Are F-measurable functions 1-to1?

[lahanadar]

Hi everybody. Can anyone help me to clarify these things? The definition of F-measurable function is as this:

f:Ω→ℝ defined on (Ω,F,P) probability space is F-measurable if f^-1(B)={ω∈Ω: f(ω)∈B} ∈ F for all B∈B(ℝ)

where B(ℝ) is Borel field over ℝ and B is any Borel subset of the Borel field.

My confusions are:

1-Is the function f:Ω→ℝ 1-to-1?
2-Is f^-1(B):B(ℝ)→F mapping to mutually exclusive and collectively exhaustive subsets of F?

Thank you for any contributions.

 

[micromass]

Hi everybody. Can anyone help me to clarify these things? The definition of F-measurable function is as this:

f:Ω→ℝ defined on (Ω,F,P) probability space is F-measurable if f^-1(B)={ω∈Ω: f(ω)∈B} ∈ F for all B∈B(ℝ)

where B(ℝ) is Borel field over ℝ and B is any Borel subset of the Borel field.

My confusions are:

1-Is the function f:Ω→ℝ 1-to-1?

No, not necessarily. The f^{-1}(B) is just an (unfortunate) notation for the preimage and has nothing to do here with the inverse of f (which does not exist necessarily).

2-Is f^-1(B):B(ℝ)→F mapping to mutually exclusive and collectively exhaustive subsets of F?

But f^{-1}(B) is a set, not a map.

 

[lahanadar]

The word mapping could be wrong maybe. I mean any Borel set in real line should go to the an element of the field F (an element is any subset of Ω). What I wonder is if that elements of the field F, that are assigned by Borel sets from real line, should be mutually exclusive and collectively exhaustive?

 

[micromass]

Not necessarily. It's not necessarily true that f^{-1}(B)\cap f^{-1}(B^\prime)=\emptyset. It is true if B\cap B^\prime=\emptyset though.

Likewise, if \bigcup_{i\in I}B_i=\mathbb{R}, then \bigcup_{i\in I} f^{-1}(B_i)=\Omega.

 

[lahanadar]

I see now, thank you for help. From this point, should I also assume that the field F to constitute a probability measure P:F→[0,1] have elements (subsets of Ω) which are not necessarily mutually disjoint?

 

[micromass]

I see now, thank you for help. From this point, should I also assume that the field F to constitute a probability measure P:F→[0,1] have elements (subsets of Ω) which are not necessarily mutually disjoint?

Right. In general, elements of \mathcal{F} are not necessarily disjoint.