# Using the concept of $G_\delta$ sets, show that the union of two measurable sets is measurable.

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.

• 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. – Math1000 Sep 9 '15 at 20:37
• @Math1000 That depends. If measurability is defined a la Caratheodory, then closure under countable unions is nontrivial. – Noah Schweber Sep 9 '15 at 20:45
• @Math1000 would like to show that there is a $G_\delta$ set $G$ that contains the $A_1\cup A_2$ and that $\mu(G-(A_1\cup A_2))=0$ – desperatemuch Sep 9 '15 at 20:59
• The last two lines: why $\mu$ of the intersection is equal to the sum and why "both are zeros"? – A.Γ. Sep 9 '15 at 21:01
• @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 – desperatemuch Sep 9 '15 at 21:14

This might help: $G - (A \cup B)=(G_1 \cup G_2) - (A \cup B)$ $=(G_1 \cup G_2) \cap A^C \cap B^C$ $= [(G_1 \cup G_2) \cap A^C] \cap B^C$ $=[(G_1 \cap A^C) \cup (G_2 \cap A^C)] \cap B^C$ $=[(G_1 \cap A^C) \cap B^C] \cup [(G_2 \cap A^C) \cap B^C]$ $\subseteq (G_1 - A) \cup (G_2 - B)$
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 $G_k-(A_1\cup A_2)\subset G_k-A_k$).