Is the set $\{(\omega, r) : f(\omega) > r\}$ measurable?

Let $(\Omega,\Sigma)$ be a measurable space and let $(\mathbb{R}, \mathcal{B})$ be the standard 1-dimensional Borel space. Let $f: (\Omega, \Sigma) \rightarrow (\mathbb{R}, \mathcal{B})$ be a measurable function from $\Omega$ to $\mathbb{R}$. Define the set $A \subseteq \Omega \times \mathbb{R}$ as follows: $$A := \{(\omega, r) \in \Omega \times \mathbb{R} \mid: f(\omega) > r\}$$

1. Is $A$ measurable, i.e. is $A \in \Sigma\otimes\mathcal{B}$ (the product $\sigma$-algebra of $\Sigma$ and $\mathcal{B}$)?

2. If the answer to the previous question is negative, is it any different when $(\Omega, \Sigma) = (\mathbb{R}^n, \mathcal{B}_n)$, i.e. when $(\Omega, \Sigma)$ is the standard $n$-dimensional Borel space?

• Try to prove 1. First, can you do it when $f$ is a simple function? Next, try approximating general measurable $f$ with simple functions. – GEdgar Sep 20 '14 at 20:42

Hint Note that the mapping $$(\mathbb{R},\mathcal{B}(\mathbb{R})) \otimes (\mathbb{R},\mathcal{B}(\mathbb{R})) \ni (x,y) \mapsto x-y \in (\mathbb{R},\mathcal{B}(\mathbb{R}))$$ is measurable since it is continuous. Conclude that $$(\mathbb{R},\mathcal{B}(\mathbb{R})) \otimes (\Omega,\Sigma) \ni (r,\omega) \mapsto f(\omega)-r \in (\mathbb{R},\mathcal{B}(\mathbb{R}))$$ is measurable.