Given that $f\leq g$ a.e then how to show that Essential sup $f\leq $ Essential sup $g$

Question

Given that $f\leq g$ a.e then how to show that Essential sup $f\leq $ Essential sup $g$?

$$\text{ess} \sup f=\inf\{b\in \mathbb R\mid \mu(\{x:f(x)>b\})=0\}$$
From the definition, it is clear that inequality holds. But how to prove rigorously given fact.

Any Help / Hint will be appreciated.

Answer 1

If $\mu (g >b)=0$ then $\mu (f>b)=0$. [ $(f >b ) \subseteq (g>b)) \cup (f>g)$ and union of two sets of measure $0$ has measure $0$].

The infimum if $A $ is $\leq$ infimum of $B$ if $B \subseteq A$.

Notation: $(f>b)=\{x\in \mathbb R: f(x) >b\}$ etc].