The complete statement of the exercise is as follows:
For $ I = [0,1]$, let be $ A \subseteq I $ the subset of points $ x \in I$ whose base 5 expansion has the form: $$ x = \sum_{k = 1}^{\infty}\frac{x_k}{5^k} $$
where $x_k \in \{0,4\}$. Show that the set $A$ is Lebesgue-Measurable and $m(A) = 0$.
I have a question about the exercise:
Is not true that all numbers in $ [0,1] $ has a 5 base expansion? In that case $I = A$ but that would contradicts the exercise becasue $m(A) = m(I) = 1$ so what us happening here?
Now my attempt to solve:
Because we already assume that $m(A) = 0$ so it is enough to show that $m^{*}(A) = 0$ and this happens if $A$ is countable, and this happensi if there is a bijection between $A$ and $\mathbb{Q}$
and that is, some idea to proceed?