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I am not sure I am using the standard definitions so I will open by defining what I need:

  1. Let $X$ be a set, $\nu:\, P(X)\to[0,\infty]$ will be called an external measure if $\nu(\emptyset)=0$ and for any $\{A_{i}\}_{i=1}^{\infty}\subseteq P(x)$ (not neccaseraly disjoint) it holds that $\nu(\cup_{i=1}^{\infty}A_{i})\leq\sum_{i=1}^{\infty}\nu(A_{i})$

  2. Let $\nu$ be an external measure on a set $X$ then we say that a set $A$ is $\nu$ measurable if for any $E\subseteq X$: $\nu(E)=\nu(E\cap A)+\nu(E\cap A^{c})$

The exercise asks to prove that the set of $\nu$ measurable sets $M\subseteq P(X)$ is an algebra.

I have proved $\emptyset,X\in M$ and that $A\in M\implies A^{c}\in M$ but I am having problems proving closer under union and intersection.

I assume that $A_{1},A_{2}$ are $\nu$ measurable so I get that for any $E$: $\nu(E)=\nu(E\cap A_{1})+\nu(E\cap A_{1}^{c})$ $\nu(E)=\nu(E\cap A_{2})+\nu(E\cap A_{2}^{c})$

And I need to prove that for any $E'$: $\nu(E')=\nu(E'\cap A_{1}\cap A_{2})+\nu(E'\cap(A_{1}\cap A_{2})^{c})$

which is the same as $\nu(E')=\nu(E'\cap A_{1}\cap A_{2})+\nu((E'\cap A_{1}^{c})\cup(E'\cap A_{2}^{c}))$

and a similar result to prove closer under union.

I guess that it all have to do with choosing the right $E$'s from knowing that $A_{i}$ are $\nu$ measurable, but I tried different options for an hour now and I don't see this going anywhere.

I need some help in showing closer under union and intersection

1 Answers 1

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If $A\in \mathcal{P}(X)$, then $E=(E\cap A)\cup (E\cap A^c)$ for alle $E\in\mathcal{P}(X)$. By 1. we have that $ \nu(E)\leq \nu(E\cap A)+\nu(E\cap A^c), $ and hence $A\in M$ if and only if $ \nu(E)\geq \nu(E\cap A)+\nu(E\cap A^c),\quad E\in\mathcal{P}(X).\qquad (*) $

Let $A,B\in M$ be given and let us show that $A\cup B\in M$ by showing that $A\cup B$ satisfies $(*)$. If $E\in\mathcal{P}(X)$, then $ \begin{align*} \nu(E)&=\nu(E\cap A)+\nu(E\cap A^c)\\ &=\nu(E\cap A \cap B)+\nu(E\cap A\cap B^c)+\nu(E\cap A^c\cap B)+\nu(E\cap A^c\cap B^c)\\ &\geq \nu\big((E\cap A\cap B)\cup(E\cap A\cap B^c)\cup(E\cap A^c\cap B)\big)+\nu(E\cap A^c\cap B^c)\\ &=\nu(E\cap (A\cup B))+\nu(E\cap (A\cup B)^c) \end{align*} $ and hence $A\cup B\in M$. To show that also $A\cap B\in M$, we just use that $(A\cap B)^c=A^c\cup B^c$ which is in $M$ and hence $A\cap B\in M$.

In the last equality we used that $ (E\cap A\cap B)\cup (E\cap A\cap B^c)\cup (E\cap A^c\cap B)=E\cap \big((A\cap B)\cup (A\cap B^c)\cup (A^c\cap B)\big) \\ =E\cap\big(A\cup (A^c\cap B)\big)=E\cap (A\cup B). $

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    I get it now, thanks!2012-12-13