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Robin Chapman introduced me to Conway's Base 13 Function. Now, my real analysis is a tiny bit rusty, so maybe my question has a really simple and quick answer, but here it goes:

Consider the support set of the base-13 function, is the set Lebesgue measurable? And if so, does it have non-zero measure?

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Willie Wong
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    (a bit sheepishly) uh, yes guys. Thanks. Got it now. At least five people have told me that the set is Borel, and at least 3 of them with proofs (which all boils down to essentially the same argument). So while I do appreciate all the attention, you don't have to keep on reminding me that I learned measure theory "funny". =) – Willie Wong Jul 20 '10 at 21:55
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    I know that this is an old question, but do you maybe know if it is true that all functions satisfying the intermediate value theorem measurable? – Marko Karbevski Dec 06 '15 at 23:43

4 Answers4

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Call the support set $S$. The answer is yes it is Lebesgue measurable and no, it has zero measure. It is even Borel measurable, which would take a tiny bit more effort to prove.

Note that $S$ is included in the set $T$ of numbers in which the "digits" '+','-', and '.' appear finitely many times in the "base 13 expansion". But almost all numbers are normal, hence have all digits appearing infinitely often. So the Borel set $T$ has measure zero, hence $S$ is Lebesgue measurable with measure zero.

Noah Stein
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  • Ah. I got as far as the lack of normality of the tredecimal expansion. But I forgot that real numbers are almost surely normal. (Now off to tracking down Borel's original proof of this fact, since I've never actually seen it.) – Willie Wong Jul 20 '10 at 15:00
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    Well we don't even really need the full version of normality, just that every digit occurs infinitely often. To see this it is enough to note if we select a number in [0,1) at random according to Lebesgue measure, then its digits in base n are i.i.d. uniform over ${0,\ldots, n-1}$. Therefore the probability that any given digit doesn't occur is zero. There are countably many digits and bases, so a.s. in any base all digits will occur. This means they will all occur infinitely often, because if say $5$ only occurred $k$ times then $5\cdots 5$ ($k+1$ times) would never occur in – Noah Stein Jul 20 '10 at 15:40
  • the base $10^{k+1}$ representation. – Noah Stein Jul 20 '10 at 15:41
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    The probability that the base-13 expansion of a uniformly distributed random number in (0,1) has contains only digits 0-9 starting at the N-th digit is $\prod_{n=N}^\infty(10/13)=0$. This is also a Borel set (intersection of a countable sequence of Borel sets defined by the $n$-th digit). A countable union of measure zero Borel sets is a measure zero Borel set. – Robin Chapman Jul 20 '10 at 17:59
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The function is easily seen to be Borel, since the graph of the function can be defined using only natural number quantifiers. In particular, a number is in the support if and only if there is a last place in its digits where $+$ or $-$ appears, followed eventually by ., but then after this, +, - and . do not appear again. That is,

  • $x\in S\iff\exists n_1\exists n_2\gt n_1\forall m\geq n_1$ the $m^{\rm th}$ digit of $x$ in base 13 is neither $+$ nor $-$ nor ., except at $n_1$, where it is either $+$ or $-$ and and $n_2$, where it is .

Any set of reals that is definable using only natural number quantification is Borel, since existential quantification over the naturals corresponds to a countable union and universal quantification corresponds to countable intersection. Such sets lie in the arithmetic hierarchy, which is a very low part of the hyperarithmetic hiearchy, which leads ultimately to the Borel sets.

The same idea shows that the graph of the function, as a set of pairs, is Borel.

Joel David Hamkins
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  • Joel: thanks for the great answer. I'm not that great with logic and set theory, but am I correct in interpreting that in this specific instance, using the fact that semi-closed intervals are Borel, and that $S$ can be defined by a countable union (over $n_1,n_2$) of countable intersections (over the tredecimal levels) of countable union (for each level there are finitely many allowed semi-closed intervals), we have that $S$ is automatically Borel. And that this notion has a proper generalization in the theorem that you stated? – Willie Wong Jul 20 '10 at 20:51
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    Yes, that's right. It is easier generally from this perspective to work with Cantor space, and think about reals explicitly as sequences. The basic finitary features of the sequences correspond to open sets, and quantification over naturals gives countable unions and intersections. Of course, this doesn't exhaust the Borel sets, but only the arithmetic sets. But this way of thinking provides a quick sufficient criterion for being Borel. – Joel David Hamkins Jul 20 '10 at 20:58
  • You also need to know that a function being Borel in the pre-image sense is the same as its graph being a Borel subset of the plane. – Joel David Hamkins Jul 20 '10 at 20:59
  • Joel: thanks again for the clarification. It is very illuminating. I am now certainly glad to have asked the question. – Willie Wong Jul 20 '10 at 21:11
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It seems that the definition of the function doesn't use the axiom of choice. This implies that the support set should be Lebesgue measurable.

Andrey Rekalo
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    I agree with that as heuristics, but is that actually a theorem? :) – Willie Wong Jul 20 '10 at 15:01
  • "In 1970, Solovay demonstrated that the existence of a non-measurable set for Lebesgue measure is not provable within the framework of Zermelo–Fraenkel set theory in the absence of the Axiom of Choice." http://en.wikipedia.org/wiki/Non-measurable_set – Andrey Rekalo Jul 20 '10 at 15:14
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    Andrey, Solovay's result does not give this implication. It is, however, a theorem that (assuming large cardinals) the fact that the function is explicitly described implies measurability. Depending on the actual large cardinals one assumes, credits may vary. It is safe to call it a theorem of Shelah-Woodin. – Andrés E. Caicedo Jul 20 '10 at 16:28
  • @Andres Caicedo: Many thanks for the comment. Could you give a reference on a penetrable book or expository article concerning the Shelah-Woodin result? – Andrey Rekalo Jul 20 '10 at 16:40
  • Ha! Hmm... Jech's book "Set Theory" discusses these matters and outlines the proof. Kanamori's book "The higher infinite" contains much more background and history, it is really nice and I highly recommend it, but it is technical. The original Shelah-Woodin paper is impossible to read if you are not a set theorist. But the argument is outlined (much more clearly, but less optimally) in "Martin's Maximum, saturated ideals, and non-regular ultrafilters", a groundbreaking paper by Foreman-Magidor-Shelah, Annals of Mathematics, 127 (1), 1-47. A very nice (technical) presentation of a better ... – Andrés E. Caicedo Jul 20 '10 at 16:50
  • (2) ... result is in Itay Neeman's "Determinacy in L(R)", a paper for the forthcoming Handbook of Set Theory (available from Itay's page). The proof requires a working understanding of forcing and large cardinal techniques, so I am afraid I cannot think of an expository paper that non-set theorists can read. But the references above should give you a good idea of what is involved even if you do not delve into the technical details. Solovay's result shows that 'automatic measurability of definable functions' is consistent, Shelah-Woodin show that it is actually true. – Andrés E. Caicedo Jul 20 '10 at 16:55
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    This previous question http://mathoverflow.net/questions/23834/drawing-conclusions-by-not-using-ac seems also relevant. – damiano Jul 20 '10 at 16:56
  • @Andres Caicedo: Awesome. Thanks again for the references and comments. This is very helpful. – Andrey Rekalo Jul 20 '10 at 17:07
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    In fact this function is Borel measurable, so no need to worry about AC and inaccessible cardinals and such. – Gerald Edgar Jul 20 '10 at 19:40
  • The questioner presumably wanted to know if there is an unconditional proof that the function is Lebesgue measurable, and this is not answered by the famous set-theoretic results. Solovay shows it's consistent that it be measurable, and Shelah-Woodin show that either it's measurable or specific large cardinal axioms are inconsistent with ZFC. This came up in the earlier question that damiano linked. Whether set theory helps in showing that given specific functions are provably Lebesgue measurable (e.g. in ZF or ZFC) is an interesting, and possibly open, question. – T.. Jul 20 '10 at 20:35
  • Awesome. I certainly learned a lot reading these answers. Thanks for all the comments! – Willie Wong Jul 20 '10 at 20:39
  • @Gerald: Sure. Very rarely I've seen explicit subsets fo ${\mathbb R}$ in an argument in analysis that are not Borel. And then, they are (typically) the continuous image of a Borel set, and therefore measurable anyway. I do not know of a single example where one needs seriously worry about the possibility of non-measurability.

    @T.: Yes, I suppose one could say that the Shelah-Woodin result only shows that either the cardinals are inconsistent or blah. If the inconsistency is the case, then there would indeed be room for exotic pathologies.

     – Andrés E. Caicedo Jul 21 '10 at 05:57
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    @Andres: why the sarcasm? When a result in number theory is shown to follow from an extremely convincing "axiom", such as the Riemann hypothesis or (cf. the $\pi(x+y)$ subadditivity conjecture) the prime k-tuplets conjecture, this is clearly distinguished from a proof in the intended/standard/tacit system, such as number theory using ordinary axioms, or ZF, or PA. And that distinction is as significant as the difference between what Solovay and Shelah+Woodin proved. – T.. Jul 23 '10 at 07:32
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The function is certainly Borel measurable, and hence so it its support set. The function $f_k(x)$ which gives the $k$th tredigit of $x$ is clearly a Borel function, and it should be a straightforward if tedious exercise to write the base-13 function in terms of the $f_k$, using only common Borel functions (arithmetic, max/min, indicators) and limits.

Of course the fact that no "nonconstructive" arguments are used in its definition is strong evidence that such a procedure should be possible, and if you know enough logic, as others have mentioned, this can even be a proof. I just wanted to point out that one can prove the measurability from first principles with a little elbow grease.

Nate Eldredge
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  • re: "this can even be a proof" --- the results from logic don't quite show that a reasonably defined $f$ has a proof (in the system that defined it) of its Lebesgue measurability. They do show consistency of $f$ being measurable, and that there are natural axioms that, when added to the system, provide a universal proof (not referring to the construction details of any specific $f$) that any reasonable function is measurable. – T.. Jul 20 '10 at 22:23