Let $(X,\Sigma,\mu)$ be a (finite) measure space, and let $g:X \to \mathbb R$ be measurable. Denote by $\text{ess} \text{ Im} g$ its essential range.
Now, let $\phi:\mathbb R \to \mathbb R$ be a continuous map.
Is it true that $\phi \circ g=0$ $\mu$-almost everywhere if and only if $\phi(\text{ess} \text{ Im} g)=0$?
What is the relation between these two notions?
Let $E$ be the essential range of $g.$ We want to show $\phi\circ g=0$ $\mu$-a.e. on $X$ iff $\phi = 0$ on $E.$
$\implies:$ Let $x\in E.$ Suppose $\phi(x)\ne 0.$ Then there is an interval $I$ around $x$ such that $\phi\ne0$ on $I,$ here using the continuity of $\phi.$ But $\mu(g^{-1}(I)) > 0$ and $\phi\circ g \ne0$ on $g^{-1}(I),$ contradiction. Thus $\phi(x)=0$ and we have $\phi = 0$ on $E.$
$\impliedby:$ Suppose $\phi = 0$ on $E.$ Let $E^c=\mathbb R\setminus E.$ Then $X=g^{-1}(E)\cup g^{-1}(E^c).$ On $g^{-1}(E),$ $\phi\circ g=0.$ Since $\mu(g^{-1}(E^c))=0,$ we have $\phi\circ g=0$ $\mu$-a.e. on $X$ as desired.