Let $(X, \mathcal B(X), \mu)$ be a $\sigma$-finite measure space with $\mathcal B(X)$ the Borel $\sigma$-algebra of a topological space $X$. The essential support of $f:X \to \mathbb R$ is defined as $$ \operatorname{ess-supp} (f) := \bigcap \{F \text{ closed} \mid f= 0 \, \mu\text{-a.e. on } F^c\}. $$
In this way, $\operatorname{ess-supp} (f)$ is always closed.
Can below theorem be generalized to spaces which are not second-countable?
Theorem: If $X$ is second-countable, then $f = 0$ $\mu$-a.e. on $(\operatorname{ess-supp} (f))^c$.
Proof: Let $S := \operatorname{ess-supp} (f)$. Then $$ S^c := \bigcup \{U \text{ open} \mid f= 0 \, \mu\text{-a.e. on } U \}. $$
Let $(O_i)_{i\in I}$ be a countable base of the topology of $X$. Let $J$ be the collection of those $j \in I$ such that $f=0$ $\mu$-a.e. on $O_i$. Then $$ S^c = \bigcup_{i \in J} O_i. $$
For each $i \in J$, there is a $\mu$-null set $N_i \in \mathcal B(X)$ such that $N_i \subset O_i$ and $f=0$ on $O_i \setminus N_i$. Let $N := \bigcup_{i \in J} N_i$. Then $N \in \mathcal B(X)$ is a $\mu$-null set such that $f = 0$ on $S^c \setminus N$. This completes the proof.