in2\Bbb N, n\gt2$. Though, admittedly, this logic may not apply to uncountable sets.
Could anyone explain or prove that either this set or its complement in $[0,1]$ has zero measure?