The essential support of a function $f:\Bbb R^n\rightarrow \Bbb R$ is defined in the following way:
Let's denote $\mathcal A_f=\{\omega \subset \Bbb R^n: \omega \quad \text{open}, \quad f(x)=0\quad \text{a.e.} \quad x\in \omega\}$ and $A_f=\cup_{\omega \in \mathcal A_f}\omega$. The essential support $supp_e f$ of $f$ is $\Bbb R^n -A_f$.
I want to show that, if $f$ is continuous, then $supp_e f=supp f$.
My attempt:
Now, let $\omega \in \mathcal A_f$. Then, there exists $N_\omega\subseteq \omega$ such that $|N_\omega|=0$, and $f(x)=0$ for each $x\in \omega- N_\omega$, and suppose that $N_\omega \neq \varnothing$. Let $x\in N_\omega$, so $f(x)\neq 0$. Since $f$ in continuous in $x$, there exists a ball $B$ centered in $x$, such that $f(y)\neq 0$ for each $y\in B$, so $B\subset N\omega$. Thus $|B|=0$, which is impossible. It follows that $N_\omega = \varnothing$. (Is it correct?)
Then $f(x)=0$ for each $x\in \omega$, for each $\omega\in \mathcal A_f$ $(*)$. From $(*)$ follows that $supp_ e f= supp f$.
In fact, if $x\in supp_e f$, then $x\notin A_f$, thus $f(x)\neq 0$, that is $x\in supp f$. On the other hand, suppose that $x\in supp f$. Then either $f(x)\neq 0$, or $x\in \overline{\{x\in \Bbb R^n:f(x)\neq0\}}-\{x\in \Bbb R^n:f(x)\neq0\}$. In the former case, we have that $x\notin A_f$, so $x\in supp_e f$. In the latter case, there are points of $\{x\in \Bbb R^n:f(x)\neq0\}$ arbitrarily closed to $x$, and again, since $f$ is continuous, $f(x)\neq 0$, so we are again in the previous case.
Is it correct? Or is there an easiest way to solve the problem?