I'm self-studying measure theory. I stumbled upon this statement:
If $X \subset \mathbb{R}$ and $f: X \to \mathbb{R}$ is a function, then $f$ is a Borel measurable function if and only if $f^{-1}((a,\infty))$ is a Borel set for all $a \in \mathbb{R}$.
The forward direction is easy ($f$ is Borel measurable $\implies$ $f^{-1}((a,\infty))$ is a Borel set for all $a \in \mathbb{R}$). My question is about the backward direction: don't we need to assume that $X$ is Borel measurable? If not, how can I prove the backward direction?
Let $f : X \to \mathbb R$ such that, for each $a$, $f^{-1}((a,\infty))$ is borel in $X$. We want to show that $f$ is borel - that is, we want to show that $f^{-1}(U)$ is borel for each $U \subseteq \mathbb R$ open.
Since every open set $U$ is a union of basic opens (the intervals), it suffices to show $f^{-1}((a,b))$ is borel for every $a,b \in \mathbb R$.
But we know that $$(a,b) = \bigcup_{n \in \mathbb N} \left ( a,b - \frac{1}{n} \right ] = \bigcup_{n \in \mathbb{N}} \bigg ( (a,\infty) \setminus (b - \frac{1}{n}, \infty) \bigg )$$
So
$$f^{-1}(a,b) = f^{-1} \bigcup_{n \in \mathbb{N}} \bigg ( (a,\infty) \setminus (b - \frac{1}{n}, \infty) \bigg ) = \bigcup_{n \in \mathbb{N}} \big [ f^{-1}(a,\infty) \setminus f^{-1}(b - \frac{1}{n}, \infty) \big ]$$
Which is borel since set operations and countable unions of borel sets are borel.
I hope this helps ^_^