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

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 ?

• No, it means that for any sequence of open intervals $I_k$ whose union contains $A$ we have $\sum_k l(I_k) = \infty$. Commented Sep 23, 2022 at 4:15
• You got your implications backwards; the infimum of the empty set is $\infty$ because by voidness any real number is less than any number in the $\varnothing$ set, which forces the infimum to be $\infty.$ Commented Sep 23, 2022 at 18:45

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 $$\infty$$.
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 $$A$$.