Why is $\{(x, y) \in [a, b] \times [a, b]: f(x) \geq y\}$ a Borel set? Suppose $f: I = [a, b] \rightarrow I$ is a Borel function. Why is $\{(x, y) \in [a, b] \times [a, b]: f(x) \geq y\}$ a Borel set?
Since $f$ is a Borel function, for any open set $U$, $f^{-1}(U)$ is Borel. However, I can't seem to be able to write $\{(x, y) \in [a, b] \times [a, b]: f(x) \geq y\}$ as the inverse image of an open set.

My first approach seems have some problem.
This is second approach:
$\{(x, y) \in [a, b] \times [a, b]: f(x) \geq y\}=\{x\times (y,+\infty): f(x)\in I\}=\cup_{y\in I} f^{-1}([y,\infty))\times y$ is a closed set hence borel.
