Examples of figures whose area depend on axiom of choice

Question

Many times I have heard that there are some 'figures' whose area is not fixed and they depend on axiom of choice.

For example,

1. In this answer.
2. In this lecture about probabilty.

I did not know what is axiom of choice, but after reading the wikipedia article and this question: Axiom of Choice Examples, I have understood a little bit about it but still I don't understand it completely.

Can anyone please provide some simple examples of these figures and why Axiom of choice is needed for them?

Answer 1

It is not so much the area of certain 'figures' that depends on AC, it is the provability of existence/nonexistence of their area, once the manner of measuring areas is defined in some way, that may be affected by the presence/absence of AC.

However, even when AC is present, there are certain 'figures' that may be given any area within certain bounds. Such figures exist only courtesy of AC: you will never see a constructive description, or be able to imagine, any such figure. Let me clarify this.

The 'usual' area of subsets of the real plane is the Lebesgue measure. This measure is intuitively 'the right one' because it gives 'correct', 'expected' areas of 'tame' figures that are bounded by finitely many straight line segments, circle arcs, or segments of some other kinds of smooth curves. Lebesgue measure is not defined for every subset of the plane, but only for subsets that belong to the so-called $\sigma$-algebra $\mathcal{M}$ of Lebesgue measurable sets. This $\mathcal{M}$ is (once more intuitively speaking) the largest $\sigma$-algebra of subsets of the plane to which the measuring of areas of 'tame' figures can be extended in a (unique) natural way.

Now consider the $\sigma$-algebra $\mathcal{U}$ of the Lebesgue measurable subsets of the unit square $U=[0,1]\!\times\![0,1]$. There are subsets of $U$ that are not Lebesgue measurable. Let $S$ be one such subset; then $0\leq\alpha:=\mu_*(S)<\mu^*(S)=:\beta\leq 1$, where $\mu_*(S)$ is the inner and $\mu^*(S)$ the outer Lebesgue measure of $S$. Now we can choose any real number $\gamma\in[\alpha,\beta]$, and extend the Lebesgue measure $\mu$ defined on the $\sigma$-algebra $\mathcal{U}$ to a measure $\mu'$ defined on the $\sigma$-algebra $\mathcal{U}_S$ generated by $\mathcal{U}\cup\{S\}$, so that we will have $\mu'(S)=\gamma$. That is, for the single subset $S$ of $U$ not belonging to $\mathcal{U}$ we are allowed to assign to $S$ any area within the bounds $\alpha$ and $\beta$, and every such choice will extend the measuring of areas from the $\sigma$-algebra $\mathcal{U}$ to the larger $\sigma$-algebra $\mathcal{U}_S$. For details of this construction of the extended measure see Theorem 1.12.14 in V.I. Bogachev, "Measure Theory, Volume I", Springer-Verlag, 2007.

Answer 2

Figure whose area depends on the axiom of choice?

Let $r=\begin{cases}1&\sf AC\\2&\sf \lnot AC\end{cases}$ then take $\{(x,y)\in\Bbb R^2\mid \max\{|x|,|y|\}\leq r\}$. Assuming the axiom of choice this is a square whose sides have length $2$, assuming the axiom of choice fails its sides have length $4$.

This seems like a strange definition of a shape, but it's a definition nonetheless.

What might be bothering you are not sets whose "area" depend on the axiom of choice. But rather sets which cannot be assigned an area to begin with (assuming that area means Lebesgue measurable, of course). Examples for such sets include:

1. The Vitali set, which is a set of representatives for $\Bbb{R/Q}$. The axiom of choice is needed here for choosing representatives.
2. Free ultrafilters on $\Bbb N$ (which can be translated nicely into subsets of $\Bbb R$). Here we use the axiom of choice to prove the existence of free ultrafilters, usually done using Zorn's lemma.
3. The graph of a discontinuous $f\colon\Bbb{R\to R}$ which satisfies $f(x+y)=f(x)+f(y)$. Here we usually use Zorn's lemma to prove the existence of a Hamel basis, and use it to construct such function as a $\Bbb Q$-linear endomorphism of $\Bbb R$.

Somewhat surprisingly, Hamel basis of $\Bbb R$ over $\Bbb Q$ can be measurable (but not Borel), but it has to have measure zero if it is measurable.

There are plenty of other examples of sets which are not Lebesgue measurable, and one can find several nice examples in Herrlich's book The Axiom of Choice.

Finally, it might be worth noting that none of these sets is really "a figure" in the intuitive sense. If a figure is any subset of the plane (or space, or whatever) then sure. But we usually expect "a figure" to have a reasonably defined boundary, something these sets do not have.