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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 $u=0$ often in $O_i$. Let be $O = \cup_{i\in I}O_i$. Then $u=0$ often in $O$.

How I can be able to do this?.

I am beginning make ...

Let be $u$ defined than $0$ in $O_i\setminus M_i$ and $\neq$ $0$ in $M_i$, then

$O = \cup_{i\in I}O_i=\cup_{i\in I}[(O_i\setminus M_i)\cup M_i]$, ...

but I don't know how find the subset of $O$ such that have measure zero.

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2  
Does "often" mean "almost always", that is "except on a set of measure $0$"? –  Alex Becker Sep 27 '12 at 19:44
    
Are we in Euclidean space here? And is $I$ supposed to be countable? –  Harald Hanche-Olsen Sep 27 '12 at 19:47
    
@AlexBecker yes –  juaninf Sep 27 '12 at 23:22
    
@HaraldHanche-Olsen yes –  juaninf Sep 27 '12 at 23:22

1 Answer 1

Let $N_i$ of measure $0$ such that $u=0$ on $O_i\setminus N_i$. Define $N:=\bigcup_{i\in I}N_i$. By sub-additivity, as $I$ is countable, $N$ has measure $0$. If $x\in O\setminus N$, $x\in O_i$ for some $i\in I$. Since $x\notin \bigcup_{j\in I}N_j$, $x\notin N_i$, so $u(x)=0$.

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sub-additivity in the function? –  juaninf Sep 28 '12 at 16:20
    
I meant of the measure: the measure of a countable union is lower than the sum of measures. –  Davide Giraudo Sep 28 '12 at 16:25

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