Could someone explain to me why the set constructed as follows:
For any positive integer $n$, let $n^*$ denote the decimal rational obtained by writing the decimal form of $n$ in reverse order and then putting a decimal point before it. For example,
$2^* = 0.2, \ \ 500^* = 0.005, \ \ 1234^* = 0.4231$,etc.
Let $A$ be the set of those elements $x$ of $[0,1]$ for which there is an infinite sequence of positive integers $k_1, k_2, ..., k_n, ... $ with $k_1^* \le k_2^* \le ... \le k_n^* \le ...$ such that $x_{k_n} = 3$ for all $n$, where $x_k$ is the $k$-th digit in some decimal expansion of $x$ (that is, $x = \sum_{k \ge 1} \frac{ x_k}{10^k}$).
is not Borel but it is Leb. measurable.
There are a few sequences in $\mathbb{N}$ which quite immediately come to mind and fulfill the condition, for example: $\{1,11,111,1111, ... \}$ or $\{1, 11, 12, 13, 14,15,16,17,18,19, 119,1119,11119,...\}$ or a mix $\{1, 11, 12, 13, 14,15,16,17,18,19,29,39,49,59,69,79,89,99,999,9999,99999,...\}$, etc., etc.
So these numbers would be $0,... \ 3 \ ... \ 3 \ ... \ 3 \ ...$ with $3$s in the special places $k_n$ as described above.
I don't see how this prevents $A$ from being Borel, but still $A$ remains measurable.
I would very much appreciate an explanation that doesn't use the fact that this set is analytic, if that's all right.
All of these sets are closed. This follows from the following: Let $\langle k_n : n \geq 1\rangle$ be any sequence of positive integers. Then the set of reals in $[0, 1]$ that have a $3$ at their $k_n$-th digits is closed. Moreover, if the set $\{ k_n: n \geq 1 \}$ is infinite then this closed set is null.