# limsup of measurable functions is not measurable?

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?

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.
• 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 $r\in(1/2^{n+1},1/2^{n}]$? – John Jan 21 '15 at 4:30