Is a countable union of measurable sets measurable? If the sum of measures of those measurable sets is finite, then their union is also measurable. But if the sum of measures of measurable sets is infinite, then the measure of their union may or may not be finite. Is their union still measurable?
[Math] Countable union of measurable sets is measurable
lebesgue-measuremeasurable-setsmeasure-theory
Related Solutions
We start with a little Lemma:
Lemma. Let $E\subseteq \Bbb R$. If $H\supseteq E$ is a $G_\delta$ set (countable intersection of open sets) such that $$m(H)=m^\ast(E),$$ then for every $C\subseteq\Bbb R$ $$m^\ast(H\cap C)=m^\ast(E\cap C).$$
Proof. Let $C\subseteq\Bbb R$. In the following the superscript $^c$ means complement. $$\begin{align*} m^\ast(H\cap C) &\leq m^\ast(H\cap C\cap E\cap C)+m^\ast((H\cap C)\setminus (E\cap C))\\ &= m^\ast(E\cap C) + m^\ast((H\cap C)\cap (E\cap C)^c)\\ &= m^\ast(E\cap C) + m^\ast(C\cap (H\setminus E))\\ &\leq m^\ast(E\cap C) + m^\ast(H\setminus E)\\ &= m^\ast(E\cap C) \end{align*}$$ The inequality $m^\ast(H\cap C)\geq m^\ast(E\cap C)$ comes free by the monotony of the outer measure since $H\supseteq E$.
Proof of $m^\ast(E\cap (A\cup B))\geq m^\ast(E\cap A)+m^\ast(E\cap B)$.
Pick $H\supseteq E$ a $G_\delta$ set so that $m(H)=m^\ast(E)$. Then $$\begin{align*} m^\ast(E\cap (A\cup B)) &= m^\ast(H\cap (A\cup B)) &&\text{by the Lemma}\\ &= m(H\cap A) + m(H\cap B) &&\text{($^\ast$)}\\ &\geq m^\ast(E\cap A) + m^\ast(E\cap B) &&\text{by the monotony of the outer measure.} \end{align*}$$ ($^\ast$) because here we are dealing with measurable sets (of finite measure).
Observation. Notice that such a $G_\delta$ set $H$ always exist even if $E$ is unbounded.
For your last Q. If $f:r\to R$ is increasing and $r\in R,$ then $f^{-1}(r,\infty)$ is $\phi, $ or is $R, $ or is $(\;\inf f^{-1}(r,\infty)\;), $ or is $[\min f^{-1}\{r\},\infty\;), $ which is in all cases a Borel set. Similarly, for $s\in R, $ the set $f^{-1}(-\infty,s)$ is Borel. So $f^{-1}(s,r)$ is Borel. Every open set of reals is a countable union of open intervals (see below); therefore $f^{-1}U$ is Borel for every open $U.$ All of this can be done in ZF.
To show in ZF that every open $U\subset R$ is a countable union of pair-wise disjoint open intervals (including possibly unbounded ones): For convenience let In $(x,y)=(x,y)\cup (y,x)$ for $x,y \in R.$ (In $(x,y)$ is the open interval between $x$ and $y.$) For any open $U\subset R$ and $x,y\in U$ let $x\equiv_U y\iff$ In$(x,y)\subset U.$ We easily show this is an equivalence relation on $U,$ and that each equivalence class $[x]_{\equiv_U}$ is an open interval. Now $U=\cup \{[x]_{\equiv_U} :x\in Q\cap V\}.$
Best Answer
Yes, it is. The measurable sets form a $\sigma$-algebra, so are closed under countable unions (and intersections). Having a finite measure is not part of the definition of being measurable; it's the reason why we allow the value $+\infty$ for measures.