Average of decimal digits

Question

I define the function $d_{\mathrm{avg}} : [0, 1]\to [0, 1]$ such that for $0.x_1x_2x_3\cdots$ the decimal expansion of $x$ (defined such that $\nexists N : x_k = 9$ for all $k \geq N$), $$d_{\mathrm{avg}} : x\mapsto \lim_{n\to \infty} \frac{1}{n}\sum_{i=1}^{\infty} x_i$$ How would I show that the set $\{x : d_{\mathrm{avg}}(x)\text{ does not exist}\}$ has Lebesgue measure $0$? It's clear to me that the set is uncountable.

Answer 1

Since almost all numbers are normal, $d_{avg}(x) =4.5 $ for almost all $x$.

Therefore the set of all $x$ for which $d_{avg}(x) $ does not exist has measure zero.