My question is on Exercise 2.24 (page 12) of Probability and Stochastics by Erhan Cinlar:
2.24 Discrete spaces. Suppose that $E$ is countable and $\mathcal{E} = 2^E$, the discrete $\sigma$-algebra on $E$. Then $(E,\mathcal{E})$ is said to be discrete. Show that every [numerical] function on $E$ is $\mathcal{E}$-measurable.
I'm unsure about my argument because it is suspiciously simple and it doesn't use the fact that $E$ is countable:
"Proof": Let $f: E \rightarrow \overline{\mathbb{R}}$ and let $B \in \mathcal{B}(\overline{\mathbb{R}})$ (the Borel $\sigma$-algebra on $\overline{\mathbb{R}}$). Then $f^{-1}(B)$ is obviously a subset of $E$. But $2^E$ is (by definition) the collection of all subsets of $E$, so $f^{-1}(B) \in 2^E$. Thus, $f$ is $\mathcal{E}$-measurable.
Am I missing something, or is the assumption that $E$ is countable irrelevant?