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let $$X=\left\{ \sum_{n=1}^{\infty}a_{n}2^{-n}:a_{n}\in\left\{ 0,1\right\} ,\liminf\frac{1}{n}\sum_{i=1}^{n}a_{i}<\limsup\frac{1}{n}\sum_{i=1}^{n}a_{i}\right\} $$

I'm studying fractal geometry and I've encountered the following question: Is there an $\alpha$-regular measure giving $X$ positive measure?

Now, I've noticed that the Lebesgue measure of $X$ must be zero, simply by looking at its complement, and by the LLN the limit $\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^{n}a_{i}$ exists almost surely, hence $X^{c}$ has Lebesgue measure 0. I'm not sure how to proceed, and would be happy for any suggestions. In general I'm trying to find the Hausdorff dimension of $X$.

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    $\begingroup$ What is an $\alpha$-regular measure? $\endgroup$
    – Christophe Leuridan
    Commented Aug 20, 2023 at 20:36
  • $\begingroup$ We say a measure $\mu$ is $\alpha$-regular if there's a constant C such that $\mu\left(B_{r}\left(x\right)\right)\leq C\cdot r^{\alpha}$ for every $x, r$ $\endgroup$
    – Simple Conjugate
    Commented Aug 20, 2023 at 21:32
  • $\begingroup$ What about the set of numbers for which for all $k$ and for all $n\in [k!,2k!]$ we have $a_n=0$, other $a_n$'s are chosen independently uniform on $\{0,1\}$? $\endgroup$
    – Fedor Petrov
    Commented Aug 8 at 4:18

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You could take a look at On simply normal numbers to different bases, by Becher, Bugeaud and Slaman (http://dx.doi.org/10.1007/s00208-015-1209-9). An instance of the main theorem of that paper is that the set of real numbers that are not simply normal in base 2 has full Hausdorff dimension. The method of the paper is to ensure failures of simple normality by introducing oscillations in the associated discrepancy function, which ensures the inequality that you requested.

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