0
$\begingroup$

This question already has an answer here:

If $\mu^*$ is an outer measure on $X$ and $\{A_j\}_{1}^{\infty}$ is a sequence of disjoint $\mu^*$-measurable sets, then $\mu^*\left(E\cap\left(\bigcup_{1}^{\infty}A_j\right)\right) = \sum_{1}^{\infty}\mu^*(E\cap A_j)$ for any $E\subset X$.

Attempted proof - From countable subadditivity, $$\mu^*\left(E\cap\left(\bigcup_{1}^{\infty}A_j\right)\right) = \mu^*\left(\bigcup_{1}^{\infty}E\cap A_j\right) \leq \sum_{1}^{\infty}\mu^*(E\cap A_j)$$ Let $B_n = \bigcup_{1}^{n}A_j$. For each $n\geq 2$ since $A_n$ is $\mu^*$-measurable we have \begin{align*} \mu^*(E\cap B_n) &= \mu^*((E\cap B_n)\cap A_n) + \mu^*((E\cap B_n)\cap A_n^c)\\ &= \mu^*(E \cap A_n) + \mu^*(E\cap B_{n-1})\\ \end{align*} (These 2 lines above me came from part of the proof in Carthedory's Theorem and I don't really understand it at all if someone would be so kind to explain that part to me.)

By induction, $\mu^*(E\cap B_n) = \sum_{1}^{n}\mu^*(E\cap A_j)$ for all $n\geq 1$. So then by monotonocity, $$\mu^*\left(E\cap\left(\bigcup_{1}^{\infty}A_j\right)\right) \geq \mu(E\cap B_n) = \sum_{1}^{n}\mu^*(E\cap A_j)$$ for all $n\geq 1$. Thus as $n\rightarrow \infty$ we have the result.

$\endgroup$

marked as duplicate by Ramiro, Giovanni, Alex Mathers, user147263, Shailesh May 19 '16 at 0:11

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ The countable subadditivity gives you a $\leqslant$ in the last relation. You need to show $\geqslant$ to complete the proof. $\endgroup$ – Daniel Fischer May 18 '16 at 14:12
2
$\begingroup$

First use $\mu^*$ measurability and induction to conclude that the relation to be proved holds for finitely many $A_j$.

For convenience, let's denote the relation to be proved by $(*)$. Say we have $n$ of $A_j$. For $n = 1$, there is nothing to prove. In the case $n = 2$, if $A_1 \cup A_2 = X$, then $(*)$ is just a restatement of $\mu^*$ measurability. If $A_1 \cup A_2$ is smaller than $X$, split $E \cap (A_1 \cup A_2)$ by $A_1$ and $A_1^c$, then since $A_1$ is $\mu^*$-measurable and $A_1 \cap A_2 = \varnothing$, \begin{align*} & \mu^*(E \cap (A_1 \cup A_2)) \\ = & \mu^*(E \cap (A_1 \cup A_2) \cap A_1) + \mu^*(E \cap (A_1 \cup A_2) \cap A_1^c) \\ = & \mu^*(E \cap A_1) + \mu^*(E \cap A_2). \end{align*}

Assume $(*)$ holds for the case of $n - 1$ sets. By the case $n = 2$, together with the induction hypothesis, $$\mu^*(E \cap (\cup_{j = 1}^n A_j)) = \mu^*(E \cap (\cup_{j = 1}^{n - 1}A_j)) + \mu^*(E \cap A_n) = \sum_{j = 1}^n \mu^*(E \cap A_j).$$

For the infinite case use monotonicity: for every $n \in \mathbb{N}$, $$\mu^*(E \cap (\cup_{j = 1}^\infty A_j)) \geq \mu^*(E \cap (\cup_{j = 1}^n A_j)) = \sum_{j = 1}^n \mu^*(E \cap A_j).$$ Let $n \to \infty$, the result follows.

$\endgroup$
  • $\begingroup$ Can you give me your reasoning for your first point? Why must we show the relation to be proved holds for finitely many $A_j$? $\endgroup$ – Wolfy May 18 '16 at 14:21
  • $\begingroup$ OK. I will add more details. $\endgroup$ – Zhanxiong May 18 '16 at 14:22
  • $\begingroup$ Also just to clairfy your second point which seems obvious to me say we have a set $A = \bigcup_{1}^{\infty} E\cap A_j$ and $B = \bigcup_{1}^{n} E\cap A_j$ then can we say that $B\subset A$ then by monotonicty $\mu(B)\leq \mu(A)$? Second questions is $B\leq A$? $\endgroup$ – Wolfy May 18 '16 at 14:32
  • $\begingroup$ @Wolfy Yes, as by definition, outer measure is monotone. $\endgroup$ – Zhanxiong May 18 '16 at 14:33
  • $\begingroup$ I don't really follow your first part at all is that necessary to have for the proof to be complete? Perhaps I do not understand what the question is asking us to do. In my proof my first statement is correct. Now from what I can tell all I have to do is show the opposite relation and then I have my result. So, using the last part of your answer it appears that I would be done. If what I am saying is not correct please let me know $\endgroup$ – Wolfy May 18 '16 at 14:38

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