Definitions of Measurable Functions

[chingkui]

I have seen a number of definitions of measurable functions, particularly, for the following two definitions, are they the same?
1) A function f:X->R is measurable iff for every open subset T of R, the inverse image of T is measurable.
2) A function f:X->R is measurable iff for every Borel set B of R, the inverse image of B is measurable.
Here, R is the real line.
I suppose they are the same, but I haven't been able to show they are equivalent (one direction is easy, but the other one, by assuming any open set has measurable inverse image and showing that the inverse image of all Borel set is measurable, doesn't seem to be that easy, since Borel sets in general cannot be easily written down in terms of open sets).

 

[Hurkyl]

since Borel sets in general cannot be easily written down in terms of open sets

Sure they can -- they're made from applying finitely many σ-algebra operations to open sets. (that is, complement and countable union)

 

[chingkui]

Sure they can -- they're made from applying finitely many σ-algebra operations to open sets. (that is, complement and countable union)

Hi Hurkyl, I am under the impression that a Borel set might not be generated under finitely many σ-algebra operations to open sets. Borel σ-algebra is defined in all the books I read as "the smallest σ-algebra containing the open sets", it is not defined as a constructive process of σ-algebra operations, so, I am not even sure whether countablely many σ-algebra operations to open sets is enough to generate all the Borel sets. That's why I am not sure how to show the two definitions of measurable functions are the same. Are you sure they all can be formed by a finite process? Thanks.

 

[Jimmy Snyder]

Is the set of inverse images of sets under f a [itex]\sigma[/itex]-algebra?

 

[Hurkyl]

Bah, you're right, I've made this mistake before, but I've fixed it before. I meant to say:

Any Borel set can be formed by γ applications of the σ-algebra operations, for some ordinal number γ. (For example, letting γ be any ordinal with cardinality greater than that of P(R) suffices)

More explicitly:

Let B(0) be the set of all open sets.

For any ordinal number α > 0:
Let C(α) be the union of all C(β) with β < α
Then, define B(α) to be the the set of all:
(1) things in C(α)
(2) complements of things in C(α)
(3) things that are countable unions of things in C(α)


You can prove there is an upper bound by some clever theorem of set theory that I don't remember, or simply noting that once you've stopped adding sets, you have the class of Borel sets, and that there are only P(R) many sets. So, B(&gamma;) is the set of Borel sets.


Then, you could do something with transfinite induction. Specifically, if you prove:

(1) If your statement holds for B(0)
(2) If your statement holds for anything in C(α) then it holds for anything in B(α)

Then, by transfinite induction, it holds for all B(α).

 

[Jimmy Snyder]

Is the set of inverse images of sets under f a [itex]\sigma[/itex]-algebra?

Bad question. Better would be:

Is the set of sets whose inverse image under f is measurable a [itex]\sigma[/itex]-algebra?

 

[Jimmy Snyder]

I think the answer is yes because the inverse image of a union of sets is the union of the inverse images and similarly for intersections. Since the set of measurable sets in the domain is a [itex]\sigma[/itex]-algebra, so is the set of sets in the range whose inverse images are measureable. Hence given (1), the set of sets whose inverse images are measureable is a [itex]\sigma[/itex]-algebra that contains the open sets.

 

[mathwonk]

i.e. one of the most useful remarks i recall from measure theory was " the process of taking inverse images is a boolean sigma homomorphism". ... i.e. it preserves all those operations.

 

[chingkui]

Thanks. Seems like transfinite induction is very useful. I don't really have any background in ordinal, cardinal, transfinite induction, etc. can anyone suggest where I can read more (preferably free online lecture note) about them?

Before I have a chance to learn them, does anyone know of any proof that avoid using transfinite induction? Thanks.

 

[matt grime]

You don't actually have to induct, since the borel sigma algebra is defined to be the smallest sigma algebra containing the open sets (strictly speaking I suppose it ought to be the borel sets, but its easy to see that they generate the same borel algebra). So the two definitions are equivalent by fiat.