No, according to my memory, open sets generate only the Borel-measurable sets of \mathbb{R}^m (which is the smallest \sigma-algebra that contains the closed sets). It is a theorem that the preimage of a Borel set under a measurable function is measurable, but what about sets that are Lebesgue...
I've encountered two definitions of measurable functions.
First, the abstract one: function f: (X, \mathcal{F}) \to (Y, \mathcal{G}), where \mathcal{F} and \mathcal{G} are \sigma-algebras respect to some measure, is measurable if for each A \in \mathcal{G}, f^{-1}(A) \in \mathcal{F}.
The...