ess-sup property

Question

Let $(X, \mathcal{A}, \mu)$ be a measured space, and let $f \in L^p(\mu)$. I would like to show that $$ |f| \leq \text{ess-sup}(|f|) \text{ a.e. on } X, $$ where $$ \text{ess-sup}(|f|) := \inf \{ \sup_{X \backslash N} |f| : N \in \mathcal{A}, \mu(N) = 0\}. $$ I'm pretty sure it must be easy to show, but unfortunately I do not know how to handle this. I thought I could try to write the set of $x \in X$ not satisfying the property as a denumerable union of sets that are readily seen to have measure 0, but I did not succeed. Any hint on how to proceed from the above definition would be very appreciated.