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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

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up vote 2 down vote accepted

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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Thanks for the answer, how did you obtain the second equality ? I think that you used that $\nu(A\cup B)=\nu(A)+\nu(B)$ if $A,B$ are disjoint, but that is actually the next part of the question so I assume its not meant to be used here (and at least its not proven at that point) (+1) – Belgi Dec 12 '12 at 21:30
You may want to change in the last equality that you are missing a complement on $A\cup B$. Am I correct that $\nu(A\cup B)=\nu(A)+\nu(B)$ only assumes that the sets are disjoint and that one of $A,B$ is in $M$ (so we can have for example $B\not\in M$ and it still true) ? – Belgi Dec 12 '12 at 21:43
@Belgi: The second equality uses the fact $B\in M$ and hence $\nu(E')=\nu(E'\cap B)+\nu(E'\cap B^c)$ for every $E'$. Use this with $E'=E\cap A$ and $E'=E\cap A^c$. Is it clear now? I'm not sure what you mean I should be changing in the last equality. – Stefan Hansen Dec 13 '12 at 6:29
Thanks for the reply. In the last eqality you add two terms that are the same, I think the right term should be $(A\cup B)^c$ instead. I think I understand what you did in the second equality then, but is it also hold since the first two terms are a partition of $A\cap E$ and the last two terms are partition of $E\cap A^c$ ? – Belgi Dec 13 '12 at 8:27
Thanks, I edited it! No, you can't say that, because you don't know that $\nu$ is additive. You don't know that $\nu(A\cup B)=\nu(A)+\nu(B)$ when $A,B$ are disjoint - all you know is that it holds with $\leq$. – Stefan Hansen Dec 13 '12 at 8:38

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