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Let be $u$ a numerical function defined over $\Omega$, with $u$ measurable, and let $(O_i)_{i\in I}$ be 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 $O = \cup_{i\in I}O_i$ (Then $u=0$ almost always except on a set of measure in $O$). How can I define in explicit form the set $\Omega\setminus O$?

I trying this ...

$\Omega\setminus O = \{x\in \Omega; (u(x)=0 \text{ and } x\in K\subset \Omega, \text{ where } K \text{ is closed}) \text{ or } (u(x)\neq 0)\}$.

This is correct?

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    @MichaelGreinecker no is duplicate2012-10-01
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    What is the difference? The problem seems to be the same.2012-10-01
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    the difference is that I want explicit that set $\Omega\setminus O$. In the last exercise I want proof a proposition.2012-10-01
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    How much more explicit than $\Omega\backslash\bigcup_{i\in I}O_i$ can you get?2012-10-01
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    yes, but I like more explicit in the form than I write iff posible2012-10-01
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    Every $x$ is in the closed set $\{x\}$, so the set defined on the right side is simply $\Omega$.2012-10-01
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    I ask this because I need show that: If $u$ is continuos then supp $u = \Omega\setminus O = \widehat{\{x \in \Omega; u(x) \neq 0\}}$2012-10-01
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    If you want to know this for continuous $u$, you should write this. This makes things much easier. You should also specify the measure you use and the topolgy. I assume you are working in Euclidean space as in the other question.2012-10-01
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    I ask this because I need show that: If $u$ is continuos then supp $u = \Omega\setminus O = closet(\{x \in \Omega; u(x) \neq 0\})$2012-10-02
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    I make a question here http://math.stackexchange.com/questions/205691/function-support, thanks @MichaelGreinecker2012-10-02

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