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Let be $u$ a numerical function defined over $\Omega$, with $u$ measurable, and let be $(O_i)_{i\in I}$ a family of all open sub-sets $O_i$ of $\Omega$, such that $u=0$ almost always except on a set of measure $0$ in $O_i$. Let be $O = \cup_{i\in I}O_i$ (Then $u=0$ almost always except on a set of measure in $O$).

I need show that: If $u$ is continuos then supp $u = \Omega\setminus O = closure(\{x \in \Omega; u(x) \neq 0\})$.

How I will be able to making this?

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I guess you want the topology to be second countable. –  André Caldas Oct 2 '12 at 0:38
    
@AndréCaldas yes countable –  juaninf Oct 2 '12 at 0:39

1 Answer 1

Let $\mathcal{B}$ be a countable base for the topology. And let $J \subset I$ be the indexes such that $O_j \in \mathcal{B}$ for $j \in J$.

First, we will to show that $$ O = \bigcup_{j \in J} O_j. $$ It is evident that $\bigcup_{j \in J} O_j \subset O$. On the other hand, for each $i \in I$, there is a subfamily $\mathcal{B}_i \subset \mathcal{B}$ such that $$ O_i = \bigcup_{U \in \mathcal{B}_i} U. $$ Of course, if $U \in \mathcal{B}_i$, then $u = 0$ a.e. in $U$, and therefore, $U = O_j$ for some $j \in J$. That is, $$ O = \bigcup_{i \in I} O_i = \bigcup_{i \in I} \bigcup_{U \in \mathcal{B}_i} U \subset \bigcup_{j \in J} O_j, $$ as we wanted to show.

Since $J$ is countable (because $\mathcal{B}$ is), we can conclude that $u = 0$ a.e. in $O$. It is evident that $O$ is the largest open set such that $u = 0$ a.e. in $O$. Therefore, $\mathrm{supp}(u)^c \subset O$. We have to show that $O \subset \mathrm{supp}(u)^c$. But this is the same as showing that $u = 0$ everywhere in $O$. In order to show this claim, we have to assume that every non-empty open set has strictly positive measure.

Suppose that $u(x) = \alpha \neq 0$. Take an open interval $A$ with $x \in A$, and $0 \not \in A$. Then, $u^{-1}(A) \cap O$ is an open subset of $O$ where $u \neq 0$ everywhere. That is, $u^{-1}(A) \cap O$ has null measure. From our assumptions, $u^{-1}(A) \cap O$ is empty. That is, $u = 0$ everywhere in $O$.

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