I've seen a couple of questions regarding compactness of the essential range of a function $f \in L^{\infty}$, but everyone seems to know that $R_{f}$ is closed. I'm pretty sure that I'm missing out on something really basic. Any help would be appreciated!
Edited : the precise definition of the essential range is given as
$$\text{the set of all} \quad z \in \mathbb{C} \quad \text{such that} \quad \{x : |f(x)-z| > \epsilon \}\quad \text{has positive measure for all}\quad \epsilon>0$$
Let $\left(z_n\right)_{n\geqslant 1}$ be a sequence of elements of $R_f$ that converges to some $z$. Let $\varepsilon_0\gt 0$ be fixed. Pick $n_0$ such that $\left\lvert z_{n_0}-z\right\rvert\lt\varepsilon_0/2$. Then use the inclusion $$\left\{x: \left\lvert f(x)-z\right\rvert\gt \varepsilon_0/2\right\}\supset\left\{x: \left\lvert f(x)-z_{n_0}\right\rvert\gt \varepsilon_0 \right\}.$$