# how to prove $m^*(E_1\cup E_2 )=m^*(E_1)+m^*(E_2)$?

$G_1,G_2$are open sets,$G_1\cap G_2=\emptyset$,$E_1\subset G_1,E_2\subset G_2$,how to prove $m^*(E_1\cup E_2 )=m^*(E_1)+m^*(E_2)$?

$m^*(E)$ means the outer measure of $E$.

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In the answer I'm giving you a list of hints to prove it in small steps.

1. From your hypothesis it follows $E_1\cap E_2=\emptyset$.

By definition, and denoting by $l(I)$ the lenght of the interval $I$:

$$m^*(E_1\cup E_2) = \inf_{(E_1\cup E_2) \subset \cup I_n} \sum l(I_n),$$

2. For any $I_n$ we can find $I_{n,1}\subset I_n$ and $I_{n,2}\subset I_n$ such that $I_{n,1}\cap I_{n,2}= \emptyset$ and $E_1 \subset \cup I_{n,1}, \: E_2 \subset \cup I_{n,2}$. You will want to use point 1.

3. Holds $l(I_n) \geq l(I_{n,1}) + l(I_{n,2})$.

4. Use points 2. and 3. to show that, since we are reduced to length of intervals: $$m^*(E_1 \cup E_2) \geq m^*(E_1) + m^*(E_2).$$

5. Conclude that equality holds.

If you have any problem just ask!

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Thank you,actually point 3 is the conclusion what we want.But it seems that point 2 and 3 is not obvious.Could you give more details? –  user39843 Oct 16 '12 at 16:37