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I recently started learning about measure theory. At the moment I am learning about measurable functions.

While doing some research, I did find two different definitions about what a measurable function is.

Definition 1: Suppose $(X,\mathcal{A})$ is a measurable space. A function $f: X \rightarrow \mathbb{R}$ is called $\mathcal{A}$-measurable if $\{x:f(x)>a\} \in \mathcal{A}$ for all $a \in \mathbb{R}$.

Definition 2: Suppose $(X,\mathcal{A})$ is a measurable space. A function $f: X \rightarrow \mathbb{R}$ is called $\mathcal{A}$-measurable if $f^{-1}(B)\in \mathcal{A}$ for every Borel set $B \subset \mathbb{R}$.

My thoughts: The first definition just states that the collection $\{x:f(x)>a\}$ for $a \in \mathbb{R}$ has to be contained in the sigma algebra. While the second definition states that the preimage of borel sets have to be contained in the sigma algebra. Why does one Definition require Borel sets while the other does not need them? Is one Definition just more general than the other?

NTc5
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    Check out Proposition 3.3 in these lecture notes. It has to do with so-called generators for the Borel $\sigma$-algebra. – Daan Mar 12 '25 at 13:45
  • Both definitions are equivalent. In fact you can restrict $a$ to be rational in your definition 1. The reason is that the $\sigma$-algebra generated by the internals $(a,\infty)$ is the Borel $\sigma$-algebra. – Mittens Mar 12 '25 at 18:20

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