# Thread: Problem dealing with probability space

1. ## Problem dealing with probability space

This is how the question reads:

Let $f$ be a function mapping $\Omega$ to another space $E$ with a $\sigma$-algebra $\mathcal{E}$. Let $\mathcal{A}=(A\subset \Omega: \text{there exists B} \in E \text{ with A}=f^{-1}(B))$. Show that $\mathcal{A}$ is a $\sigma$-algebra on $\Omega$.

How do I go about mapping the function? Sorry, I am not very sure with the concept of probability space

2. Originally Posted by noob mathematician
This is how the question reads:
Let $f$ be a function mapping $\Omega$ to another space $E$ with a $\sigma$-algebra $\mathcal{E}$. Let $\mathcal{A}=(A\subset \Omega: \text{there exists B} \in E \text{ with A}=f^{-1}(B))$. Show that $\mathcal{A}$ is a $\sigma$-algebra on $\Omega$.
How do I go about mapping the function? Sorry, I am not very sure with the concept of probability space
This problem has nothing to do with properties of probability spaces.
It has everything to do properties of functions.

You need to know that $f^{ - 1} \left( {\bigcup\limits_\alpha {B_\alpha } } \right) = \bigcup\limits_\alpha {f^{ - 1} \left( {B_\alpha } \right)} \;\& \;f^{ - 1} \left( {B^c } \right) = \left( {f^{ - 1} (B)} \right)^c$

The apply the definition of $\sigma$-algebras.

3. Originally Posted by Plato
This problem has nothing to do with properties of probability spaces.
It has everything to do properties of functions.

You need to know that $f^{ - 1} \left( {\bigcup\limits_\alpha {B_\alpha } } \right) = \bigcup\limits_\alpha {f^{ - 1} \left( {B_\alpha } \right)} \;\& \;f^{ - 1} \left( {B^c } \right) = \left( {f^{ - 1} (B)} \right)^c$

The apply the definition of $\sigma$-algebras.
Hi thanks for the reply.

From the definition, given that If $A \in \mathcal{A}$, then the complement $A^c \in \mathcal{A}$.

From here, since I know that $A \subset \Omega \text{ and } A=f^{-1}(B)$, the complement $( {f^{ - 1} (B)} \right))^c$ will be a $\sigma$-algebra on $\Omega$?