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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?

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  • $\begingroup$ If a continuous function is a.e. zero on an open set it is in fact zero because open sets with measure zero are empty. $\endgroup$ – Jochen Aug 6 '14 at 8:32
  • $\begingroup$ That is the fact in $(*)$, isn't it? $\endgroup$ – batman Aug 6 '14 at 8:40
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Everything is good up to "In the latter case, ...". It may well be that $x$ is a limit point of $\{f \neq 0\}$ while $f(x) = 0$. The identity map on $\mathbb R$ at $0$ is an example of this.

To finish off the proof, all that's needed is that $\overline{\{f \neq 0\}}^c \in \mathcal A_f$. Clearly $\overline{\{f \neq 0\}}^c$ is open and since $\{f \neq 0\} \subset \overline{\{f \neq 0\}}$, we have that $\overline{\{f \neq 0\}}^c \subset \{f \neq 0\}^c = \{f = 0\}$.

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Try the following. Instead of prooving directly that $supp\: f\subset supp_{e}f$ and that $supp_{e}\: f\subset supp\: f$, you will find easier to proove that $(supp_{e}\: f)^{c}\subset (supp\: f)^{c}$ and that $(supp\: f)^{c}\subset (supp_{e}f)^{c}$, which is equivalent, since that a theorem is equivalent to it's counter positive. Actually, I did not understand your point, so I tried this way and I found it. I think it's a better way for a reason: the definition of the essential support is made by taking a complement with respect to the domain. So, it's 'natural' to think about complements. Good work to you.

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