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

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 ?

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.

• After thinking it through I think I understood! I will write it out more thoroughly in my proof as I guess this would a bit too short. – babemcnuggets Jan 9 '19 at 1:47

$$\Rightarrow:$$ 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 $$0. By density of $$\mathbb{Q}$$, there exists $$r_{0}\in\mathbb{Q}$$ such that $$y. 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 $$0 and hence $$(\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.

• Making use of rational numbers and the density of them in the real numbers was never my strong point. However I understand your proof. Would you say my approach of the implication is wrong? – babemcnuggets Jan 9 '19 at 1:10
• @babemcnuggets It is hard to directly prove that $A=\cup_n A_n$. (In fact, I don't know if we really have $A=\cup_n A_n$.) For the converse, you need the fact that any section of a measurable set $A\in\mathcal{F}\otimes\mathcal{B}$ is measurable. I include a detailed proof below. – Danny Pak-Keung Chan Jan 9 '19 at 2:20

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 $$\begin{eqnarray*} \iota_{t}^{-1}(A) & = & \{\omega\mid(\omega,t)\in A\}\\ & = & \{\omega\mid0 Moreover, $$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.