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Let $\Omega$ and $\Gamma$ be two nonempty sets and $\mathscr{A}$ and $\mathscr{B}$ be $\sigma$-algebras over $\Omega$ and $\Gamma$, respectively. The product $\sigma$-algebra of $\mathscr{A}$ and $\mathscr{B}$ is defined to be the smallest $\sigma$-algebra over $\Omega\times\Gamma$ that contains the sets of the forma $A\times B$, for $A\in\mathscr{A}$ and $B\in\mathscr{B}$, and it is denoted by $\mathscr{A}\times\mathscr{B}$. When referring to measurable subsets of $\Omega$, $\Gamma$ or $\Omega\times\Gamma$, I will mean elements of the respective $\sigma$-algebras$.

Given a set $L\subseteq\Omega\times\Gamma$ and elements $\omega\in\Omega$ and $\gamma\in\Gamma$, the sections of $L$ by $\omega$ and $\gamma$ are, respectively, $$L_\omega=\left\{y\in\Gamma:(\omega,y)\in L\right\}\qquad\text{and}\qquad L^\gamma=\left\{w\in\Omega:(w,\gamma)\in L\right\}.$$

The following fact is very easily proved:

Fact: If $L\subseteq\Omega\times\Gamma$ is measurable, then all its sections are measurable.

Now, the converse is not true: If $\Omega$ is a set with cardinality $>2^{\aleph_0}$ equipped with the sigma algebra $\mathscr{P}(\Omega)$, the power set of $\Omega$, then the diagonal $\Delta_\Omega=\left\{(\omega,\omega):\omega\in\Omega\right\}$ is not in the product $\sigma$-algebra $\mathscr{P}(\Omega)\times\mathscr{P}(\Omega)$, but all its sections are (obviously) measurable. For a proof, see Does $\mathcal P ( \mathbb R ) \otimes \mathcal P ( \mathbb R ) = \mathcal P ( \mathbb R \times \mathbb R )$? and the reference given in the question.

What I want is a simpler example of it. Summarizing:

Question: Give an (simple) example of a non-measurable set in the product space s.t. all its sections are measurable.

All I could find out is that the product $\sigma$-algebra cannot be finite, so neither $\mathscr{A}$ nor $\mathscr{B}$ can.

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Luiz Cordeiro
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2 Answers2

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Pick $\Omega$ to be any uncountable set and let $\mathcal{A}$ be the $\sigma$-algebra of subsets which are countable or have countable complement. Then there exists a non-measurable set $E\subset \Omega$, i.e., an uncountable set with uncountable complement. Let $L:=\{(e,e) : e\in E\}$ (i.e., the "$E$-diagonal"). The set $L$ has measurable sections in $\mathcal{A} \otimes \mathcal{A}$, because they are all either empty or points. The diagonal embedding $\omega \mapsto (\omega,\omega)$ of $\Omega$ into $\Omega \times \Omega$ is measurable (since preimages of measurable rectangles under this embedding are easily seen to be measurable), and since the preimage of $L$ under this embedding is the non-measurable set $E$, we see that $L$ itself is not measurable.

Lukas Geyer
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  • Perfect. This kind of non-separable space seems to appear a lot in counter-examples. – Luiz Cordeiro Dec 03 '13 at 00:47
  • Actually, there is a little problem with the argument since the projection of a measurable set is not always measurable. Not sure how to save the proof in the general case, but it can be modified by using the Borel $\sigma$-algebra instead of the abstract one given. – Lukas Geyer Dec 03 '13 at 03:51
  • I think the general argument is perfectly fine. The mapping $\omega\mapsto (\omega,\omega)$ from $\Omega$ into the diagonal of $\Omega\times \Omega$ is certainly measurable, so the inverse image of any measurable subset of the diagonal must be measurable. But this is just the projection. –  Dec 03 '13 at 04:13
  • @ByronSchmuland: Thanks, I got confused by overdosing on counterexamples, your argument works nicely, so I should probably edit it back. – Lukas Geyer Dec 03 '13 at 04:21
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Here is an example on real line: There is a bijection $f: \mathbb{R} \rightarrow \mathbb{R}$ whose graph has full outer measure in plane. Such a function can be constructed by a simple transfinite induction.

Contrast this with the following: If $F: \mathbb{R}^2 \rightarrow \mathbb{R}$ is such that for each fixed $a$, the functions $y \mapsto F(a, y)$ and $x \mapsto F(x, a)$ are continuous, then $F$ is Borel. By previous example, one cannot replace "continuous" by "Borel" here.

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