A problem on the essential range has been solved here. In the solution, it's been stated that if $\lVert f \rVert_{\infty}$ is finite, then it trivially follows that the essential range $R_f$ is bounded. It's presumably easy but, unfortunately I can't see why. here's the precise definition of the essential range of $f$: $$R_f :=\left\{z \in \mathbb{C} \mid \text{for all}\ \varepsilon > 0: \mu(\{x : |f(x) - z| < \varepsilon\}) > 0\right\}$$
I know that if $\lVert f \rVert_{\infty} < \infty$ then $\lvert f \rvert \leq \lVert f \rVert_{\infty}$ a.e but I can't see why this immidiately implies that $R_f$ is bounded, as the definition of $R_f$ looks formidable and I can't seem to parse it. Any help will be appreciated.
Edit: I know $R_f$ is closed. If it's proved that $R_f$ is bounded then by the Heine-Borel Theorem it follows that $R_f$ is compact. But how does this imply that $\lVert f \rVert_{\infty} = \max \{ \lvert z \rvert : z\in R_f \}$?
Here is the formal proof you want: let $|z|>||f||_\infty$ and let $0<\epsilon <|z|-|f||_\infty$. Consider the open ball $B(z,\epsilon)$ with center z and radius $\epsilon$. If $f(x) \in B(z,\epsilon)$ then $|f(x)| \geq |z| -|f(x)-z|>|z|-\epsilon>||f||_\infty$. By definition of $||f||_\infty$ this implies that $\mu (f^{-1} (B,x,\epsilon))=0$ which implies that z is not in $R_f$. In other words $|z|\leq ||f||_\infty$ whenever $z \in R_f$. As you have observed this implies that $R_f$ is compact. So $max\{|z|:z\in R_f\}$ is attained and if this maximum is less than $||f||_\infty$ we see that $|f| \leq r$ a.e. for some $r<||f||_\infty$ contradicting the definition of $||f||_\infty$.
Some details for the last part: suppose $max\{|z|:z\in R_f\}<||f||_\infty$. Consider $A=\{z: max\{|z|:z\in R_f\}<|z|<||f||_\infty$. This annulus can be written as a countable union of open balls. Also $\mu(f^{-1}(A))>0$ (by definition of $||f||_\infty$). Hence there is an open ball $B(z_0,r)$ in $A$ such that $\mu (f^{-1}(B(z_0,r)))>0$. But then $z_0 \in R_f$ and this contradicts the fact that $|z_0| >max\{|z|:z\in R_f\}$.