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Following the nice answer to Do the Lebesgue-null sets cover "all the sets can naturally be regarded as sort-of-null sets"?, the particular situation that I am especially interested in (which is a kind of "advanced version" of https://math.stackexchange.com/questions/35606/lebesgue-measure-of-the-graph-of-a-function) is:

Let $F$ be the set of bijective Borel-measurable functions $f \colon [0,1] \times [0,1] \to [0,1] \times [0,1]$ that preserve the Lebesgue measure.

  1. Is it the case that for any sequence $(K_n)_{n \in \mathbb{N}}$ of functions $K_n \colon [0,1] \to [0,1]$ and any sequence $(f_n)_{n \in \mathbb{N}}$ in $F$, the set $$ B \ := \ \bigcup_{n \in \mathbb{N}} f_n(\mathrm{graph}\,K_n) $$ does not contain a set of positive Lebesgue measure?

  2. If the answer to the above is yes: what if we allow the functions $K_n$ to be set-valued functions, with $K_n(x)$ being a Lebesgue-null subset of $[0,1]$ for all $x \in [0,1]$?

[Intuitive motivation: Under suitable conditions, the "disintegration theorem" allows one to regard a measure on a product space $X \times Y$ as an $X$-based random measure on $Y$. From this point of view, one might wish to regard a set whose $Y$-sections are null (under the disintegrated measure) as being itself "effectively null" under the original measure; but if the answers to the above questions are no, then we cannot really do this.]

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    $\begingroup$ See math.stackexchange.com/questions/836856/… $\endgroup$ – Ashutosh Jun 21 '17 at 13:02
  • $\begingroup$ (Okay, I now see that the answer given there assumes CH, but that's okay.) $\endgroup$ – Julian Newman Jun 21 '17 at 20:04
  • $\begingroup$ Davies' proof does not use CH. The reference is Davies, Roy O., Covering the plane with denumerably many curves, J. London Math. Soc. 38 1963, 433–438 $\endgroup$ – Ashutosh Jun 21 '17 at 21:17

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