# Thread: Prove measurability

1. ## Prove measurability

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!

2. ## Re: Prove measurability

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?

3. ## Re: Prove measurability

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.