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I'm considering the following situation. Suppose $\mathcal{A}$ is a semiring of sets in $\mathbb{R}^n$ of the form $(a_1,b_1]\times\cdots\times(a_n,b_n]$. Then there is the unique Lebesgue measure $\lambda$ such that $\lambda((a_1,b_n]\times(a_n,b_n])=\prod_{i=1}^n (b_i-a_i)$ for all the sets in the semiring. I'll denote the completion of $\lambda$ as $\lambda$ as well.

If I denote by $\mathcal{C}$ the subfamily of boxes with equal side lengths, i.e., elements of form $(c_1,c_1+L]\times\cdots\times(c_n,c_n+L]$. Then for any $C\subseteq \mathbb{R}^n$, why is it that $$\lambda^*(C)=\inf\left\{\sum_i \lambda(A_i)\mid C\subseteq\bigcup_i A_i, \ A_i\in\mathcal{C}\right\}?$$

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For $A\subseteq \mathbb{R}^n$, write $$ \mu ^*(A)\equiv \inf \left\{ \sum _i\lambda (A_i)|\, A\subseteq \bigcup _i,A_i\in \mathcal{C}\right\} $$

You know right away that $\lambda ^*(A)\leq \mu ^*(A)$ (because you are taking the infimum over more sets in the former case).

For the other direction, let $\varepsilon >0$ be arbitrary, and let $A_i\in \mathcal{A}$ be such that $A\subseteq \bigcup _iA_i$. Then, we can find finitely many $C_{i,k}\in \mathcal{C}$ such that $A_i\subseteq \bigcup \limits_{k=1}^{n_i}C_{i,k}$ and $\sum\limits_{k=1}^{n_i}\lambda (C_{i,k})-\lambda (A_i)<\varepsilon/2^i$. (That is, each $A_i$ is almost a finite disjoint union of squares). It follows that (after reindexing the $C_{i,k}$’s) $$ \sum _j\lambda (C_j)<\sum _i\lambda (A_i)+2\varepsilon . $$ It follows that $\mu ^*(A)\leq \lambda ^*(A)$, and hence $\mu ^*(A)=\lambda ^*(A)$.

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Are you sure about finitely many ‘squares’? How will you write a $1\times\sqrt 2$ rectangle as a union of finitely many pairwise disjoint squares? (By the way, hi.) – Brian M. Scott Nov 19 '11 at 2:51
@BrianM.Scott Of course. That was dumb. In any case, you can still approximate each "rectangle" with finitely many "squares" with error less than $\varepsilon$, in which case, essentially the same argument should go through. I'll modify the answer to reflect this. (Hi!) – Jonathan Gleason Nov 19 '11 at 3:00
Hey Brian I can't help but notice that your definition of $\mu$* is exactly the same as $\lambda^{}$* if we just replace your $A$ by $C$. Can you please explain more? Thanks. – Xiaowen Li Nov 22 '11 at 4:00
@XiaowenLi That is exactly correct. You might say that $\lambda ^*$ is the outer-measure generated by $\mathcal{A}$ and $\mu ^*$ is the outer-measure generated by $\mathcal{C}$. The content of the statement is that these two outer-measures are the same. – Jonathan Gleason Nov 22 '11 at 14:34

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