# Examples of properties that hold almost everywhere, but such that explicit examples are not known

In measure theory one makes rigorous the concept of something holding "almost everywhere" or "almost surely", meaning the set on which the property fails has measure zero.

I think it is very interesting that there are some properties which hold almost surely for which it is either very difficult to construct an example, an example requires the axiom of choice, or for which no known examples exist.

I was looking to put together a list of some of these properties. More precisely, I am looking for (interesting) examples of measure spaces $(X,\mathscr{A},\mu)$ and $S\subseteq X$ such that there exists $N \in \mathscr{A}$ with $S^c \subseteq N$ and $\mu(N)=0$, but that it is hard to find/construct an explicit example of an element of $S$.

A good place to start is of course $(\mathbb{R}^n,\mathcal{B}^n,\lambda)$.

Here's one to get us started:

Theorem. Take $X=(0,1]$ with Lebesgue measure. Writing $\frac{p_n}{q_n}$ to be the $n$th continued fraction approximate of $x\in(0,1]$ in reduced terms, we have $$\lim_{n \to \infty} \frac{\log q_n}{n} = \frac{\pi^2}{12 \log 2}$$ almost surely. Note: The result fails for all rationals.

• How about the set of undefinable real numbers. – Abel May 17 '13 at 16:44
• Shouldn't this be made community wiki? – Najib Idrissi May 17 '13 at 16:59

The set of normal numbers in base $b$ has full Lebesgue measure, for each integer $b\geqslant2$. Hence the set of numbers simultaneously normal in every base has full Lebesgue measure. Examples of numbers normal in a given base $b$ are easy to exhibit, à la Champernowne or à la Copeland-Erdős.

But no number is known to be normal in two given different bases simultaneously (for example in base $b=2$ and in base $b=10$).

• That is, as far as I know. – Did May 17 '13 at 16:56
• But no number is known to be normal in two given different bases simultaneously Maybe that depends on what you mean is a specific well-defined number. According to the paper sciencedirect.com/science/article/pii/S0304397501001700, Sierpiński gave an absolutely normal number in 1917, and the article proves that number is computable. – Jeppe Stig Nielsen May 17 '13 at 23:06
• @JeppeStigNielsen Thanks for the reference. I guess the degree of "effectiveness" of Sierpinski's construction depends on the meaning one gives to the word. As regards Becher and Figueira's paper, note that their section 3.1. "Determination of the first digit" does NOT determine the first bit $b_1$ of the desired number, oddly, since it ends with the assertion that $b_1=0$ if [some condition] is fulfilled and $b_1=1$ otherwise. – Did May 18 '13 at 7:52

One of my favorite examples is more number theoretic: Artin's conjecture

A weak form of it, paraphrased from Wikipidia, is the following. Consider the set $\mathbb{Z}^\square$, consisting of all integers except squares and $-1$.

Pick any integer $a\in\mathbb{Z}^\square$ and let $S(a)$ denote the collection of all primes for which $a$ generates the mulitplicative group of $\mathbb{Z}/p$ (with $0$ removed). Then $S(a)$ is infinite.

Heath-Brown showed that Artin's conjecture is true except for possibly at most 2 counterexamples. But his proof is not constructive, and there is no known examples of an $a$ for which $S(a)$ is infinite.

In other words, Artin's conjecture holds almost everywhere on $\mathbb{Z}^\square$ (and possibly everywhere), but we don't know of a single point in $\mathbb{Z}^\square$ for which it holds.

Almost all real numbers are undefineable. But by definition we can't exhibit an example.

(Somewhat) more precisely: Fix a symbol set $\Sigma$. The set of finite length sequences of symbols from $\Sigma$ is countable, hence the set of finite length sequences which define a unique real number is countable. Hence there are only countably many such real numbers, so the set of defineable real numbers has measure $0$.

• I can "exhibit" an example by diagonalizing over all definable real numbers. That isn't so bad. – Qiaochu Yuan May 17 '13 at 18:54
• @QiaochuYuan I'm not sure that's "exhibiting". – Alex Becker May 17 '13 at 18:56
• Why not? If your symbol set $\Sigma$ is finite (or more generally recursively enumerable?) then I can write down a Turing machine that goes through all descriptions of real numbers, so this is a well-defined mathematical construction, you just don't get a description in the same language as you started with. – Qiaochu Yuan May 17 '13 at 18:57
• @QiaochuYuan True, but even if we allow arbitrary (finite) iteration of such a construction, only countably many numbers are so described. In general we could let $\Sigma$ be a countable set including the symbols needed to describe such turing machines to arbitrary levels of recursion. – Alex Becker May 17 '13 at 19:31

Almost all continuous functions are nowhere differentiable for the classical Wiener measure (see Wikipedia for example), but it's not easy to construct a counterexample. The set of nowhere differentiable functions is also comeager (for a more topology-oriented point of view).

• This may be not easy but explicit examples exist. – Did May 17 '13 at 17:23
• @Did: Yes, and the Wikipedia article I linked is about the Weierstrass function... But the OP asked "for which it is either very difficult to construct an example". – Najib Idrissi May 17 '13 at 18:22

The Wikipedia article on transcendental numbers lists $15$ different classes. An example of a 'class' of transcendental numbers is all numbers of the form $\ln(a)$, where $a$ is an algebraic number different from $0$ and $1$. The moral is that transcendental numbers are very hard to write down.

Then again, pick a random real number, and it will be transcendental.

• Well, depends on what you mean by "write down": you can diagonalize over all algebraic numbers. – Qiaochu Yuan May 17 '13 at 18:54