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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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    I'm not sure I understand what "almost" means, mathematically.2011-09-17
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    @Peter, "P is true almost everywhere" is the same as saying "The set of points where P is false is a measure-zero set".2011-09-17

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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    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 $C$.]2011-09-17
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    That was basically going to be my suggestion,Mike-I was going to suggest taking the intersection of all subsets of R where m ({ x: f(x) = 0}. This subset is clearly in the complement of rationals in R.You need to identify all these subsets and see if this intersection contains any open balls in the usual topology on R-if so, clearly it's not of measure 0 and f(x) is not positive a.e. Great counterexample.2011-09-17
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    I did it for you,Sriv.You owe me an upvote......LOL Any analysts want to chime in on this to make sure we're ok here?2011-09-17
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    @Srivatsan: But I can. :-)2011-09-17
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    No problem Srivatsan! All the best.2011-09-17
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    @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.2011-12-04
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    @Jonas The measure of any open subset of R must be positive. Therefore, any subset of R where the measure is not 0 must contain some open subset. Recall the definition of a condition being true almost everywhere on R-it means that the set where it's false must have measure 0. Therefore, the set where f(x) is NOT greater then 0 must be subset of R which contains no open subsets. In addition, in the example above,it cannot contain any rational numbers.It CAN be countable,but it doesn't have to be(i.e. the Cantor set). (continued)2011-12-04
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    @ Jonas (cont) Therefore, if f is indeed positive almost everywhere in R, the set of all points where f(x) is not positive must be a nonempty subset of the non-rationals where there are no open subsets contained in it.So what I suggested was a possible way to construct this set if it exists. If such a set cannot be constructed, the statement is false.Analysts in here-is my reasoning sound,yes or no?2011-12-04
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    @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.2011-12-04
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    @Jonas My error-I'd forgotten about the quotient space counterexample and others. I hate when I forget counterexamples,it makes me look like an idiot and I'm not. I just should double check in Gelbaum and Olmstead before posting anything in analysis. : (2011-12-04
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    @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.2011-12-04
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    @Adam You mean the set-theoretic complement of the rationals in R. The irrationals are uncountable,though,so I wouldn't expect the irrationals to have measure zero, Adam. (Of course,that's no guaruntee they DON'T,is it?) (Cont.)2011-12-04
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    @Adam (cont.) That raises an interesting question: Which dense subsets of R have positive measure when R has the usual topology? Suppose we take the rationals alone and insert a single irrational between every 2 rationals (which of course is easy to do). This set is clearly dense in the usual topology on R.But it is a union of 2 countable sets since there is a one to one correspondence between each irrational and each pair of rationals.So I'd expect this set to have measure zero. My question: What conditions have to be met for a dense subset of R to have positive measure?2011-12-04
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    @Mathemagician1234: No one said the set of irrational numbers has measure zero, and it doesn't. The complement, $\mathbb Q$, does. Adam was saying that your "quotient space" comment is irrelevant. I don't know what "measure when R has the usual topology means" exactly; here we're just talking about Lebesgue measure. Yes, countable sets have measure zero. (This is true for any measure space in which there are no atoms.) I don't know what you are asking about dense subsets having positive measure. Some do, some don't, and if you have a specific question, you could ask it as a new question.2011-12-05
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    @Jonas When I talked about quotient spaces,I misunderstood what you meant by "R/Q",which looks like the notation for quotient spaces in either algebra or topology to me. "R - Q" or "comp(Q) in R" I would have understood. Obviously,the irrationals have to have positive measure because otherwise R would have measure 0. (Obviously I mean Lebesgue measure-when you say "measure on the real line" without qualification, I was taught that it's understood you usually mean Lebesgue measure.) I was asking because you gave a dense set with positive measure as an example set with no open subintervals.2011-12-05
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    @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.2011-12-05
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    One way to construct it is to enumerate the rationals and put a small interval around each. Take the union of the intervals (open!) and subtract that from $\mathbb{R}$. Then use the hint.2012-12-20
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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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    This is also a nice solution. Thank you.2011-09-19
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    +1 for this "complementary" solution to the approaches suggested above.2011-12-04