# Show that $\mu^*(\mathcal{U})=\sum_n \ell(I_n).$

Suppose $\mathcal{U}$ is an open set which is the union of a sequence of disjoint open intervals, $I_1,I_2,I_3,\dots$. Show that $$\mu^*(\mathcal{U})=\sum_n \ell(I_n).$$

Proof: Let $\mathcal{U}=\cup_{n=1}^\infty I_n$ be an open set such that $\cap_{n=1}^\infty I_n=\emptyset$ where $I_n$ is an open set for all n. We need to show $$\mu^*(\mathcal{U})=\sum_n \ell(I_n).$$ By subaddivitity, we have $$\mu^*(\mathcal{U}) \leq \sum_n \mu(I_n)=\sum_n \ell(I_n).$$ We need to show that $$\mu^*(\mathcal{U}) \geq \sum_n \ell(I_n).$$ By Caratheodory's Construction, \begin{equation*} \begin{aligned} \mu^*(\mathcal{U}) & =\mu^*\left(\cup_{n=1}^\infty I_n\right) \\ & =\mu^*\left(\cup_{n=1}^\infty I_n\cap I_1 \right) + \mu^*\left(\cup_{n=1}^\infty I_n \cap I_1^c\right) \\ & =\mu^*\left( I_1 \right) + \mu^*\left(\cup_{n=2}^\infty I_n\right) \\ & =\ell\left( I_1 \right) + \mu^*\left(\cup_{n=2}^\infty I_n\right) \\ \end{aligned} \end{equation*}

If we repeat the process enough times we get that \begin{equation*} \begin{aligned} \mu^*\left(\mathcal{U}\right) & =\sum_{n=1}^i \mu^*\left( I_n \right) + \mu^*\left(\cup_{n=i+1}^\infty I_n\right) \\ & =\sum_{n=1}^i \ell\left( I_n \right) + \mu^*\left(\cup_{n=i+1}^\infty I_n\right) \\ & \geq \sum_{n=1}^i \ell\left( I_n \right) \end{aligned} \end{equation*} Hence, $$\mu^*(\mathcal{U})=\sum_n \ell(I_n).$$

• There doesn't seem to be anything to prove here, if my interpretation of your notation is correct ($\mu^*$ being Lebesgue outer measure and $\ell$ being the length of an interval). – Math1000 Feb 29 '16 at 3:50
• Why do you say that there is nothing to prove? – Username Unknown Feb 29 '16 at 4:34
• You assumed $\mathcal U$ is a pairwise disjoint union of open intervals $I_n.$ Why then in the first step of the proof do you repeat that assumption with the indices now represented by $i$ rather than $n?$ Thi – zhw. Mar 2 '16 at 5:31
• I don't believe that I am. – Username Unknown Mar 2 '16 at 20:33
• You did at first, then edited it. $\cap_{n=1}\infty I_n = \emptyset$ is not what you want above. Do you know $\mu^*(\cup_{n=1}^N I_n )= \sum_{n=1}^Nl(I_n)$ for finitely many pairwise disjoint open intervals $I_n?$ – zhw. Mar 5 '16 at 18:48

If $\mathcal{U}$ is the disjoint union of countably many open intervals. Since each $I$ is measurable, and since $\mathcal{M}$ is a $\sigma-\textrm{algebra}$, it is closed under countable union. Then $\mathcal{U}$ is measurable, and hence $$m^*(\mathcal{U})=m(\mathcal{U})=m(\bigsqcup_{I \in \mathscr{C}} m(I))=\sum_{I \in \mathscr{C}} l(I)$$

However, we can go ahead and try to prove something much stronger:

Lemma 0 Every open subset of $\mathbb{R}$ is a countable disjoint union of open intervals, and is hence measurable.

First seperate the open subset $U \subset \mathbb{R}$ into components by imposing the equivalence relation $x~y$ iff $(x,y) \in U$. Use this to prove that each component is either an interval or a ray.

hint: define $a= \inf(U)$, $b=\sup(U)$. Let $x \in (a,b)$. Then find an open ball about $x$ so that it is between $a$ and $b$. The result will follow shortly thereafter.

Then utilize the countable basis for $\mathbb{R}$ defined by $$\mathbb{B}= \{\beta(q, 1/n) \mid q \in \mathbb{Q}, n \in \mathbb{N}\}$$, and use the fact that the rationals are dense in $\mathbb{R}$, and hence there are only countably many disjoint components.

Lemma 1 is due to James Belk:

$L(\mathcal{I})$ denotes the length of the interval.

Lemma 1: Let $S \subseteq \mathbb{R}$. Then $m^*(S)=\textrm{inf} \{ m(U) \mid U \textrm{is open}, S \subseteq U\}$.

proof: Let $x$ be the value of the infimum. Clearly $m^*(S) \leq m(U)$ by monotonicity, implying that $m^*(S) \leq x$. For the other side of the inequality: let $\epsilon>0$, and let $\mathscr{C}$ be an open cover of $S$ so that $$\sum_{I \in \mathscr{C}} L(I) \leq m^*(S)+\epsilon$$ Then $U \bigcup \mathscr{C}$ contains $S$, so

$$x \leq m(U) \leq \sum_{I \in \mathscr{C}} m(I)=\sum_{I \in \mathscr{C}} m(I) \leq m^*(S)+ \epsilon.$$

The result follows.

Lemma 2: Let $(x,\mathcal{M},\mu)$ be a measure space, and define $\mu^{*}:\mathbb{P}(x) \to [0,\infty]$ by $$\mu^{*}(s)= \textrm{inf}\{\mu(E) \mid E \in \mathcal{M}, \, S \subseteq E\}$$. Then for each $S \in \mathbb{P}(x)$, there exists $E \in \mathcal{M}$ so that $S \subseteq E$ and $\mu(E)=\mu^{*}(S)$.

Let $S \in \mathbb{P}(x)$. let $\{E_n\}$ be a sequence of measurable sets in $\mathcal{M}$ so that $S \subseteq E_n$ and $\mu(E_n) \leq \mu^{*}(S)+1/n$ for each $n \in \mathbb{N}$. Let $E= \cap_{n \in \mathbb{N}}E_n$. Then $E \in \mathcal{M}$, $S \subseteq E$, and $\mu(E) \leq \mu^{*}(S)+1/n$, implying that $\mu^{*}(S)=\mu(E)$.

Corollary: For all $S \subseteq \mathbb{R}$, $m^*(S)=\sum_{I \in \mathscr{C}} l(I)$ for some disjoint union of open intervals.

This follows by countable additivity. This result is a little bit stronger, and is pretty cool.

For measures in general, the components proof will suffice to show the measurability of open sets for all measure spaces that have a countable basis.