Property of Lebesgue Measure. Monotonicity of measure

Question

I have been reading that a property of Lebesgue Measurable set $E$ is that, for each $\epsilon \lt 0$, there exist a family of open and disjoint intervals such that $$E \subseteq \cup_{n=1}^{\infty} I_n \space,\space m(\cup_{n=1}^{\infty} I_n ) \leq m(E) +\epsilon$$

Now, it’s well know that there exist a covering of $E$ by open intervals , not necessarily disjoint, such that the proposition above holds. It’s the disjoint part due to the monotonicity of measure?

That is, if we “eliminate” intervals of {$I_n$} that intersect each other, we can form a family {$I_k$} of disjoint open intervals with $\cup_{k=1}^{\infty}I_k \subseteq \cup_{n=1}^{\infty} I_n $. Then monotonicty of measure applies.

It’s my understanding correct? Thanks.

Answer 1

An important theorem of Real Analysis is that any open subset of the real line consists of a disjoint union of at most countably-infinitely-many open intervals. By "intervals," I mean any subset $I$ of $\Bbb R$ with the following property: $$\forall x\in I,\forall y\in I,\forall z\in\Bbb R,(x<z<y)\implies(z\in I).\tag{$\star$}$$ Open intervals then include the empty set, bounded open intervals of the form $(a,b)$ for some $a,b\in\Bbb R$ with $a<b,$ open rays of the form $(-\infty,c)$ or $(c,\infty)$ for some $c\in\Bbb R,$ and the whole real line.

Since an arbitrary union of open sets is necessarily open, then any union (countable or otherwise) of open intervals will necessarily be equal to a disjoint union of at most countably-many open intervals. That is, if we start with any family open intervals which aren't necessarily pairwise disjoint, we're guaranteed to be able to replace it with a family of at most countably-infinitely-many open intervals which are pairwise disjoint and which have the same union as the original family.

It's worth noting, though, that the two families may have no intervals in common, so the idea of "eliminating intervals that intersect" need not work. Consider for example the family of intervals $I_n:=\left(\frac{1}{n},2\right)$ to see why.

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If you aren't familiar with the important theorem I mentioned above, here's an outline of how one can prove it for an arbitrary open subset $O$ of $\Bbb R.$

• Define a relation $\sim$ on $O$ by $x\sim y$ if there exist some $a,b\in\Bbb R$ such that $x,y\in(a,b)$ and $(a,b)\subseteq O.$
• Show that $\sim$ is reflexive, symmetric, and transitive on $O,$ so is an equivalence relation on $O.$
• Observe that since $\sim$ is an equivalence relation on $O,$ then denoting the $\sim$-equivalence classes by $\langle x\rangle_\sim:=\{y\in O\mid x\sim y\},$ we have that the set $\mathcal{E}_\sim:=\bigl\{\langle x\rangle_\sim\mid x\in O\bigr\}$ of all $\sim$-equivalence classes is a partition of $O.$
• Show that for any $x\in O$ and any $y\in \langle x\rangle,$ there is some $\varepsilon>0$ such that $(y-\varepsilon,y+\varepsilon)\subseteq \langle x\rangle_\sim,$ meaning that each $\sim$-equivalence class is open.
• Show that each $\sim$-equivalence class is an interval--that is, if $I=\langle x\rangle_\sim$ for some $x\in O,$ then $(\star)$ holds.
• Use the fact that the set $\Bbb Q$ of rational numbers is countably-infinite to note that there is a bijection $f$ from the set $\Bbb N$ of positive integers to $\Bbb Q$.
• Use the fact that any non-empty open interval has a rational element to show that for each $x\in O,$ there is some $n\in\Bbb N$ such that $f(n)\in\langle x\rangle_\sim.$ Thus by the well-ordering property of $\Bbb N,$ there is some least such $n\in\Bbb N,$ say $n_{\langle x\rangle_\sim}.$
• Show that the mapping $\langle x\rangle_\sim\mapsto n_{\langle x\rangle_\sim}$ is an injection $\mathcal{E}_\sim\to\Bbb N,$ whence $\mathcal{E}_\sim$ is at most countably-infinite.
• Note that since $\mathcal{E}_\sim$ is a partition of $O,$ then $O$ is the union of all elements of $\mathcal{E}_\sim,$ which are pairwise disjoint, completing the proof.