Q: Normal numbers have a measure of 1 So, I have a problem with the following proof from my course notes:
"Let us say a number $x \in [0, 1]$ is (weakly) normal if any finite sequence of digits occurs infinitely often in its decimal expansion. We claim the set E of all such numbers has $m(E) = 1$. To see this, let $E_i$ be the set of $x \in [0, 1]$ such that the $i$-th digit of $x$ is $1$. Then
the $E_i$ are independent, and $m(E_i) = 1/10$, so $\sum_i E_i = \infty$. Thus the digit 1 occurs infinitely often for almost every $x \in [0, 1]$. The same reasoning applies to any finite sequence of digits. Intersecting these countably many sets of measure one again yields a set of full measure."
I understand everything in this proof except for the fact that $E_i$ and $E_j$ for $i \neq j$ are indeed independent (which was defined as $m(E_i \cap E_j) = m(E_i) m(E_j)$). Intuitively, it is very clear, however, I don't know how I could prove that rigorously. I see that $E_i$ is measurable as a countable union of intervals, and I also intuitively understand that the measure of $E_i \cap E_j$ should be equal to the measure of $E_i \cap F_j$ (where $F_j$ represents the set of numbers with, say, 2 at the $j$-th place after comma), since we have a bijective mapping between intervals which unions constitute $E_j$ and $F_j$, however, I don't know how to make this rigorous.
Could someone please help me with that?
