Prove the union of uncountably many closed intervals is measurable Problem

$\{I_\alpha\}_{\alpha\in A}$ are closed intervals with positive length, prove $E=\cup_{\alpha\in A}I_\alpha$ is measurable.

Since $E$ can be written as $(\cup_{\alpha\in A}I_\alpha^\circ)\cup(\cup_{\alpha\in A}\partial I_\alpha\setminus\cup_{\alpha\in A}I_\alpha^\circ)$ and $\cup_{\alpha\in A}I_\alpha^\circ$ is measurable, it suffices to show $\cup_{\alpha\in A}\partial I_\alpha\setminus\cup_{\alpha\in A}I_\alpha^\circ$ is measurable. My intuition is that $\cup_{\alpha\in A}\partial I_\alpha\setminus\cup_{\alpha\in A}I_\alpha^\circ$ is just the union of countably many points and is therefore measurable, but I don't know how to prove this.
 A: You want to show that
$$\cup_{\alpha\in A}\partial I_\alpha\setminus\cup_{\alpha\in A}I_\alpha^\circ$$
is a countable set. If we call that set $S$, then $S=L\cup R$ where $L$ is the set of all points $x$ such that $x$ is the left endpoint of some $I_\alpha$ but $x$ is not in the interior of any $I_\alpha$, and $R$ is defined similarly but with "right" instead of "left".
To see that $L$ is countable, define a mapping $f:L\to\mathbb Q$ so that, for $x\in L$, $f(x)$ is some rational number $q\gt x$ such that $[x,q]$ is contained in some $I_\alpha$. Observe that $f$ is injective. Since $\mathbb Q$ is countable, this shows that $L$ is countable.
The proof that $R$ is countable is similar.
Therefore $\bigcup_{\alpha\in A}I_\alpha$ is the union of an open set and a countable set, so it's an $F_\sigma$ set.
In fact it can be shown that $\bigcup_{\alpha\in A}I_\alpha=\bigcup_{\alpha\in B}I_\alpha$ for some countable set $B\subseteq A$. Namely, for each point $x\in S$ choose some $I_\alpha$ having $x$ as an endpoint; and for each interval with rational endpoints which is covered by some $I_\alpha$, choose some $I_\alpha$ which covers it.
