# Proving that $\nu(E\cap(A_{1}\cup A_{2}))=\nu(E\cap A_{1})+\nu(E\cap A_{2})$

I am going over my tutorials in my real analysis course and there is an unproved statement that I am having difficulty to verify, and I could use some help with.

Definition:

$\nu:\, P(X)\to[0,\infty]$ s.t $\nu(\emptyset)=0$, $\nu(\cup_{i=1}^{\infty}A_{i})\leq\sum_{i=1}^{\infty}\nu(A_{i})$ and s.t if $A\subseteq B$ then $\nu(A)\leq\nu(B)$ is called an outer measure.

Definition:

A set $A$ is called $\nu$ measurable if $\forall E\subseteq X:\,\nu(E)=\nu(E\cap A)+\nu(E\cap A^{c})$.

Statement:

If $\{A_{i}\}_{i=1}^{n}$ are disjoint $\nu$ measurable sets (where $\nu$ is an outer measure) then $$\nu(E\cap(\cup_{i=1}^{n}A_{i}))=\sum_{i=1}^{n}\nu(E\cap A_{i})$$ The proof is by induction.

I have tried proving the last statement, the case $n=1$ is trivial since the expression on the LHS is exactly the expression on the LHS, after this I only tried to prove the claim for $n=2$, just to get the idea, but I didn't manage to prove that.

If $A_{1},A_{2}$ are $\nu$ measurable then so is $A_{1}\cap A_{2}$ and $A_{1}\cup A_{2}$ and I hoped that I could find some good $E$ and a choise of $\nu$ measurable sets (like the two examples I gave) that will yield the desired result, but I couldn't find any good choice for $E$ and the $\nu$ measurable set.

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@Did - thanks for the correction! I edited accordingly – Belgi Mar 4 '13 at 11:05
Interesting question (+1) [joke]We can prove this easily on $\mathfrak{M}_{\nu}$, because $\nu$ is measure on $\mathfrak{M}_{\nu}$ :)[/joke] – Cortizol Mar 4 '13 at 11:14

For any two $A$ and $B$ and some measurable $M$ we have

$$\nu(A \cup B) = \nu\big((A \cup B) \cap M\big) + \nu\big((A \cup B) \cap M^c\big).$$

Let $A \cap B = \varnothing$, and suppose there exists $M$ that separates $A$ and $B$, that is, $M \cap A = A$ and $M \cap B = \varnothing$. It follows that

\begin{align} \nu(A \cup B) &= \nu\big((A \cap M) \cup (B \cap M)\big) + \nu\big((A \cap M^c)\cup(B \cap M^c)\big) \\ &= \nu(A \cup \varnothing) + \nu(\varnothing \cup B) \\ &= \nu(A) + \nu(B). \end{align}

The above is enough to conclude

\begin{align} \nu\big(E \cap (A \cup B)\big) &= \nu\big((E \cap A) \cup (E \cap B)\big) \\ &= \nu(E \cap A) + \nu(E \cap B) \end{align}

for any measurable disjoint $A$ and $B$ (just set $M = A$).

I hope it helps ;-)

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Thanks for the asnwer! Can you please explain how did you use $\nu(A\cup B)=\nu(A)+\nu(B)$ to get the conclusion in the last equality ? – Belgi Mar 4 '13 at 12:34
@Belgi Set $A := E \cap A$ and $B := E \cap B$. – dtldarek Mar 4 '13 at 12:57

Since $A_1$ is measurable, for all $S\subseteq X$ we have $\nu(S)=\nu(S\cap A_1)+\nu(S\cap {A_1}^\complement)$.

Use it for $S=E\cap(A_1\cup A_2)$. Then, $S\cap A_1=E\cap A_1$ and, since $A_2\subseteq {A_1}^\complement$, we also have $S\cap {A_1}^\complement =E\cap A_2$.

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