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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.

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    $\begingroup$ Why is $a$ obviously $0$? Informally it looks like you are asking for the average value of a digit in a randomly chosen decimal. As each decimal is equally probable I'd have thought that was $\frac {0+1+2+\cdots +9}{10}=4.5$ $\endgroup$ – lulu Dec 18 '17 at 16:34
  • $\begingroup$ For $ x \in Q$ the number of $x_i $ unequal to 0 is finite. Since we view for the limit of n this must be 0 right? $\endgroup$ – Andreas Wicher Dec 18 '17 at 16:38
  • $\begingroup$ In other words, $\lim_{n\to\infty} \frac{constant}{n}$ = 0 right? $\endgroup$ – Andreas Wicher Dec 18 '17 at 16:41
  • $\begingroup$ Oh, but the rationals have measure $0$. $\endgroup$ – lulu Dec 18 '17 at 16:50
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    $\begingroup$ @AndreasWicher : 1/3 = 0.3333333.... $\endgroup$ – Michael Dec 18 '17 at 17:00
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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.

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  • $\begingroup$ Okey thank you very much!! I will try :) $\endgroup$ – Andreas Wicher Dec 18 '17 at 16:46
  • $\begingroup$ So, pretty sure $x \mapsto 10x $ is also ergodic and lebesque measure presurving. And I guess I should use the interval [0.1, 0.2) but i don't understand right now, why the orbit should correspond to the number of digits? And do i need to use ergodicity somewhere? $\endgroup$ – Andreas Wicher Dec 18 '17 at 17:23
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    $\begingroup$ Hint2: Try with an observable that equals 0 on [0,0.1), 1 on [0.1,0.2), ... 9 on [0.9,1) $\endgroup$ – H. H. Rugh Dec 18 '17 at 17:32
  • $\begingroup$ What is a observable here exactly, it's a function right? The only definition i find is, that it's a property on a non-zeromeasure set... $\endgroup$ – Andreas Wicher Dec 18 '17 at 17:43
  • $\begingroup$ It is a function. You know the Birkhoff ergodic theorem, I presume? It deals with averages of a function (or observable) $A$ along the orbit of $x$ under an ergodic transformation $T$, i.e. of the (possible) limit of $(A(x)+A(Tx)+...+A(T^{n-1}x))/n$ as $n$ goes to infinity. $\endgroup$ – H. H. Rugh Dec 18 '17 at 18:12
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Finally I understood :D So I take the MPS $[0,1) -> [0,1)$, $x \mapsto 10x $ with the lebesque 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!!

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  • $\begingroup$ Yes correct, except that the sum along the orbit is the sum of digits (not just the number) but this is certainly also what you meant. This holds for Lebesgue a.e. point. $\endgroup$ – H. H. Rugh Dec 20 '17 at 14:25

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