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The Borel number theorem states that with respect to Lebesgue measure, almost all real numbers are normal numbers. It is sometimes stated in the context of the compact interval $[0,1]$, where one replaces the term "Lebesgue" with "uniformly distributed".

  1. What if I sampled from a Cauchy distribution on the real line or a Beta distribution on $[0,1]$? My hunch is that the result still holds because the absolute continuity of these distributions (might) ensure that the set of measurable sets, $\mathcal{A}$, is the same set of measurable sets used in Lebesgue measure, and that Lebesgue measure 0 sets are also measure 0 under the new measure or distribution. But if I had specified a discrete distribution or a singular measure (e.g. Cantor's singular measure) then I might change the result.

    So the general question for part 1. is "what is the most general class of measures or probability distributions on either $\mathbb{R}$ or $[0,1]$ for which the Borel number theorem holds?"

  2. What transformations of real numbers preserve the normal number property? I have read that multiplication by a rational number preserves this property. Clearly translation by an arbitrary number does not, but I imagine translation by a rational does.

    So in general, the second question is what is the subset of functions $T:\mathbb{R}\to\mathbb{R}$ (or $T:[0,1]\to[0,1]$) such that if $x$ is a normal number then $T(x)$ is a normal number?

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  • $\begingroup$ I beleive it is unknown even whether $x$ normal implies $x^2$ normal, so I expect the second question is out of reach. $\endgroup$ Commented Apr 17, 2016 at 22:53
  • $\begingroup$ @GerryMyerson Thank you. I've realized now that the class of normal numbers may be quite a bit "smaller" than the class of numbers that pertain to the thing that inspired the question (symbol sequences in which every finite sequence appears at least once). I will rephrase in separate question. $\endgroup$
    – cantorhead
    Commented Apr 18, 2016 at 18:06

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Any set of measure $0$ with respect to Lebesgue measure also has measure $0$ with respect to a measure absolutely continuous with respect to Lebesgue measure: that's just the definition of absolute continuity of measures. The non-normal numbers $N^c$ form an uncountable Borel subset of $\mathbb R$, and this will have positive measure for some measures not absolutely continuous wrt Lebesgue measure. I doubt that there is a simple characterization of the measures $\mu$ such that $\mu(N^c) > 0$ that isn't essentially a restatement of the definition.

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I can give a partial answer to your second question. Partial because a good amount of work has been done on this, but I think there is little chance that this questions will ever be fully answered. I will give a few papers that discuss this with the admission that I'm somewhat biased towards my own work and interests in this direction. "Normality preserving operations" have been studied for the $b$-ary expansions, continued fraction expansion, and the Cantor series expansions. I will state some of the results and literature here. Please let me know if you want more information.

I first want to remark that D. D. Wall first proved in his dissertation that rational multiplication and addition preserve normality in base $b$. You can get this over ILLIAD if you want, but please message me if you want an electronic copy. G. Rauzy fully classified the set of real numbers that preserve normality under addition. See: Nombres normaux et processus déterministes. Acta Arith. 29 (1976), no. 3, 211–225. Available for free here: http://matwbn.icm.edu.pl/ksiazki/aa/aa29/aa2931.pdf. More on this topic from a descriptive set theoretic perspective can be found here: https://arxiv.org/abs/1609.08702.

J. Vandehey considered this problem for continued fraction expansions. See https://arxiv.org/abs/1504.05121. Also this was considered for Cantor series expansions in two papers: https://arxiv.org/abs/1407.0777 and Airey, Dylan; Mance, Bill; Vandehey, Joseph Normality preserving operations for Cantor series expansions and associated fractals. II. New York J. Math. 21 (2015), 1311–1326. (which can be found at http://nyjm.albany.edu/j/2015/21-60v.pdf). Also the intro in the last two papers contain much of what is currently known about normality preserving operations.

Lastly, I wanted to talk about operations that form real numbers by sampling along digits of special sequences. See Furstenberg, Harry Disjointness in ergodic theory, minimal sets, and a problem in Diophantine approximation. Math. Systems Theory 1 1967 1–49. which can be found here http://www.math.tau.ac.il/~barakw/seminar/furstenberg_disjointness.pdf. Also, I think this paper of T. Kamae was important: Kamae, Teturo Subsequences of normal sequences. Israel J. Math. 16 (1973), 121–149. But further work has been done on this.

For a negative result for continued fraction expansions, see Heersink, Byron; Vandehey, Joseph Continued fraction normality is not preserved along arithmetic progressions. Arch. Math. (Basel) 106 (2016), no. 4, 363–370. And for Cantor series expansion, see Airey, Dylan; Mance, Bill Unexpected distribution phenomenon resulting from Cantor series expansions. Adv. Math. 279 (2015), 372–404 and https://arxiv.org/abs/1607.07164.

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  • $\begingroup$ Thank you! This literature review seems a very good start toward answering the more general question and suggests new directions that had not occurred to me. $\endgroup$
    – cantorhead
    Commented Dec 21, 2016 at 14:12
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In my paper with M. Hochman "Equidistribution from fractals", we give a geometric condition on a probability measure $\mu$ on the real numbers which ensures that $\mu$ almost every point is normal to a given base $m\ge 2$. The condition is fairly technical to state precisely, but it roughly says that if the measure $\mu$ does not exhibit any almost-periodic behaviour under the process of ``magnifying by a factor $m$ around a typical point'', then $\mu$ almost all points are normal to base $m$.

Our condition is invariant under $C^1$ maps. This does not say that normal numbers are invariant under $C^1$ maps, which is trivially false. It does say that some geometric features that ensure normality are, in a global sense, invariant under $C^1$ maps.

As pointed out by Robert Israel and Christian Remling, this is of interest only when $\mu$ is a singular measure with respect to Lebesgue measure. The good news it that there are plenty of natural singular measures to which our results apply. In particular, absolute continuity is definitely not the only reasonable condition that ensures that almost all points are normal (although it is the only condition if one considers ``size'' only).

For example, suppose $\mu$ is a measure on $[0,1)$ which is invariant under multiplication by $p$ on the circle, i.e. $\mu(A)=\mu(T_p^{-1}A)$ where $T_p(x)=px\bmod 1$. Then $\mu$ almost every point is normal to base $m$, for any $m$ such that $\log p/\log m$ is irrational (if $\log p/\log m$ is rational, this is not true). This extends previous results by Cassels, B. Host, E. Lindenstrauss and others.

For $T_p$-invariant measures $\mu$, it is not true that $\mu$ almost all points are normal to all bases. We do get many examples where this is the case.

Let $B\subset\mathbb{N}$ be a finite set with at least two elements, and let $A$ be the set of all points whose continued fraction expansion has only digits in the set $B$. It is well known that $A$ has positive and finite Hausdorff measure in its dimension, let $\mu$ be the restriction of the corresponding Hausdorff measure to $A$ (alternatively, there is a geometric Gibbs measure on $A$ for the Gauss map which is equivalent to $\mu$). Then $\mu$ almost all points are normal. Related results have been obtained by Kaufman, and by T. Jordan and T. Sahlsten.

To give a final example, let $A$ be the self-similar set obtained by replacing $[0,1]$ with $[0,1/2]\cup [2/3,1]$ and continuing inductively on each interval, always keeping intervals of relative lengths $1/2$ and $1/3$. Again, $A$ supports a natural measure $\mu$, and $\mu$ almost all points are normal, even though $A$ has Hausdorff dimension less than $1$. The same holds for a much wider class of self-similar measures, as long as there are two contraction ratios $r_1,r_2$ in the construction such that $\log r_1/\log r_2\notin\mathbb{Q}$.

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  • $\begingroup$ Thank you @PabloShmerkin, this is definitely solid progress. This is a really nice condition regarding the dynamics of multiplying by $p$ on the unit circle. $\endgroup$
    – cantorhead
    Commented Dec 21, 2016 at 14:15
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As already pointed out, almost certainly the only possible answer to (1) is the tautological "exactly those measures that give zero weight to the set $N^c$ of all non-normal numbers."

However, let me perhaps point out that $\dim (N^c\cap I)=1$ (Hausdorff dimension) for any open interval $I\not=\emptyset$, so absolute continuity is close to what you need. That $\dim (N^c\cap I)=1$ is essentially obvious: Lebesgue measure can be viewed as tossing a fair coin to produce the individual digits. A bias in your coin, no matter how small, will produce a non-normal number with probability one, and these measures will give zero weight to sets of Hausdorff dimension $<1-\epsilon$, if the bias is sufficiently small (depending on $\epsilon$).

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  • $\begingroup$ the statement that a biased coin produces a non-normal number with probability 1 is not intuitive to me. I believe the biased process can still be represented as an irreducible Markov Process. For the probabilities of seeing $...a_1a_2a_3...a_m...$ can we not "lift" to $2^m$ dimensional Markov Process? And if the original transition probabilities were positive wouldn't the lifted transitions also be positive and the process irreducible? $\endgroup$
    – cantorhead
    Commented Apr 18, 2016 at 15:59
  • $\begingroup$ I produce a number $x=0.x_1x_2x_2\ldots$ by making the $x_j$ iid random variables. If the individual distribution is biased, then (law of large numbers) $x$ will not be normal with prob $1$. $\endgroup$ Commented Apr 18, 2016 at 16:34
  • $\begingroup$ I don't think you mean "support": the support of the measure will be all of $[0,1]$. $\endgroup$ Commented Apr 18, 2016 at 17:20
  • $\begingroup$ @RobertIsrael: I do mean support, defined as any set $S$ with $\mu (S^c)=0$ (not closed). $\endgroup$ Commented Apr 18, 2016 at 17:24
  • $\begingroup$ The terminology I'm used to would be "$\mu$ is concentrated on $S$." But do you know that every set on which $\mu$ is concentrated has Hausdorff dimension $\ge 1-\epsilon$? $\endgroup$ Commented Apr 18, 2016 at 17:53

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