# Equivalence between definitions of measurable function

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?

• It does. Given any function $f : X \to Y$, prove that $\{S \subseteq Y \mid f^{-1}(S)$ is measurable$\}$ is a $\sigma$-algebra. Note that if this algebra contains all the half-lines, it must contain all Borel sets. Commented Oct 28, 2021 at 20:59

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.

• Is this the proof of the implication 1 -> 2?. So is my proof of implication 2 -> 1 correct or is there anything missing? Commented Oct 28, 2021 at 21:30
• Is $f^{-1}(\mathcal{H}) \subseteq \Sigma$ the hypothesis ?(that is definition 1?) Commented Oct 28, 2021 at 21:32
• @J.C.VegaO Yes, you were correct that 2 -> 1 is trivial. Yes, $f^{-1}(\mathcal{H}) \subseteq \Sigma$ is the hypothesis in 1. Commented Oct 28, 2021 at 21:37
• I m copying here part of the prove in the linked post: The "$\subseteq$" is not immediately obvious. Let us denote by ${\cal G}$ the collection of all sets $G\subseteq \Omega_2$ such that $f^{-1}(G) \in \sigma(f^{-1}({\cal C}))$. We note that ${\cal G}$ is a $\sigma$-algebra and that ${\cal C}\subseteq {\cal G}$, so $\sigma({\cal C})\subseteq {\cal G}$ (again by minimality of generated $\sigma$-algebra). This means $f^{-1}(\sigma({\cal C})) \subseteq \sigma(f^{-1}({\cal C}))$. Commented Oct 28, 2021 at 22:27
• Could you please explain it in more detail ?, I don't understant at all the last two sentences. Like ${\cal G}$ is a $\sigma$-algebra, ${\cal C}\subseteq {\cal G}$ and $f^{-1}(\sigma({\cal C})) \subseteq \sigma(f^{-1}({\cal C}))$ Commented Oct 28, 2021 at 22:28