2
$\begingroup$

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.

  • 0
    @HaraldHanche-Olsen yes2012-09-27

1 Answers 1

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 $u(x)=0$.

  • 0
    I meant of the measure: the measure of a countable union is lower than the sum of measures.2012-09-28