This question intuitively sounds obvious at first glance. Surprisingly I am unable to provide a correct argument.
Here is the matter we consider a measurable function $u\Bbb R^d\to\Bbb R$. Then the support of $u$ is defined as the complement of its open for cancellation set. More precisely,
\begin{split} \operatorname{supp} u &= \Bbb R^d\setminus\bigcup\big\{O\,:\, \text{$O$ is open and $u=0$ a.e. on $O$}\big\}\\ &= \bigcap\big\{ \Bbb R^d\setminus O\,:\, \text{$O$ is open and $u=0$ a.e. on $O$}\big\}. \end{split} Then here comes my question: is it true that, the set $ A= \{x\in \operatorname{supp}u,\,:\, u(x)=0\}$ is a null set? That is, do we have $$|\operatorname{supp}u \cap \{ u=0\}|=0?$$
To glimpse this question, the simple $u(x)= \mathbf{1}_{(-1, 1)}(x)$ gives $\operatorname{supp} u=[-1,1]$ and hence $|\operatorname{supp}u \cap \{ u=0\}= |\{-1,1\}|=0.$
If $u$ is a continuous then $\operatorname{supp}u= \overline{\big\{h\in \Bbb R^d\,:\,u(h)\neq 0\big\}}.$ The latter is not true in general. For instance,for $u = \mathbf{1}_{\mathbb{Q}}$ we have $\operatorname{supp}u=\emptyset$ and $\overline{\big\{h\in \Bbb R^d\,:\,u(h)\neq 0\big\}}=\Bbb R.$
This is false, even for continuous $u$.
Outline of a construction of a counterexample:
Say $U$ is an open set such that the boundary of $U$ has positive measure. Choose a continuous function $u\ge0$ so that $U=\{x:u(x)>0\}$. Then $\text{supp}\, u=\overline U$ and $u=0$ on $\partial U\subset\overline U$.
Details:
First, an open set with large boundary. Say $K$ is a compact set with empty interior but $|K|>0$ (for example a fat higher-dimensional Cantor set). Say $V$ is open with $K\subset V$ and let $U=V\setminus K$. Then $K\subset\partial U$ because $K$ has empty interior.
A continuous function $u\ge0$ that is strictly positive exactly on $U$: Say $$U=\bigcup_n B_n$$where each $B_n$ is an open ball. You can easily write down a formula for a continuous function $u_n\ge0$ such that $B_n=\{u_n>0\}$. If needed divide by a constant so $0\le u_n(x)\le 1/n^2$ for every $x$ and let $$u=\sum_n u_n.$$