I am looking at the proof that the outer Lebesgue measure that is defined by $$m^*(A)=inf\{\sum_{n=1}^{\infty}v(R_n):A \subset\cup_{n=1}^{\infty}R_n , ( \text{ where } R_n \text{ are open rectangles}) \}$$ is actually an outer measure.

($v(R_n)$, the volume of R is the product of the lengths $I_j$.)

To show that it is an outer measure, we have to show the monotonicity and the subadditivity.

For the subadditivity:

If $A_n \subset \mathbb{R}$ then $$m^* \left ( \cup_{n=1}^{\infty} A_n\right ) \leq \sum_{n=1}^{\infty}m^*(A_n)$$

We suppose that $m^*(A_n)<+\infty, \forall n$.

Let $\epsilon>0$. Then for each $n$ there are rectangles $R_j^n,j=1,2,...$ such that $A_n \subset \cup_{j=1}^{\infty}R_j^n$ and $\sum_{j=1}^{\infty}v(R_j^n)<m^*(A_n)+\frac{\epsilon}{2^n}$.

Then $\cup_{n=1}^{\infty}A_n\subset\cup_{j,n}R_j^n$


$$m^* \left ( \cup_{n=1}^{\infty}A_n \right ) \leq \sum_{j,n}v(R_j^n)<\sum_{n=1}^{\infty}m^*(A_n)+\epsilon \\ \Rightarrow m^* \left ( \cup_{n=1}^{\infty}A_n\right ) \leq \sum_{n=1}^{\infty}m^*(A_n)$$

Could you explain me this proof??


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