Let $(X,\tau)$ be a topological space. We call $S\subseteq X$ *saturated* if $S=\bigcap\{U\in\tau: U\supseteq S\}$. Let $\sigma(X,\tau)$ be the $\sigma$-algebra generated by $\tau\cup\{K\subseteq X: K \text{ is compact and saturated}\}$.


Let $(X\_i, \tau\_i)$ be topological spaces for $i=1,2$. A function $f:X\_1\to X\_2$ is said to be *measurable* if $S\in \sigma(X\_2,\tau\_2)$ implies $f^{-1}(S)\in\sigma(X\_1,\tau\_1)$.


Are there topological spaces $X,Y$ and a continuous function $f: X\to Y$ such that $f$ is not measurable in the above sense?



Measurability and continuity for general topological spaces
I'm assuming that by "compact" you only mean which satisfies the finite cover properties, and not compact and Hausdorff. in this case the following produces a counterexample:


Let $X\_1$ be $\mathbb{R}$ with the ordinary topology, (and assume that there is at least one non Borel subset of $\mathbb{R}$...).


Let $X\_2$ be $\mathbb{R}$ with the co-finite topology. then the identity map from $X\_1$ to $X\_2$ is continuous, but any subset of $X\_2$ is saturated and compact hence in the tribe you define.


