I was reading the proposition
Let ${A_n}$ be a countable collection of sets of real numbers. Then $m^*(\cup A_n)\leq \sum m^*(A_n)$.
Proof: fix $\epsilon >0$ and for every n choose a set of open intervals $\{I_m^n\}_{m=1}^\infty$ that cover $A_n$ and such that $\sum_m l({I_m^n}) \leq m^*(A_n) - \frac{\epsilon}{2^n}$...
What does the notation with dual indexes $n$ and $m$ of $\{I_m^n\}_{m=1}^\infty$? mean?
I think it means, for every set $A_n$, there is a corresponding $I_n$ and in total there are $m$ open intervals $I_n$ that whose union cover $A_n$?
The set with these $m$ numbered from one to infinity open intervals $I_n$ is $\{I_m^n\}_{m=1}^\infty$?
You want to remember for the remainder of the proof that the the $I^n_m$ "belong to" $A_n$, the $m$ is just the index of the countable family of intervals. $I(n)_m$ would have been just as valid, or some other notation. You can do it for each fixed $n$ and for that $n$ we call the intervals $I^n_m$, where $m=1,2,3,\ldots$ etc.
Of course, in total we are choosing countably many such intervals, as a countable union of countable sets is still countable, and all of them together can now be denoted $\{I^n_m: n,m \in \Bbb N\}$, e.g.