Sierpinski's example under the continuum hypothesis

Assume the continuum hypothesis. Then there is a set $E \subseteq [0,1]^2$ such that

• for every $x \in [0,1]$ the vertical section $E_x = \{ y : (x,y) \in E \}$ has countable complement in $[0,1]$, and
• for every $y \in [0,1]$ the horizontal section $E^y = \{ x : (x,y) \in E \}$ is countable.

Consequently every section is Lebesgue measurable, $\lambda(E_x) = 1$ for every $x$ and $\lambda(E^y) = 0$ for every $y$, so both iterated integrals of $\mathbf{1}_E$ exist and

$\int_0^1 \! \Big( \int_0^1 \mathbf{1}_E(x,y) \, dy \Big) dx = 1, \qquad \int_0^1 \! \Big( \int_0^1 \mathbf{1}_E(x,y) \, dx \Big) dy = 0 .$

By Fubini's theorem $E$ is therefore not measurable in $[0,1]^2$. Sierpinski's construction uses a well ordering of $[0,1]$ in order type $\omega_1$, which is where the continuum hypothesis is spent, and sets $E = \{ (x,y) : y \not\preceq x \}$ for that well ordering $\preceq$.

Not proved in this library. It is recorded with a citation to Sierpinski's 1920 paper and used in no proof here.

What would prove it. A well ordering of $[0,1]$ of order type $\omega_1$, which under the continuum hypothesis (The continuum hypothesis, and what this page does not prove ↗) exists by the well ordering theorem, plus the observation that each initial segment of such a well ordering is countable. The measure theory needed is then only that countable sets are null and that Fubini's theorem (Fubini-Tonelli theorem and the $\sigma$-finiteness hypothesis ‡) forbids the displayed pair of values for a measurable set.

Which page it serves. The Fubini and change of variables page, beside Failure of Tonelli without $\sigma$-finiteness: the diagonal under Lebesgue times counting measure ‡. The two together answer the question "which hypothesis of Fubini's theorem is doing the work": the diagonal example kills $\sigma$-finiteness, this one kills the idea that measurability of every section is enough.

What it costs, exactly. More than ZFC: the statement is conditional on the continuum hypothesis, which is independent of ZFC, so this is not a ZFC counterexample but a consistency result. The dependence is an equivalence and not merely a sufficient condition. A set $E$ with the two section properties above splits the square into $E$, whose horizontal sections are countable, and its complement, whose vertical sections are countable; and by Sierpinski's decomposition theorem such a splitting exists, for the square or equally for the plane, if and only if the continuum hypothesis holds. So under the negation of the continuum hypothesis this construction is unavailable outright.

What is not thereby restored. The conclusion, unlike the construction, does not go away. Martin's axiom also implies that there is a function on the unit square whose two iterated integrals are defined and unequal, so Martin's axiom together with the negation of the continuum hypothesis still produces one, by a different route. What is consistent with ZFC, by a theorem of Friedman (1980), is the opposite: there are models of ZFC in which no such function exists at all, so that whenever both iterated integrals of a function on the unit square exist they agree. The strong Fubini statement for non-measurable functions is therefore independent of ZFC, and this item is one half of that independence. The ordinals and transfinite recursion machinery the construction uses is itself only ordinary ZFC, and this library plans that page; what is deferred is the independence apparatus and the measure theory, not the well ordering.