- #1
Turambar
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I've encountered two definitions of measurable functions.
First, the abstract one: function [tex]f: (X, \mathcal{F}) \to (Y, \mathcal{G})[/tex], where [tex]\mathcal{F}[/tex] and [tex]\mathcal{G}[/tex] are [tex]\sigma[/tex]-algebras respect to some measure, is measurable if for each [tex]A \in \mathcal{G}, f^{-1}(A) \in \mathcal{F}[/tex].
The more concrete definition: [tex]f: \mathbb{R}^n \to \mathbb{R}^m[/tex] is measurable if for each open set [tex]A \in \mathbb{R}^m, f^{-1}(A)[/tex] is Lebesgue-measurable.
So my question is why in the latter definition is the set [tex]A[/tex] defined to be open? In the sense of the first definition the set [tex]A[/tex] was only measurable, not necessarily open. I can see that the latter definition can be seen as a generalization of the definition of continuity, but still why isn't [tex]A[/tex] just taken to be Lebesgue-measurable, not necessarily open or some other Borel set? Maybe it's because many often encountered sets are Borel sets.
First, the abstract one: function [tex]f: (X, \mathcal{F}) \to (Y, \mathcal{G})[/tex], where [tex]\mathcal{F}[/tex] and [tex]\mathcal{G}[/tex] are [tex]\sigma[/tex]-algebras respect to some measure, is measurable if for each [tex]A \in \mathcal{G}, f^{-1}(A) \in \mathcal{F}[/tex].
The more concrete definition: [tex]f: \mathbb{R}^n \to \mathbb{R}^m[/tex] is measurable if for each open set [tex]A \in \mathbb{R}^m, f^{-1}(A)[/tex] is Lebesgue-measurable.
So my question is why in the latter definition is the set [tex]A[/tex] defined to be open? In the sense of the first definition the set [tex]A[/tex] was only measurable, not necessarily open. I can see that the latter definition can be seen as a generalization of the definition of continuity, but still why isn't [tex]A[/tex] just taken to be Lebesgue-measurable, not necessarily open or some other Borel set? Maybe it's because many often encountered sets are Borel sets.