1
$\begingroup$

Let $f\colon \mathbb R\to (0,\infty)$ be a function taking positive values. Does there exist a Borel measurable function $g\colon \mathbb R\to (0,\infty)$ taking positive values as well such that $g(x)\leq f(x)$ for all $x\in\mathbb R$?

$\endgroup$

2 Answers 2

4
$\begingroup$

The answer is No.

Let us assume that the claim you are asking about is true and let us try to arrive at a contradiction.

Cardinality of the set of all Borel measurable functions is $\mathfrak c=2^{\aleph^0}$; see here.

Let us well-order all positive Borel functions as $\{g_\alpha; \alpha<\mathfrak c\}$. Let us consider some well-ordering $\mathbb R=\{x_\alpha; \alpha<\mathfrak c\}$ of real numbers. Let us define a new function $f\colon\mathbb R\to(0,\infty)$ as $$f(x_\alpha)=\frac{g_\alpha(x_\alpha)}2.$$ Clearly, there is no $\alpha$ such that $g_\alpha\le f$. This contradicts the given claim.

$\endgroup$
3
  • $\begingroup$ Still it might be interesting to see whether somebody comes up with a more direct/explicit approach. In general, I think that it is advised a bit before accepting an answer. Just to see whether you get some better answers or whether somebody points out a mistake in the answers which were posted so far. $\endgroup$ Jul 7, 2016 at 13:02
  • $\begingroup$ And since I am already giving unsolicited advice, I will also mention that registering would probably make easier for you to follow your questions, comment on them, etc. (It is easier to lose access to an unregistered account than to a registered account. If you do not have access to your account, you can't comment on your question or edit your question.) $\endgroup$ Jul 7, 2016 at 13:04
  • 1
    $\begingroup$ It's hard to imagine a very explicit approach, since it's consistent with ZF (without Choice) that there are no non-Borel functions at all. $\endgroup$ Jul 7, 2016 at 22:38
3
$\begingroup$

For this construction we need a countable disjoint family of sets $A_n \subset [0,1]$ of (Lebesgue) outer measure $1$. (See note below.)

Define function $f$ so that $f(x) = 1/n$ for $x \in A_n$. We claim there is no Lebesgue measurable $g$ with $0 < g \le f$. Indeed, $g(x) \le 1/n$ on set $A_n$, which has outer measure $1$, and $g$ is Lebesgue measurable, so $g(x) \le 1/n$ a.e. on $[0,1]$. This is true for all $n$, so $g(x) \le 0$ a.e. on $[0,1]$.

note
How to construct sets $A_n$? Follow the usual transfinite recursion construction for the Bernstein set. (For example, in my answer here.) But instead of choosing two points at each stage, choose countably many.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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