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.

Solution 1:

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 $b=10$).

Solution 2:

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.

Solution 3:

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 $0$.

Solution 4:

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).

Solution 5:

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.