I have difficulties proving the following :
Let A be the set of all real x between [0 , 1], such that x has a decimal representation and each digit has infinite occurence. Now, what i do understand is, that a borel set is a set that can be constructed using countable union or intersection. But i dont really see a way to construct this set, since there are uncountably many of such x. Any guesses?
First note that the numbers with non-unique decimal representation are not in $A$, and neither of their two representations (one ending in $000\ldots$ and one in $999\ldots$ has infinitely many of any digit other than $0$ or $9$. So we may restrict our attention to the representation that, in case of non-uniqueness, ends in $0$'s. Let $D_n(x)$ be the $n$'th digit of $x$ in this representation. We have $A = A_0 \cap \ldots \cap A_9$, where $A_d$ is the set of numbers whose decimal representation has infinitely many $d$'s. Now $x \in A_d$ iff for every positive integer $m$ there is integer $n > m$ such that the $n$'th digit of $x$ is $d$. That says $$ A_d = \bigcap_{m=1}^\infty \bigcup_{n=m+1}^\infty \{x\in (0,1): D_n(x) = d\}$$ And finally, write $\{x \in (0,1): D_n(x) = d\}$ as a finite union of intervals.