3
$\begingroup$

Suppose $f_r(x)$ is measurable for any fixed $r>0$, I was wondering whether $\limsup_{r\to 0^+}f_r(x)$ is measurable.

I know the limsup of sequence of measurable functions is measurable, and I also know for each $x$, $\limsup f_r(x)$ will be achieved by a sequence of $\big(f_{r_i}(x)\big)_i$, with $r_i\to 0$, but the problem is that I still don't know how to write $\{x|\limsup_{r\to 0^+}f_r(x)\le a\}$ as a countable union of measurable sets since it seems the sup is taken in an uncountable set.

I know if $f_r(x)$ is continuous, then we have $$\{x|\limsup_{r\to 0^+}f_r(x)\le a\}=\{x|\lim_{k\to \infty}\sup_{0<r<\frac{1}{k},r\in \mathbb{Q}}f_r(x)\le a\}$$ which is countable, hence $\limsup_{r\to 0^+}f_r(x)$ is measurable. Is it hold for general $f_r$? How about $f_r$ be semi-continuous?

$\endgroup$
2
$\begingroup$

Let $A$ be a non-measurable set. Assume that for each $x\in A$ $f_r$ is the indicator function of $\{x\}$ for infinitely many $r$.

Since $A$ is uncountable there is a bijection $g$ from $(1/2,1]$ to $A$. Then take $f_r=\chi_{\{g(2^{n}r)\}}$ for $r\in(1/2^{n},1/2^{n-1}]$, for $n=1,2,...$

Then $\limsup_{r\to0^+} f_r=\chi_A$ which is not measurable.

$\endgroup$
  • $\begingroup$ Thanks. I get your idea. Any $x\in A$ corresponds to infinity many $y$ with $y\to 0$, so $\limsup_{r\to0^+} f_r(x)=1$. And if $x\not\in A$, since $g$ has range $A$, all $f_r(x)=0$. Is it correct? Btw, shall it be $r\in(1/2^{n+1},1/2^{n}]$? $\endgroup$ – John Jan 21 '15 at 4:30
  • $\begingroup$ @JohnZHANG That's it. $\endgroup$ – Pp.. Jan 21 '15 at 4:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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