# Prove measurability

• Sep 8th 2011, 02:26 PM
TheProphet
Prove measurability
Let $\displaystyle f = (f_1, f_2): \Omega \to E \times F$.
Show that $\displaystyle f : (\Omega,\mathcal{A}) \to (E \times F, \mathcal{E} \otimes \mathcal{F})$ is measurable if and only if $\displaystyle f_1$ is measurable from $\displaystyle (\Omega,\mathcal{A})$ to $\displaystyle (E, \mathcal{E})$ and $\displaystyle f_2$ is measurable from $\displaystyle (\Omega,\mathcal{A})$ to $\displaystyle (F,\mathcal{F})$.

Attempted solution.

"$\displaystyle \Rightarrow$"
$\displaystyle \mathcal{E} \times \mathcal{F} \subset \mathcal{E} \otimes \mathcal{F}$. Set $\displaystyle \mathcal{B} = \{ C \in \mathcal{E} \otimes \mathcal{F} : C = \Lambda \times F, \Lambda \in \mathcal{E} \}$. Then $\displaystyle \mathcal{B} \subset \mathcal{E} \otimes \mathcal{F}$, so $\displaystyle f^{-1}(\mathcal{B}) \subset \mathcal{A}$. But $\displaystyle f^{-1}(\Lambda \times F) = \{ \omega \in \Omega : f_{1}(\omega) \in \Lambda \}$. Similarily, $\displaystyle f^{-1}(E \times \Gamma) = \{ \omega \in \Omega : f_{2}(\omega) \in \Gamma\}$.

"$\displaystyle \Leftarrow$"

$\displaystyle \{ \omega : f \in A \times B, A \in \mathcal{E}, B \in \mathcal{F} \} =$
$\displaystyle \{ \omega : f_{1}(\omega) \in A, f_{2}(\omega) \in B, A \in \mathcal{E}, B \in \mathcal{F} \}$, which is in $\displaystyle \mathcal{A}$.
Since $\displaystyle f^{-1}(\mathcal{E} \times \mathcal{F}) \subset \mathcal{A}$ and $\displaystyle \sigma(\mathcal{E} \times \mathcal{F}) = \mathcal{E} \otimes \mathcal{F}$, it follows that $\displaystyle f^{-1}(\mathcal{E} \otimes \mathcal{F}) \subset \mathcal{A}$.

Appreciate any input/corrections!
• Sep 8th 2011, 05:26 PM
Drexel28
Re: Prove measurability
Quote:

Originally Posted by TheProphet
Let $\displaystyle f = (f_1, f_2): \Omega \to E \times F$.
Show that $\displaystyle f : (\Omega,\mathcal{A}) \to (E \times F, \mathcal{E} \otimes \mathcal{F})$ is measurable if and only if $\displaystyle f_1$ is measurable from $\displaystyle (\Omega,\mathcal{A})$ to $\displaystyle (E, \mathcal{E})$ and $\displaystyle f_2$ is measurable from $\displaystyle (\Omega,\mathcal{A})$ to $\displaystyle (F,\mathcal{F})$.

Attempted solution.

"$\displaystyle \Rightarrow$"
$\displaystyle \mathcal{E} \times \mathcal{F} \subset \mathcal{E} \otimes \mathcal{F}$. Set $\displaystyle \mathcal{B} = \{ C \in \mathcal{E} \otimes \mathcal{F} : C = \Lambda \times F, \Lambda \in \mathcal{E} \}$. Then $\displaystyle \mathcal{B} \subset \mathcal{E} \otimes \mathcal{F}$, so $\displaystyle f^{-1}(\mathcal{B}) \subset \mathcal{A}$. But $\displaystyle f^{-1}(\Lambda \times F) = \{ \omega \in \Omega : f_{1}(\omega) \in \Lambda \}$. Similarily, $\displaystyle f^{-1}(E \times \Gamma) = \{ \omega \in \Omega : f_{2}(\omega) \in \Gamma\}$.

"$\displaystyle \Leftarrow$"

$\displaystyle \{ \omega : f \in A \times B, A \in \mathcal{E}, B \in \mathcal{F} \} =$
$\displaystyle \{ \omega : f_{1}(\omega) \in A, f_{2}(\omega) \in B, A \in \mathcal{E}, B \in \mathcal{F} \}$, which is in $\displaystyle \mathcal{A}$.
Since $\displaystyle f^{-1}(\mathcal{E} \times \mathcal{F}) \subset \mathcal{A}$ and $\displaystyle \sigma(\mathcal{E} \times \mathcal{F}) = \mathcal{E} \otimes \mathcal{F}$, it follows that $\displaystyle f^{-1}(\mathcal{E} \otimes \mathcal{F}) \subset \mathcal{A}$.

Appreciate any input/corrections!

Whoah, overusage of fancy letters. What is the tensor product? Just the usual product measure?
• Sep 8th 2011, 11:44 PM
TheProphet
Re: Prove measurability
Quote:

Originally Posted by Drexel28
Whoah, overusage of fancy letters. What is the tensor product? Just the usual product measure?

This was not my intention, I used the same the letters my book uses =)

Yeah, its the usual product measure. $\displaystyle \mathcal{E} \otimes \mathcal{F}$ is the $\displaystyle \sigma$-algebra generate by $\displaystyle \mathcal{E} \times \mathcal{F}$,
where $\displaystyle \mathcal{E}$ and $\displaystyle \mathcal{F}$ are $\displaystyle \sigma$-algebras on $\displaystyle E$ and $\displaystyle F$ recpectively.