$f$ is $\mathcal{F} - \mathcal{B}$ measurable $\iff$ $A \in \sigma(\mathcal{F} \times \mathcal{B})$

Question

Let $(\Omega,\mathcal{F})$ be a measure space and $f: \Omega \rightarrow [0,\infty]$ a nonnegative function and let
$A$ :={$(\omega,y) \in\Omega\times\mathbb{R} | 0 < y < f(x)$}. Show:

$f$ is $\mathcal{F} - \mathcal{B}$ measurable $\iff$ $A \in \sigma(\mathcal{F} \times \mathcal{B})$
Where $\mathcal{B}$ denotes the Borel-$\sigma$-Algebra generated from the open sets on $[0,\infty)$ and {$\infty$}.

I think I am going in the correct direction of one implication: Let $f$ be measurable $\rightarrow$ $f$ is a nonnegative measurable function, we have shown there exist simple function $f_n$ with $f_n \rightarrow f $ for $n \rightarrow \infty$ where $f_n =\sum_{k=1}^{n2^n} (k-1)2^{-n} \mathcal{1}_{(k-1)2^{-n}\le f < k2^{-n}} + n\mathcal{1}_{n \le f}$ with $\mathcal{1}$ being the indicator-function.

Define

$A_n$ :={$(\omega,y) \in\Omega\times\mathbb{R} | 0 < y < f_n(x)$}= $\bigcup_{k=1}^{n*2^n}$ {$f\in[(k-1)2^{-n},k2^{-n}),y\in[0,(k-1)2^{-n})$}. $A$ is the union of those $A_n$. Is it correct, that the set where f is in is the inverse image of an interval under a measurable function, hence measurable and the set in which y is an element of is in $\mathcal{B}$, which would imply $A_n$ in $\sigma(\mathcal{F} \times \mathcal{B})$, so ultimately $A$ because it is the countable union of those $A_n$.

For the other implication I could not come up with anything good so far. Because I do not really know what $\mathcal{F}$ is, I will have to use

$f$ is $\mathcal{F} - \mathcal{B}$ measurable if $f^{-1}(\mathcal{B})\subset \mathcal{F}$.

Maybe I am overlooking something, but how would one show this ?

Answer 1

If $A$ is measurable (meaning it belongs to the product sigma algebra) then $A\cap (\Omega \times (t,\infty))$ is measurable . Its section $\{\omega:(\omega,y) \in \Omega \times (t,\infty) \text {for some}\, y\}$ is measurable. Hence $f^{-1}(t, \infty)$ is measurable for each $t$ and $f$ is therefore measurable.

Answer 2

$\Rightarrow:$ Suppose that $f$ is $\mathcal{F}/\mathcal{B}$-measurable. We assert that $$ A=\bigcup_{r\in\mathbb{Q}}f^{-1}\left((r,\infty]\right)\times(0,r). $$ For, let $(\omega,y)\in A$, then $0<y<f(\omega)$. By density of $\mathbb{Q}$, there exists $r_{0}\in\mathbb{Q}$ such that $y<r_{0}<f(\omega)$. Then $\omega\in f^{-1}\left((r_{0},\infty]\right)$ and $y\in(0,r_{0})$. That is, $(\omega,y)\in f^{-1}\left((r_{0},\infty]\right)\times(0,r_{0})\subseteq\bigcup_{r\in\mathbb{Q}}f^{-1}\left((r,\infty]\right)\times(0,r)$. Conversely, let $(\omega,y)\in\bigcup_{r\in\mathbb{Q}}f^{-1}\left((r,\infty]\right)\times(0,r)$, then there exists $r_{0}\in\mathbb{Q}$ such that $(\omega,y)\in f^{-1}\left((r_{0},\infty]\right)\times(0,r_{0})$. Therefore $f(\omega)>r_{0}$ and $0<y<r_{0}$ and hence $(\omega,y)\in A$.

For each $r\in\mathbb{Q}$, $f^{-1}\left((r,\infty]\right)\times(0,r)\in\mathcal{F}\otimes\mathcal{B}$, where $\mathcal{F}\otimes\mathcal{B}$ is the product $\sigma$-algebra. Now it is clear that $A\in\mathcal{F}\otimes\mathcal{B}$ because the union is a countable union.

Answer 3

For the converse and Murphy's proof, let me elaborate it a little bit, especially about the measurability of sections.

Let $(X_{1},\mathcal{F}_{1})$ and $(X_{2},\mathcal{F}_{2})$ be measurable spaces. Let $X=X_{1}\times X_{2}$. For $i=1,2$, let $\pi_{i}:X\rightarrow X_{i}$ be the canonical projection map defined by $\pi_{i}(x_{1},x_{2})=x_{i}.$ Let $\mathcal{F}=\mathcal{F}_{1}\otimes\mathcal{F}_{2}$ be the product $\sigma$-algebra. It is fundamental that $\mathcal{F}$ is the smallest $\sigma$-algebra on $X$ such that for each $i$, $\pi_{i}$ is $\mathcal{F}/\mathcal{F}_{i}$-measurable. Moreover, if $(Y,\mathcal{G})$ is a measurable space and $f:Y\rightarrow X$ is a map. Then $f$ is $\mathcal{G}/\mathcal{F}$-measureble iff for each $i$, the composited map $\pi_{i}\circ f:Y\rightarrow X_{i}$ is $\mathcal{G}/\mathcal{F}_{i}$-measurable. This is known as the universal property of product $\sigma$-algebra.

Now let $a_{1}\in X_{1}$ and $a_{2}\in X_{2}$. Define $\iota_{a_{1}}:X_{2}\rightarrow X$ and $\iota_{a_{2}}:X_{1}\rightarrow X$ by $\iota_{a_{2}}(x_{1})=(x_{1},a_{2})$ and $\iota_{a_{1}}(x_{2})=(a_{1},x_{2})$. We can verify that $\iota_{a_{1}}$ is $\mathcal{F}_{2}/\mathcal{F}$-measurable and $\iota_{a_{2}}$ is $\mathcal{F}_{1}/\mathcal{F}$-measurable. For example, note that $\pi_{1}\circ\iota_{a_{2}}(x_{1})=x_{1}$, i.e., $\pi_{1}\circ\iota_{a_{2}}=id_{X_{1}}$, the identity map on $X_{1}$, which is clearly $\mathcal{F}_{1}/\mathcal{F}_{1}$-measurable. $\pi_{2}\circ\iota_{a_{2}}:X_{1}\rightarrow X_{2}$ is the constant map $x_{1}\mapsto a_{2}$, which is also $\mathcal{F}_{1}/\mathcal{F}_{2}$-measurable. By the universial property of product $\sigma$-algebra, it follows that $\iota_{a_{2}}:X_{1}\rightarrow X$ is $\mathcal{F}_{1}/\mathcal{F}$-measurable. Similarly, $\iota_{a_{1}}:X_{2}\rightarrow X$ is $\mathcal{F}_{2}/\mathcal{F}$-measurable.

If $A\in\mathcal{F},$ then $\iota_{a_{1}}^{-1}(A)\in\mathcal{F}_{2}$ and $\iota_{a_{2}}^{-1}(A)\in\mathcal{F}_{1}$. However, by writing out $\iota_{a_{1}}^{-1}(A)$, we have $\iota_{a_{1}}^{-1}(A)=\{x_{2}\mid(a_{1},x_2)\in A\}$. Similarly, $\iota_{a_{2}}^{-1}(A)=\{x_{1}\mid(x_{1},a_{2})\in A\}$.

We are now ready to prove the converse $\Leftarrow:$ Suppose that $A\in\mathcal{F}\otimes\mathcal{B}$. Let $t>0$ be arbitrary. Define $\iota_{t}:\Omega\rightarrow\Omega\times\mathbb{R}$ by $\iota_{t}(\omega)=(\omega,t)$. By the above discussion, $\iota_{t}^{-1}(A)\in\mathcal{F}$. But \begin{eqnarray*} \iota_{t}^{-1}(A) & = & \{\omega\mid(\omega,t)\in A\}\\ & = & \{\omega\mid0<t<f(\omega)\}\\ & = & f^{-1}\left((t,\infty]\right). \end{eqnarray*} Moreover, $f^{-1}\left((0,\infty]\right)=\bigcup_{n=1}^{\infty}f^{-1}\left((\frac{1}{n},\infty]\right)\in\mathcal{F}$. Finally, if $t<0$, we have $f^{-1}\left((t,\infty]\right)=\Omega\in\mathcal{B}$ because $f$ is non-negative.