2
$\begingroup$

$\textbf{Definition:}$ Let $A\in \mathcal{P}(\mathbb{R})$. In the real analysis book I am using by Royden, he says consider $\lbrace I_n \rbrace$ to be a countable collection of open intervals which cover $A$ (aka $A\subseteq \bigcup I_n$) and denotes the outer measure to be as follows: $m^*(A)=\underset{A\subseteq\cup I_n}{\inf}\sum l(I_n)$.

Both the notation he uses and the wording (particularly with the word $\textbf{consider}$) is confusing me. Does he mean let $\lbrace I_n \rbrace$ cover $A$ by an arbitrary countable set of open intervals? Does he mean $\lbrace I_n \rbrace$ is associated with set builder notation which is left out? I tried to rewrite his definition below as best I could below and am wanting to see if the definition I am using is correct.

$\textbf{Question:}$ Will the definition I made up below match up with his definition?

$\textbf{Made Up Definiton:}$ Let $A\in \mathcal{P}(\mathbb{R})$. Then, we denote the outer measure of $A$ as follows: $m^*(A):= \inf\{{\sum_{j\in J}{l(I_j)|\exists \text{ a countable collection of open intervals }\lbrace I_j \rbrace_{j\in J}: A \subseteq \bigcup_{j\in J} I_j}\}}$.

$\endgroup$

2 Answers 2

2
$\begingroup$

They're equivalent.

The pedantic way of expressing this is the following:

Define $m^*:\wp(\mathbb R)\to[0,\infty]$ by

$$A\mapsto\inf\left\{\sum_{j\in J}l\left(I_j\right):\left\{I_j\right\}_{j\in J}\text{ is a countable collection of open intervals that covers }A.\right\}.$$

$\endgroup$
1
$\begingroup$

Observe that there always exist a countable collection of open intervals covering $A$. The existential qualifier in your definition does not make much sense to me. What the definition does is consider all countable collections of open intervals $\lbrace {I_j}\rbrace$ covering $A$ and take the infimum of the sum of the lengths of the intervals in the collection. This may be what you are looking for:

$$ m^*(A)=\inf\Bigl\lbrace\sum_{j\in J}l(I_j):\lbrace I_j\rbrace\text{ is a countable collection of open intervals covering }A\Bigr\rbrace. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.