pachaquis3s

2022-06-20

Additivity of outer measure if one of the sets is open.
Suppose $A$ and $G$ are disjoint subsets of $\mathbb{R}$ and $G$ is open. Then $|A\cup G|=|A|+|G|$.
*Note that: |$A$| and l($I$) denote the outer measure of $A$ and the length of interval $I$ respectively.
I wonder that, for the part of proof to the case $G=\left(a,b\right)$ and the deduced inequality:
$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}l\left({I}_{n}\right)& =\sum _{n=1}^{\mathrm{\infty }}\left(l\left({J}_{n}\right)+l\left({L}_{n}\right)\right)+\sum _{n=1}^{\mathrm{\infty }}l\left({K}_{n}\right)\\ & \ge |A|+|G|\end{array}$
How can the above inequality show that $|A\cup G|\ge |A|+|G|$?
I can only deduce from $\left(A\cup G\right)\subset \bigcup _{n=1}^{\mathrm{\infty }}{I}_{n}$, that:
$|A\cup G|\le |\bigcup _{n=1}^{\mathrm{\infty }}{I}_{n}|\le \sum _{n=1}^{\mathrm{\infty }}l\left({I}_{n}\right)\phantom{\rule{0ex}{0ex}}$
However, along with:
$\sum _{n=1}^{\mathrm{\infty }}l\left({I}_{n}\right)\ge |A|+|G|$
then both the inequalities are on the same side, so I cannot link them to prove what I desire. How did author reach his conclusion?

marktje28

Recall the definition of outer measure:
$|A\cup G|=inf\left\{\sum l\left({I}_{n}\right):A\cup G\subset \cup {I}_{n}\right\}.$
Then you already know for all $\left\{{I}_{n}\right\}$ covering $A\cup G$,
$\sum l\left({I}_{n}\right)\ge |A|+|G|,$
after taking infimum over these intervals you get the result.

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