MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is a reference request.

The phrase in the title is, if I remember correctly, how Eli Stein described the following set (the definition may be faulty, but I think it is right):

There exists a set $S$, which is a subset of the unit square $[0,1]^2\subset\mathbb{R}^2$ with full Lebesgue measure, with the property that for any point $p\in S$ there exists a line $\ell \subset \mathbb{R}^2$ such that $\ell \cap S = \{ p\}$.

I vaguely remember the theorem being attributed to one of the Rieszes (not sure M. or F.; I might also be completely wrong about that).

Can someone tell me whether my rough statement of the theorem above is correct, and better yet, can someone give a reference to the paper which proved it? (I know a paper exists; I remember digging it up in the bowels of the library when I was a beginning graduate student. I am sad to say that two moves later I can no longer find my photocopy of that article.)

share|cite|improve this question
up vote 12 down vote accepted

This is a Nikodymn set. I haven't seen a citation to Nikodymn's original paper, but the history is breifly discussed (with references) in Stein's survey article Singular Integrals: The Roles of Calderón and Zygmund.

Edit: Stein's Harmonic Analysis cites this to: "O. Nikodym, Sur les ensembles accessibles, Fund. Math. 1927 10:116-168"

share|cite|improve this answer
Okay, now I feel silly. I actually did pull down Harmonic Analysis from the bookshelf to see if there is any citation to Riesz. It just didn't occur to me to check the table of contents. Thanks for the memory jot. – Willie Wong Apr 23 '11 at 1:43
The actual title is "Sur la mesure des ensambles plans dont tous les points sont rectilinéairement accessibles". – Andrés E. Caicedo Apr 23 '11 at 4:35

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.