# Borel Measurability of Certain Type of Function

$\newcommand{\RR}{\mathbb{R}} \newcommand{\O}{\mathcal{O}}$ I'm looking through some practice qualifying exams (real analysis), and came across this one that's bugging me:

Suppose $f: \RR \to \RR$ satisfies $$f(x) \geq \limsup_{y \to x} f(y)$$ for all $x \in \RR$. Prove that $f$ is Borel measurable.

There are two techniques I've been working with:

(1) Directly from the definition. I'd show $f^{-1}(\O)$ or $f^{-1}([-\infty,a))$ is measurable for each open set $\O$ or $a \in \RR$. The problem with this is that I don't really have a solid understanding of what Borel sets look like; yes, the Borel $\sigma$-algebra is the smallest one that contains all open and closed sets. The best bet (in my mind) would be to use a sort of limiting argument (find open / closed sets to show $f^{-1}(\O)$ is $G_\delta$ or $F_\sigma$). But, I can't think of what those sets would be.

(2) By contraposition. If $f$ is not Borel measurable, then there is an open set $\O$ such that $f^{-1}(\O)$ is not Borel measurable. But this leads me to think, "What do non-Borel sets look like?" but I don't really know. They could act very sporadically, or they could be relatively tame (like Lebesgue sets) or they could be somewhere in the middle. There are just so many possibilities!

So, any thoughts / hints? I'm not looking for a complete solution, just a push in the right direction.

• Take route (1). The preimages of a certain family of generators of the Borel $\sigma$-algebra on $\mathbb{R}$ are quite nice. Commented Jul 7, 2014 at 21:48
• Thanks, @DanielFischer! I'll give that a go Commented Jul 7, 2014 at 21:50
• @DanielFischer, I'm still having some trouble. What was it that you had in mind? Commented Jul 8, 2014 at 0:22
• Nevermind, I've got it now. Commented Jul 8, 2014 at 2:07

$\newcommand{\RR}{\mathbb{R}} \newcommand{\BB}{\mathbb{B}} \newcommand{\ep}{\varepsilon}$Okay, I think I've got it:

We will prove $f^{-1}(-\infty,a)$ is open for each $a \in \RR$. Indeed, suppose that were not the case. Then, there exist real numbers, $x$ and $a$ such that $f(x) < a$ and for every $\ep > 0$, $f(\BB_\ep(x)) \not\subseteq (-\infty,a)$. Now, we may take a sequence $\{x_n\}$ such that $x_n \in \BB_{1/n}(x)$ but $f(x_n) \geq a$. Thus, any limit point (one has to exist even if it's $\infty$) is greater than or equal to $a$ (since $(-\infty,a)^c = [a,\infty)$ is closed), and $$\limsup_{y \to x} f(y) > f(x)$$ which contradicts the hypothesis.

Therefore, $f^{-1}(-\infty,a)$ is open for every real number $a$, and $f$ is a Borel measurable function.

• I haven't checked the details, but I suspect this isn't quite right, because this actually implies $f$ is continuous.
– Ian
Commented Jul 8, 2014 at 3:08
• @Ian, this actually only proves upper semicontinuity; since $f^{-1}(a,\infty)$ is not necessarily open. (Although that doesn't mean the details are correct.) Thanks for the feedback Commented Jul 8, 2014 at 3:42
• Ah, you are right.
– Ian
Commented Jul 8, 2014 at 4:04
• One small point: You never said that you have $f(x) < a$. You use that (not unimportant) assumption in the proof, so you need to state it. Yup, the condition is an equivalent way of describing upper semicontinuity. Commented Jul 8, 2014 at 10:52