Let me quote en.wikipedia about the original Nikodym's example (the definitions above I've written myself just for MO):

a Nikodym set is a subset of the unit square in $\ \mathbb R ^2\ $ with the complement of Lebesgue measure zero, such that, given any point in the set, there is a straight line that only intersects the set at that point. The existence of a Nikodym set was first proved by Otto Nikodym in 1927.

End of quote. See https://en.wikipedia.org/wiki/Nikodym_set.

In the above example, we can talk about the open square (or else, we may remove the border points of the square from the set and the things would still work). The $\ L\ $ would be the intersections of the straight lines with the open square. This would define a little bit nicer Nikodym set.

I am surprised that the original Nikodym set was provided in $\ (0;1)^2\ $ instead of $\ \mathbb R^2,\ $hence

Question 1: Let $\ L\ $ be the set of all straight lines in $\ \mathbb R^2\ $ Does there exist a respective Nikodym set, i.e. $\ A\subseteq \mathbb R^2\ $ such that the complement of $\ A\ $ has measure $0\,$ and $$ \forall_{x\in A}\exists_{\ell\in L}\quad \ell\cap A=\{x\} $$ ?

Question 2: Does there exist $\ A\ $ as in  Question 1  and such that $\ A = \{2\cdot x: x\in A\}\ $ ?

I'll stop now while additional natural questions come to one's mind; some of them after reading the whole quoted en.wikipedia article.


1 Answer 1


The answer to Question 1 is yes. This is a result of Falconer, Corollary 6.6 in

K. J. Falconer, Sets with prescribed projections and Nikodým sets. Proc. London Math. Soc. (3) 53 (1986), no. 1, 48–64.

The result states as follows:

Theorem. Let $1\leq m<n$. Then there exists a set $K\subset\mathbb{R}^n$ with $\mathcal{L}^n(K)=0$, such that for each $x\in\mathbb{R}^n\setminus K$, there is a $m$-plane $P$ such that $P\cap (\mathbb{R}^n\setminus K)=\{ x\}$.

This answers positively Question 1 with $A=\mathbb{R}^n\setminus K$.

  • $\begingroup$ Thank you, Piotr, for an extra general answer. My Q2 was initially a bit different -- I asked about a Nykodym set in $\ A\subseteq(-1;1)^2\ $ such that $\ (-1;1)^2\cap B = A,\ $ where $\ B:=\{2\cdot x:x\in A\}.\ $ Answer Yes to this modified Q2 could provide Yes to Q1. $\endgroup$
    – Wlod AA
    Commented Jan 31, 2020 at 22:40
  • $\begingroup$ * Nikodym (sorry for my typo). $\endgroup$
    – Wlod AA
    Commented Jan 31, 2020 at 22:52
  • $\begingroup$ (It seems, Piotr, that you have 2 typos -- the last two $\ \mathbb R^n\setminus K\ $ should be simply $\ K,\ $ I guess). $\endgroup$
    – Wlod AA
    Commented Jan 31, 2020 at 22:57
  • $\begingroup$ @WlodAA No typo. I followed Falconer's notation. The set $\mathbb{R}^n\setminus K$ has full measure so in your notation $A=K\setminus\mathbb{R}^n$. $\endgroup$ Commented Jan 31, 2020 at 23:00
  • 1
    $\begingroup$ @WlodAA You are right. There was a typo in Falconer's paper. Dziękuję! $\endgroup$ Commented Feb 1, 2020 at 0:27

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