Prove measurability

**TheProphet** · September 8th 2011, 02:26 PM

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

Attempted solution.

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

" $\Leftarrow$ "

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

Appreciate any input/corrections!

**Drexel28** · September 8th 2011, 05:26 PM

Originally Posted by TheProphet

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

Attempted solution.

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

" $\Leftarrow$ "

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

Appreciate any input/corrections!

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

**TheProphet** · September 8th 2011, 11:44 PM

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

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