I’m traveling this week, so we have a guest post from the inimitable Ashutosh Kumar. Enjoy!

In Measuring II: The limits of measurability, we saw that there is no measure satisfying properties 1-6 that could be defined on all subsets of $\mathbb{R}$. Relaxing one or more of these properties leads to some very interesting problems. Well discuss one such variation today which is obtained by dropping the shift-invariance requirement.

The measure problem: Is it possible to assign to every subset $A$ of the unit interval $[0, 1]$ a value $m(A)$ such that the following requirements hold?
(1)  For every $A \subseteq [0, 1]$, $0 \leq m(A) \leq 1$
(2)  $m([0, 1]) = 1$
(3)  For every $a \in [0, 1], m(\{a\}) = 0$
(4)  For every sequence $\langle A_n : n \geq 1 \rangle$ of pairwise disjoint subsets of $[0, 1]$, $m(\bigcup_{n \geq 1} A_n) = \sum_{n \geq 1} m(A_n)$

Banach and Kuratowski made an important early contribution to this problem in 1928. They showed that if we assume that the Continuum Hypothesis is true, then there is no such measure. We have seen the Continuum Hypothesis before, but we will use it here in a the following form:

Continuum Hypothesis: There is a linear ordering $\prec$ of $[0, 1]$ that satisfies: For every $x \in [0, 1]$, the set $\{y \in [0, 1]: y \prec x\}$ of $\prec$-predecessors of $x$ is countable.

Kuratowski later writes:

As I recall, it was sometime around 1928/1929, during one of my many discussions with Banach in the Scottish Cafe, that we got interested in the so called measure problem stated many years before by Hausdorff and still unresolved at the time. During a many-hour discussion we tried to tried to attack the problem in various ways, unsuccessfully. I got home around midnight. But I couldnt fall asleep until I had found a solution (to my great joy). I met Banach the following morning. “You know Stefan, Ive solved Hausdorff’s problem.” “So have I,” said Banach. What is more, it turns out my reasoning and Banach’s were almost identical: it was essentially a continuation of our discussion in the cafe. Obviously, we decided to publish it in a joint paper (in “Fundamenta”).

A decade later, Gödel showed that the Continuum Hypothesis is not outright false in the sense that if the usual axioms of mathematics are consistent, then they remain so if we additionally assume the Continuum Hypothesis.

Let us prove Banach-Kuratowski’s theorem. Towards a contradiction, let us assume that there is a measure $m$ that satisfies properties (1-4) above. Define $P = \{(x, y) : y \prec x \}$. $P$ is a subset of the unit square $[0, 1] \times [0, 1]$. It has the following property: For every $x \in [0, 1]$, the vertical section $P_x = \{y : (x, y) \in P\}$ is countable. It follows that we can write $P$ as a countable union of subsets $Q$ of the unit square, each of whose vertical sections contains at most one point.

Claim: We can write each $Q$ as the complement of a countable union of sets of the form $A \times B$ where $A, B \subseteq [0, 1]$.

Proof of Claim: For each subinterval $J$ of $[0, 1]$ whose endpoints are rational, let $A_J = \{x \in [0, 1] : Q_x \cap J = \emptyset\}$. Then it is easy to check that $Q = [0, 1] \times [0, 1] \setminus \bigcup_J A_J \times J$ where the union runs over every subinterval $J$ with rational endpoints. The number of such intervals is countable, since it is possible to list all pairs of rationals in a sequence.

It follows that $P$ is a countable union of sets each of which is the complement of a countable union of sets of the form $A \times B$ where $A, B \subseteq [0, 1]$. This allows us to use a basic fact from measure theory (Fubinis theorem) which yields the following.

Consequence of Fubini’s Theorem: If every vertical section of $P$ has $m$-measure zero, then some horizontal section $P^y = \{x \in [0, 1]: (x, y) \in P\}$ also has $m$-measure zero.

Note that every vertical section of $P$ is countable and therefore has  $m$-measure zero (use properties (3) and (4)). So there must be a $y \in [0, 1]$ such that $m(P^y) = 0$. But note that $[0, 1] = \{z \in [0, 1]: z \prec y\} \cup \{y\} \cup P^y$. The set $\{z \in [0, 1]: z \prec y\}$ is countable, so it must have $m$-measure zero. But this means $m([0, 1]) = 0 + 0 + 0 = 0$ which contradicts property (2).

This leaves us with the question: Can we show the non-existence of a measure satisfying properties (1-4) without any extra assumption? The answer lies in the hierarchy of large cardinals and we hope to get back to it soon.

Cover image: Postcard depicting the Scottish Café in Lwów.