1
$\begingroup$

Let $f_s: \mathbb R \to \mathbb R$ a be family of Borel measurable functions parameterized by $s\in \mathbb R$. Consider the limit function $$ F(t)=\limsup_{s\to 0} f_s(t). $$ Is the function $F$ Borel measurable. This seems to be not true in general.

Consider a locally finite Borel measure $\mu$ on $\mathbb R$. Is the function $$F(x)=\liminf _{r\to 0+} \frac{log (\mu([x-r, x+r]))}{\log r} $$ Borel measurable?

$\endgroup$
5
  • 2
    $\begingroup$ Yes, because you can take the $\liminf$ along the rationals. $\endgroup$ Oct 8, 2017 at 3:43
  • $\begingroup$ @ Christian Remling: You are right. Thanks. $\endgroup$
    – ronggang
    Oct 8, 2017 at 5:21
  • 3
    $\begingroup$ @ChristianRemling What if $t\mapsto f_s(t)$ is $1$ if $s$ is irrational and $0$ otherwise? Or is your remark only meant to apply to the example given? $\endgroup$ Oct 8, 2017 at 9:11
  • $\begingroup$ @MichaelGreinecker: Just for this example, surely. $\endgroup$ Oct 8, 2017 at 13:45
  • 1
    $\begingroup$ @MichaelGreinecker: Yes, of course the remark was about the "example" (if you want to call it that), which I thought was the actual question, since the OP answers the first part him/herself. I have made many silly mistakes on MO in the past, but I think I'd have to get quite a bit more senile still before I could think that $\limsup_{s\to 0} \ldots = \limsup_{s\to 0, s\in\mathbb Q} \ldots$ in general. $\endgroup$ Oct 8, 2017 at 16:04

2 Answers 2

1
$\begingroup$

The function $F$ need not be Borel, even if $(s,t) \mapsto f_s(t)$ is a Borel function on $\mathbb{R}^2$. I wrote down a counterexample on Math.SE.

$\endgroup$
0
$\begingroup$

This will depend on how $F_s$ depends on $s$. If there is no restriction, then every function $F : \mathbb{R} \to \mathbb{R}_{>0}$ can be obtained that way:

Partition $\mathbb{R}$ into countable $R_s$ such that each $R_s$ has $0$ as accumulation point. Now let $F_s(x) = 0$ if $x \notin R_s$ and $F_s(x) = F(x)$ otherwise.

$\endgroup$

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.