$\overline{\mathbb{R}}$ stands for $\mathbb{R}\cup \{\infty\} \cup \{-\infty\}$.
Define $f:\mathbb{R^n} \to \overline{\mathbb{R}}$ by $f(x)=\infty$ for all $x\in \mathbb{R^n}$.
Is $f$ a Lebesgue measurable function ?
This may be a foolish question, but I'm not sure my idea is correct.
For all $a\in \mathbb{R},$ $\{ f>a\}=\{\infty >a \}=\mathbb{R^n}$ since $\infty$ is bigger than all real numbers. And $\mathbb{R^n}$ is a Lebesgue measurable set hence $f$ is Lebesgue measurable.
Is my idea correct ?
It is indeed correct. However, in this case maybe another approach is preferable. It would be more convenient that you explicitly state that $f^{-1}(\infty) = \mathbb{R}^n$, $f^{-1}(-\infty) = \emptyset$ and $f^{-1}(G) = \emptyset$ for any $G\subset \mathbb{R}$ open. All these preimages are measurable sets, so $f$ is (by definition) measurable.