About the definition of measurable functions

[Turambar]

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 more concrete definition: f: \mathbb{R}^n \to \mathbb{R}^m is measurable if for each open set A \in \mathbb{R}^m, f^{-1}(A) is Lebesgue-measurable.

So my question is why in the latter definition is the set A defined to be open? In the sense of the first definition the set A 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 A 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.

 

[mathman]

I believe it has to do with the fact that the sigma algebra in R^m is gnerated by the open sets.

 

[Turambar]

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, but not Borel?

The reason behind the definition might be the following. Clearly we want every continuous function to be measurable. Now in the case of continuous function the preimage of every open set is open, ie. measurable in Lebesgue sense. So by the latter definition continuous functions are measurable. Now if the former definition is used, then there might be some measurable set whose preimage under a continuous function is not measurable, and thus continuous functions wouldn't necessarily be measurable. I don't know if continuity implies that this is impossible, but I guess it doesn't. Thoughts?

 

[g_edgar]

In the definition, when we talk about measurability of a function f \colon \mathbb{R}^n \to \mathbb{R}^m without specifying sigma-algebras, we mean that you should use the Lebesgue measurable sets in \mathbb{R}^n and the Borel sets in \mathbb{R}^m. Why that choice? Because it is the most often used.

 

[dimitri151]

Measurability of functions is often stated as X-measurability where X is the \sigma-algebra defining the measurable space, so I think you need to specify a \sigma-algebra to talk about a functions measurability. Usually if measurability is used without specifying ( as when you say X-measurable) Lebesgue measurability is implied. If you look around in your book you'll find some statement like "from here on when we say measurable we'll mean Lebesgue measurable..." Basically what Edgar said. There's an assumption of the \sigma-algebra being used.