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?