I noticed something strange. If we look at a function $f \in L^p$, then this is an equivalence class. Hypothetically: $\operatorname{supp}(f) = \overline{\{f\neq 0\}}$. But this is strange, as $f$ is only defined modulo the rationals. But if $f\neq 0$ on all rationals, then the support would be automatically the whole space.
Therefore, the definition support is highly dependent on the choice of the functions in the equivalence class and my question would be: How is the support for $L^p$ functions defined?
If anything is unclear, please let me know.
The Wikipedia article makes it clear that the definition of support as $\overline{\{f\ne 0\}}$ for meant for continuous functions. It is not a natural concept for functions that are not necessarily continuous. Why take closure with respect to some topology if the function does not care about that topology? I suggest abandoning the idea that $\overline{\{f\ne 0\}}$ is somehow a canonical notion of support for "ordinary" functions.
Essential support is the natural concept to use for $L^p$ elements, just as closed support is the natural concept for continuous functions.
Also, it is often preferable to think of elements $f\in L^p$ as distributions, i.e., functionals $\varphi\mapsto \int f\varphi$. Then the issue of null sets and ambiguous point-wise evaluation disappears; we see only what's essential. There is a natural concept of support of a distribution: the complement of the largest open set $U$ such that the distribution kills all test functions which are compactly supported in $U$.
In the Euclidean setting, the distributional support of a continuous function is the closed support $\overline{\{f\ne 0\}}$; and the distributional support of an $L^p$ element is its essential support. Thus, the distributional support unifies the two notions. If an element $f\in L^p$ has a continuous representative $\tilde f$, then all three notions of support are the same.