The limit inferior of Borel functions

Question

Suppose $X$ is a separable metric space and $F \colon X \times ℝ_+→[0,1]$ is Borel. Let $f(x) = \liminf_{ε→0} F(x,ε)$. Is $f$ Borel?

Answer 1

No, $f$ is not necessarily Borel.

Let's make the stronger assumption that $X$ is Polish. Then I claim:

For each $a \in [0,1]$, the set $\{f \le a\}$ is analytic, but need not be Borel.

To see it is analytic, note that we can write $$\{f \le a\} = \bigcap_{n=1}^\infty \bigcap_{k=1}^\infty \pi_1\left(\left\{F \le a + \frac{1}{k}\right\} \cap \left(X \times \left(0, \frac{1}{n}\right)\right)\right).$$ Here $\pi_1 : X \times \mathbb{R}_+ \to X$ is the projection onto the first coordinate. This is the set version of the statement

$\liminf_{\epsilon \to 0} F(x,\epsilon) \le a$ iff for every $n$ and every $k$ there exists $\epsilon < \frac{1}{n}$ such that $F(x, \epsilon) \le a + \frac{1}{k}$.

(I had this wrong in a previous version of this answer, hopefully it is correct now.)

The set $\{F \le a + \frac{1}{k}\}$ is Borel in $X \times \mathbb{R}_+$ since $F$ is Borel, and so is $X \times (0, \frac{1}{n})$. The continuous image of a Borel subset of a Polish space is analytic, and a countable intersection of analytic sets is analytic.

To see it need not be Borel, suppose $A \subset X$ is analytic but not Borel (such a set exists whenever $X$ is an uncountable Polish space). Then there is a set $E \subset X \times [0,1)$ such that $E$ is Borel and $\pi_1(E) = A$. Define $F$ by $$F(x,t) = \begin{cases} 1, & t \ge 1 \\ 1 - 1_E(x,2^n t -1), & 2^{-n} \le t < 2^{-(n-1)}, n \ge 1 \end{cases}$$ Clearly $F$ is Borel.

I claim $f = 1 - 1_A$. If $x \notin A$ then $1_E(x,t) = 0$ for all $t$, and hence $F(x,t) = 1$ for all $t$, so $f(x) = 1$. If $x \in A$ then there exists $t_0 \in [0,1)$ such that $(x, t_0) \in E$, so if we take $t_n = 2^{-n} (t_0 + 1)$ then we have $t_n \to 0$ and $F(x,t_n) = 0$, so $f(x) = 0$.

In particular $\{f \le 0\} = A$.