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 $G=(a,b)$ and the deduced inequality:

$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty}}l({I}_{n})& =\sum _{n=1}^{\mathrm{\infty}}(l({J}_{n})+l({L}_{n}))+\sum _{n=1}^{\mathrm{\infty}}l({K}_{n})\\ & \ge |A|+|G|\end{array}$

How can the above inequality show that $|A\cup G|\ge |A|+|G|$?

I can only deduce from $(A\cup G)\subset \bigcup _{n=1}^{\mathrm{\infty}}{I}_{n}$, that:

$|A\cup G|\le \left|\bigcup _{n=1}^{\mathrm{\infty}}{I}_{n}\right|\le \sum _{n=1}^{\mathrm{\infty}}l({I}_{n})\phantom{\rule{0ex}{0ex}}$

However, along with:

$\sum _{n=1}^{\mathrm{\infty}}l({I}_{n})\ge |A|+|G|$

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?

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 $G=(a,b)$ and the deduced inequality:

$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty}}l({I}_{n})& =\sum _{n=1}^{\mathrm{\infty}}(l({J}_{n})+l({L}_{n}))+\sum _{n=1}^{\mathrm{\infty}}l({K}_{n})\\ & \ge |A|+|G|\end{array}$

How can the above inequality show that $|A\cup G|\ge |A|+|G|$?

I can only deduce from $(A\cup G)\subset \bigcup _{n=1}^{\mathrm{\infty}}{I}_{n}$, that:

$|A\cup G|\le \left|\bigcup _{n=1}^{\mathrm{\infty}}{I}_{n}\right|\le \sum _{n=1}^{\mathrm{\infty}}l({I}_{n})\phantom{\rule{0ex}{0ex}}$

However, along with:

$\sum _{n=1}^{\mathrm{\infty}}l({I}_{n})\ge |A|+|G|$

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

Beginner2022-06-21Added 22 answers

Recall the definition of outer measure:

$|A\cup G|=inf\{\sum l({I}_{n}):A\cup G\subset \cup {I}_{n}\}.$

Then you already know for all $\{{I}_{n}\}$ covering $A\cup G$,

$\sum l({I}_{n})\ge |A|+|G|,$

after taking infimum over these intervals you get the result.

$|A\cup G|=inf\{\sum l({I}_{n}):A\cup G\subset \cup {I}_{n}\}.$

Then you already know for all $\{{I}_{n}\}$ covering $A\cup G$,

$\sum l({I}_{n})\ge |A|+|G|,$

after taking infimum over these intervals you get the result.