Let $D$ be a domain in $\mathbb{R^d}$ and denote the continuous function space on $D$ as $X := C(\overline{D})$ where we can define the $\sigma$-algebra $\mathscr{B}(X)$ of $X$, that is sets in $X$ containing all open sets of $X$ and closed under countable unions and taking complements. We define the zero set of $u$ as $Z_u = \{x\in \overline{D}: u(x) = 0\}$ and denote $m(A)$ as the Lebesgue measure of $A\subset \overline{D}$. If $F: = \{u\in C(\overline{D}): m(Z_u) = 0\}$, can we prove that $F\in \mathscr{B}(X)$?
It is easy to see that $F$ is not a closed set in $X$. But can we write $F$ as countable union of closed set or other form of Borel set? Otherwise, is there any counterexample to make that false?
Your set $F$ is a $G_\delta$ set (countable intersection of open sets).
I assume that your domain $D$ is bounded, so that $K:=\overline D$ is compact; and that the topology of $X=\mathcal C(K)$ is accordingly given by the sup norm.
Observe first that a set $A\subset K$ has Lebesgue measure $0$ if and only if, for any $\varepsilon >0$, one can find an open set $O$ containing $A$ such that $m(O)<\varepsilon$. Denoting (for each $\varepsilon >0$)by $\mathcal O_\varepsilon$ the family of all open sets $O\subset \mathbb R^d$, it follows that $$F=\bigcap_{k\in\mathbb N} \bigcup_{O\in\mathcal O_{1/k}}\mathbf U_{O}\, , $$ where $$\mathbf U_{O}=\{ u\in X;\; Z_u\subset O\}\, . $$ So, the only thing to check is that for any fixed ope set $O$, the set $\mathbf U_O$ is open in $X=\mathcal C(K)$.
Let us fix an open set $O$, and let $u\in\mathbf U_O$. Then $u(x)\neq 0$ for all $x\in K\setminus O$. Since $u$ is continuous and $K\setminus O$ is compact, it follows that one can find $\alpha >0$ such that $u(x)\geq\alpha$ for all $x\in K\setminus O$. Then, any function $v\in X$ such that $\Vert v-u\Vert_\infty<\alpha/2$ will satisfy $v(x)>\alpha/2$ on $K\setminus O$, so that in particular $Z_v\subset O$. In other words, the open ball $B(u,\alpha/2)$ in $X$ is contained in $\mathbf U_O$. This shows that $\mathbf U_O$ is open in $X$, as required.