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?

  • $\begingroup$ Unless I'm missing something, If your topological spaces are separated the answer is clear: adding closed subset to $\tau$ will not change the $\sigma$-algebra it generates. So I guess the topological spaces are not assumed to be separated, and in this case I need to ask: does compact mean compact & separated or just compact ? $\endgroup$ – Simon Henry May 20 '15 at 6:56
  • $\begingroup$ In the non-$T_1$ context, there may be compact sets that are not closed. $\endgroup$ – Dominic van der Zypen May 20 '15 at 7:05
  • $\begingroup$ possible duplicate of Is every continuous function measurable? $\endgroup$ – Johannes Hahn May 20 '15 at 10:57
  • $\begingroup$ By separated I mean Hausdorff.I wouldn't say it is a duplicate, the linked question ask for conditions, this one for counterexamples. $\endgroup$ – Simon Henry May 20 '15 at 11:39

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.


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