# Are F-measurable functions 1-to1?

1. Jul 24, 2012

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.

2. Jul 24, 2012

### micromass

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).

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

3. Jul 24, 2012

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?

4. Jul 24, 2012

### 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 $\bigcup_{i\in I} f^{-1}(B_i)=\Omega$.

5. Jul 25, 2012

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