Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

Can someone please help me out ?

share|improve this question
    
@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
add comment

2 Answers

up vote 2 down vote accepted

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 ;-)

share|improve this answer
    
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
add comment

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 $S\cap {A_1}^\complement =E\cap A_2$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.