I found in a book that they define a measurable function in the following way:
Definition 1:
Consider the measure space $(X, \Sigma, \mu)$.
A function $f : X \to \mathbb{R}$ is said to be measurable if, for every $α \in \mathbb{R},
\{x \in X : f(x) > \alpha\} \in \Sigma$.
The definition I knew was that
Definition 2 :
A function is measurable if the pre-image of every measurable set is measurable.
How come these definitions are equivalent?
I suspect it has something to do with the fact that half-lines are generators of the Borel sigma algebra, but I haven't been able to prove it.
I guess the implication 2 -> 1 is trivial. If a function is measurable, half lines being borel sets, are measurable sets on $\mathbb{R}$, therefore their preimages are measurable
Can someone tell me if I am on the right track and complete the proof?
Best Answer
Let $\mathcal{H}$ denote the collection of half lines.
We know they generate the Borel sigma-algebra, i.e. $\sigma(\mathcal{H}) = \mathcal{B}$, so the preimages of measurable sets are $$f^{-1}(\mathcal{B}) = f^{-1}(\sigma(\mathcal{H})).$$
We also know $f^{-1}(\mathcal{H}) \subseteq \Sigma$, which implies $$\sigma(f^{-1}(\mathcal{H})) \subseteq \Sigma.$$
Thus, if we show $$f^{-1}(\sigma(\mathcal{H})) = \sigma(f^{-1}(\mathcal{H})), \tag{$*$}$$ then we are done, since we would then have $f^{-1}(\mathcal{B}) \subseteq \Sigma$. Try showing this last claim yourself; if you are stuck, see this answer.