1
$\begingroup$

Let $\mu: $ outer measure.
Given the fact that if $A_1$ and $A_2$ are measurable then for $G_\delta$ sets $G_1$ that contains $A_1$ and $G_2$ that contains $A_2$, we have $\mu(G_1 - A_1)=0$ and $\mu(G_2-A_2)=0$. I want to show that $\exists G$, a $G_\delta$ set such that $\mu(G - (A_1 \cup A_2))=0$.

Can I start of with supposing $G=G_1\cup G_2$. Since each of $G_1$ and $G_2$ contains $A_1, A_2$, respectively, then $G$ should also contain both $A_1$ and $A_2$, which implies that $A_1 \cup A_2 \subset G$. Hence, $\mu(G-(A_1\cup A_2))=\mu((G-A_1)\cap(G-A_2))=\mu(G-A_1)+\mu(G-A_2)$ since both are zeros thus $A_1\cup A_2$ is measurable.

$\endgroup$
  • $\begingroup$ By definition, the countable union of measurable sets are measurable. You might want to rephrase your question title to make it a bit more clear what you're asking. $\endgroup$ – Math1000 Sep 9 '15 at 20:37
  • $\begingroup$ @Math1000 That depends. If measurability is defined a la Caratheodory, then closure under countable unions is nontrivial. $\endgroup$ – Noah Schweber Sep 9 '15 at 20:45
  • $\begingroup$ @Math1000 would like to show that there is a $G_\delta$ set $G$ that contains the $A_1\cup A_2$ and that $\mu(G-(A_1\cup A_2))=0$ $\endgroup$ – desperatemuch Sep 9 '15 at 20:59
  • $\begingroup$ The last two lines: why $\mu$ of the intersection is equal to the sum and why "both are zeros"? $\endgroup$ – A.Γ. Sep 9 '15 at 21:01
  • $\begingroup$ @A.G. that is the thing that is confusing me because I couldn't make it out either. but I assumed that since $G=G_1\cup G_2$, then $G$ is also a $G_\delta set$ that contains both $A_1$ and $A_2$. Hence the measure of addend in the last line is both zero $\endgroup$ – desperatemuch Sep 9 '15 at 21:14
1
$\begingroup$

This might help: $G - (A \cup B)=(G_1 \cup G_2) - (A \cup B)$ $=(G_1 \cup G_2) \cap A^C \cap B^C$ $= [(G_1 \cup G_2) \cap A^C] \cap B^C$ $=[(G_1 \cap A^C) \cup (G_2 \cap A^C)] \cap B^C$ $=[(G_1 \cap A^C) \cap B^C] \cup [(G_2 \cap A^C) \cap B^C]$ $\subseteq (G_1 - A) \cup (G_2 - B)$

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

You can work out the last two lines like the following: prove first that $$ G-(A_1\cup A_2)=(G_1-(A_1\cup A_2))\cup(G_2-(A_1\cup A_2)) $$ and then do the estimate \begin{align} \mu(G-(A_1\cup A_2))&=\mu((G_1-(A_1\cup A_2))\cup(G_2-(A_1\cup A_2)))\le \\ &\le \mu(G_1-(A_1\cup A_2))+\mu(G_2-(A_1\cup A_2))\le\\ &\le \mu(G_1-A_1)+\mu(G_2-A_2)=0+0=0 \end{align} where in the first inequality you use the subadditivity and in the second one the monotonicity of the outer measure $\mu$ (since $G_k-(A_1\cup A_2)\subset G_k-A_k$).

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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