1

I have no clue about this question, so here are some questions that may have answers or counter-examples. I hope this leads to more relevant results. If you know how to deal with any of them, or know of any related content, welcome to reply.

The following measures are all Lebesgue measures.

Q1: Consider the set $E$ on the square $\left[0,1\right]^2$, satisfied that for any $x_0\in\left[0,1\right]$, the intersection of straight line $\{x=x_0\}$ with $E$ (something we called truncated set $E(x)$) is zero-measured in one dimension, that is $m(\{x=x_0\} \cap E)=0$. Is $E$ a zero-measured set in two dimension?

Q2: Consider the set $E$ on the square $\left[0,1\right]^2$, satisfied that for any $x_0,y_0\in\left[0,1\right]$, the intersection of straight line $\{x=x_0\},\{y=y_0\}$ with $E$ is zero-measured in one dimension, that is $m(\{x=x_0\} \cap E)=0$ and $m(\{y=y_0\} \cap E)=0$. Is $E$ a zero-measured set in two dimension?

Having to do with the Fubini & Tonelli theorem, the problem arises from finding a measurable function for which both kinds of repeated integrals exist and are equal, but for which multiple integrals do not exist. I am trying to construct an unmeasurable set, but any of its truncated sets are zero-measured, so that its representation function $\chi_E$ is satisfactory. But I ran into difficulties.

daidaitx
  • 307
  • 1
    I think this is a counterexample for both. It's quite a strong counterexample - $f$ being bijective means every horizontal section has cardinality exactly $1$ and every vertical section has cardinality exactly $1$. Of course here $E$ is not Lebesgue-measurable - and if you assume that $E$ is measurable, then it does have measure $0$ by Fubini's theorem. – Izaak van Dongen Nov 22 '23 at 13:06
  • @IzaakvanDongen That's a great example! But I may have limited knowledge and not be able to understand much about how it is constructed. But your comments are useful, thanks! I'll wait and see if someone can come up with some simpler counter-examples. – daidaitx Nov 22 '23 at 15:35
  • Another funny example for Q1 is to assume CH and let $E$ be a well-ordering of $[0, 1]$ of minimal order type. (then because every element has countably many predecessors, the horizontal sections are all countable but the vertical ones are cocountable). See here. I don't know anything less set-theoretic than this I'm afraid. – Izaak van Dongen Nov 23 '23 at 00:01

0 Answers0