Limit of average of decimal digits: $\lim\frac{1}{n} (x_1 + \dots +x_n) = constant$

Question

I have a problem to solve in Ergodic Theory, but I am stuck and have no idea how to procedure. The problem is the following.

Prove that there exists a constant α such that for Lebesgue a.e. x∈[0,1] $\lim_{n\to\infty} \frac{1}{n} (x_1 + \dots + x_n) = \alpha$ where $x_1 ,...,x_n$ are digits of the decimal expansion of x meaning $x_i \in $ {0,...,9}.

I have, that if $x \in Q$, $\alpha$ is obviously 0.

So if $x \in $ R\Q we can bound the limit by above by 9 and below by 1 e.g. $\lim_{n\to\infty} \frac{1}{n} (x_1 + \dots + x_n) \leq \lim_{n\to\infty} \frac{9n}{n} = 9$.

Right? But now I still have to prove it exists, how can I do that? Thanks a lot already.

Answer 1

Hint: If you have been following a course on Ergodic theory you have most certainly encountered the map $x\mapsto 2 x$ (mod 1) and the fact that it preserves and is ergodic with respect to Lebesgue measure?

If you consider the indicator function on $[1/2,1)$ as an observable then the sum along an orbit of a number $x$ corresponds to the number of binary digits in the expansion of $x$. For Lesbesgue a.e. point the average therefore converges to the integral of the observable, i.e. 1/2.

Redo this exercise but for the map $x\mapsto 10 x$ (mod 1) and figure out the right observable to use.

Answer 2

Finally I understood :D So I take the MPS $[0,1) -> [0,1)$, $x \mapsto 10x $ with the Lebesgue measure. This is ergodic. I take as my function $f(x) = 0 on [0,0.1), 1 on [0.1,0.2) ... $. Then the sum along an orbit of a number $x$ correspond to the number of 10 digits.

So we get by Birkhoff $\frac{1}{n} \sum_{i=1}^{n-1} f (10^{-1}(x)) -> \int_{[0,1)} f d\mu =4.5$. Right? Thank you so much!!