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I made up this question, but unable to solve it:

Let $f : \mathbb R \to \mathbb R$ be a continuous function such that $f(x) > 0$ for all $x \in \mathbb Q$. Is it necessary that $f(x) > 0$ almost everywhere?

This is my attempt.

  1. It is easy to show that $f(x) \geq 0$ everywhere, so the real question is whether $f$ can be zero at an "almost all" irrational points.

  2. The function can become $0$ at isolated points, e.g., $f(x) = (x - \sqrt{2})^2$. In particular, the qualification "almost" is necessary for the question to be nontrivial.

  3. Every rational point has an open neighborhood where $f$ is positive. Hence at least know that the set $\{ x \,:\, f(x) > 0 \}$ is not a measure-zero set.

  4. I first mistakenly assumed that Thomae's function provides a counter-example to this. Indeed, it is positive at all rationals and zero at all irrationals, but the function is continuous at only the irrationals, not everywhere.

Then I tried to prove that the question has an affirmative answer, but do not have much progress there. Please suggest some hints!

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  • $\begingroup$ I'm not sure I understand what "almost" means, mathematically. $\endgroup$
    – Peter Olson
    Commented Sep 17, 2011 at 15:44
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    $\begingroup$ @Peter, "P is true almost everywhere" is the same as saying "The set of points where P is false is a measure-zero set". $\endgroup$
    – Srivatsan
    Commented Sep 17, 2011 at 16:18

2 Answers 2

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Here's a hint: If you can think of closed set $C \subset \mathbb{R}$ of positive measure containing no rationals, then the function sending $x$ to its distance from $C$ is an example.

(I changed this because I just realized you were looking for a hint)

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    $\begingroup$ Very nice (and also natural) example. Thanks! Unfortunately, I already ran out of my votes; I cannot upvote this today. :-) [Added: Yes, I saw the full solution inadvertently, but I guess your hint would have worked for me as well. I know how to construct the requisite $C$.] $\endgroup$
    – Srivatsan
    Commented Sep 17, 2011 at 4:24
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    $\begingroup$ @Mathemagician1234: I do not understand what "intersection of all subsets of R where m ({ x: f(x) = 0}" means. It is incorrect that you need to find open balls. Every open interval in $\mathbb R$ contains rational numbers, so every open interval will contain points where $f$ is positive. $\endgroup$
    – Jonas Meyer
    Commented Dec 4, 2011 at 5:29
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    $\begingroup$ @Mathemagician1234: That is incorrect, and is confusing a correct statement with its incorrect converse. If a set is open, or contains an open subset, then it has positive measure. The converse is false. For example, there are dense sets with measure 0. Observe that $\mathbb R\setminus \mathbb Q$ contains no open set, but has positive measure. Also, please take a look at fat Cantor sets, for examples of sets of positive measure that do not differ from an open set by merely a null set. Here, however, the point is that there are closed subsets of irrationals with positive measure. $\endgroup$
    – Jonas Meyer
    Commented Dec 4, 2011 at 18:15
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    $\begingroup$ @Mathemagician1234 : $\mathbb{R} \setminus \mathbb{Q}$ is the set-theoretic difference, not the quotient space. In other words, $\mathbb{R} \setminus \mathbb{Q}$ is the set of irrational numbers, not the result of squashing all the rational numbers in $\mathbb{R}$ to a single point. $\endgroup$
    – Adam Smith
    Commented Dec 4, 2011 at 21:44
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    $\begingroup$ @Math: Here is some standard notation: $\mathbb R/\mathbb Q$ is a quotient, while $\mathbb R\setminus \mathbb Q=\{x\in\mathbb R:x\not\in\mathbb Q\}$, the latter being the same thing you denoted "$\mathbb R-\mathbb Q$" or "$\mathrm{comp}(\mathbb Q)$ in $\mathbb R$". I used a backslash, meaning the set difference, rather than a forward slash, which would mean a quotient. $\endgroup$
    – Jonas Meyer
    Commented Dec 5, 2011 at 2:58
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Here's another approach to the question, in hint form:

(1) Construct a function $f(x)$ that is continuous, nonnegative, positive at $x=0$, bounded above by 1, and supported on an interval of length at most 1.

(2) Deduce that if $a,b>0$ and $c$ are real numbers, then $a f(b(x-c))$ is continuous, nonnegative, positive at $x=c$, bounded above by $a$, and supported on an interval of length at most $1/b$.

(3) Let $q_1,q_2,q_3,\dots$ be an enumeration of the rationals. Show that the function $$\sum_{j=1}^\infty 2^{-j} f\big( 2^j(x-q_j) \big)$$ is continuous, nonnegative, positive at every rational number, but equal to zero on a set of infinite measure.

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    $\begingroup$ +1 for this "complementary" solution to the approaches suggested above. $\endgroup$
    – Mathemagician1234
    Commented Dec 4, 2011 at 17:19

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