The meaning of $\mu^\ast(A)=\infty$, where $\mu^\ast$ is outer measure.

Question

In the measure theory written by Axler, the outer measure $\mu^\ast : \mathcal P(\mathbb R)\to [0,\infty]$ is defined by $$\mathcal P(\mathbb R)\ni A\mapsto \inf \left\{ \sum_{k=1}^\infty L(I_k) \ \middle| \ \{I_k\}_{k=1}^\infty \mathrm{is \ a \ sequence \ of \ open \ intervals \ s.t. }A\subset \displaystyle\bigcup_{k=1}^\infty I_k \right\}\in[0,\infty]$$, where $\mathcal P(\mathbb R)$ is the set of all subsets of $\mathbb R$, $L(\cdot)$ represents the length.

I'm considering about the case of $\mu^\ast(A)=\infty.$

Seeing What does $\inf A=\infty$ mean?, I learned that if the infimum of the subset of $\mathbb R$ is $\infty$, then the set is empty.

So, does $\mu^\ast(A)=\infty$ mean $$\left\{ \sum_{k=1}^\infty L(I_k) \ \middle| \ \{I_k\}_{k=1}^\infty \mathrm{is \ a \ sequence \ of \ open \ intervals \ s.t. }A\subset \displaystyle\bigcup_{k=1}^\infty I_k \right\}=\emptyset\ ?$$

And if so, I think that the meaning is ; there doesn't exist the sequence of open intervals whose union contains $A$.

Is my interpretation correct ?

Answer 1

No, there always exists such a sequence of open intervals: $(-1,1),\,(-2,2)\,,\dots$.

Given $A$, we consider a set of sums $\sum_{k=1}^\infty L(I_k)$ any one of which may be real, or may take the value $\infty$.

The meaning is:

There doesn't exist a sequence of open intervals $I_k$ whose union contains $A$ and at the same time satisfies $\sum_{k=1}^\infty L(I_k) \in \mathbb{R}$.

Remark: What Ian has written in the answer to which you have linked above is correct, but is not applicable to our situation with the outer measure. Ian wrote: "For subsets of $\mathbb{R}$ the only way for the $\inf$ to be $\infty$ is for the set to be empty." But here we consider subsets of $\mathbb{R} \cup \{\infty\}$. That is,

$$\emptyset \;\; \not= \;\; \left\{ \sum_{k=1}^\infty L(I_k) \right\}_{\displaystyle \cup_k I_k \supset A} \;\; \subset \;\; \mathbb{R} \cup \{\infty\}$$ holds for each $A$.