Prove $\{(\omega,y) \in \Omega \times [0,\infty) \ | \ y \leqslant f(\omega) \}$ is in the product $\sigma$-algebra $\mathcal{A} \otimes \mathcal{B}$

Question

We have a $\sigma$-finite measure space $(\Omega,\mathcal{A},\mu)$, a measurable nonnegative function $f:\Omega \to \mathbb{R}$ and

\begin{equation} G_f := \{(\omega,y) \in \Omega \times [0,\infty) \ | \ y \leqslant f(\omega) \}. \end{equation}

The Borel $\sigma$-algebra on $[0,\infty)$ is denoted by $\mathcal{B}$. The aim is to prove that

\begin{equation} G_f \in \mathcal{A} \otimes \mathcal{B} \quad \text{(the product $\sigma$-algebra of $\mathcal{A}$ and $\mathcal{B}$).} \end{equation}

My current approach:

Since $f$ is measurable, it follows that ${G_f}^y := \{\omega \in \Omega \ | \ f(\omega) \geqslant y\} \in \mathcal{A}$ since $f$ is measurable (this follows from an elementary theorem we are allowed to use). Moreover, since $\{y\}$ is closed for any $y\in[0,\infty)$, we also have that $\{y\}\in\mathcal{B}$, since $\mathcal{B}$ is generated by all closed subsets of $[0,\infty)$. Consequently,

\begin{equation} {G_f}^y \times \{y\} \in \mathcal{A} \otimes \mathcal{B} \quad \text{for each } y\in [0,\infty). \end{equation}

Now since

\begin{equation} G_f = \bigcup_{y\in[0,\infty)}{G_f}^y \times \{y\}, \end{equation}

it would be tempting to conclude that also $G_f\in \mathcal{A} \otimes \mathcal{B}$. However, this union is not countable, hence it is not guaranteed that it is included in $\mathcal{A} \otimes \mathcal{B}$.

I am kind of stuck at this point, any help would be much appreciated!

Answer 1

By definition, $ G_f $= { $(\omega, Y) \in \Omega \times [0, \infty) | 0 \leq f(\omega) - y$}

Define $ g : \Omega \times [0, \infty) \rightarrow \mathbb{R} $ as $ g( \omega, y) = f(\omega) -y$ and show that $g$ is a measurable function. Here, $[0, \infty)$ is the measure space with the Borel sigma algebra. Next, since $g ^{-1} ( [0, \infty) ) $ is measurable in $\Omega \times [0, \infty)$, and this set equals $G_f$, we have the result.

Answer 2

Define $F\colon \Omega\times \left[0,+\infty\right)\to \mathbb R$ by $F\left(\omega,y\right)=f\left(\omega\right)-y$. If we endow $ \Omega\times \left[0,+\infty\right)$ with the $\sigma$-algebra $\mathcal A\otimes \mathcal B$, the function $F$ is measurable. Since $G_f:=F^{-1}\left(\left[0,+\infty\right)\right)$, we conclude.