I have the following question:
Let $\Omega_1, \Omega_2$ be some spaces (don't have to be necessairly equal) endowed with the "corresponding" borel-$\sigma$-algebra. Then every continuous function $$f:(\Omega_1, \mathfrak{B}(\Omega_1))\rightarrow (\Omega_2, \mathfrak{B}(\Omega_2))$$ is borel mesurable.
I thought about it and I'm not sure if I can simply say that this is true since the preimage of an open set under a continuous function is again open and since the borel-$\sigma$-algebra is generated by open sets we are done. More precisly we use the fact that mesurability only need to be checked for the generator sets of the borel algebra in the codomain (In our case this are the open sets since the borel-$\sigma$-algebra is generated by them).
Thanks for your help.