How do I prove that counting measure is subadditive?

Asked Nov 07 '19 at 20:21

Active Nov 07 '19 at 20:21

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Particularly, I'm confused about, what cases do I need to explore?

$$\mu^* \bigg(\bigcup_{i=1}^{\infty} E_i \bigg)$$

then since each $E_i$ is either some finite natural number or $\infty$, then do I need to induct or something?

Or perhaps one considers:

asked Nov 07 '19 at 20:21

mavavilj

@JoséCarlosSantos Not exactly since that proof proves additivity, not subadditivity. – mavavilj Nov 07 '19 at 20:26
@mavavilj I wonder though if the cases given there are the cases that I would explore as well. – mavavilj Nov 07 '19 at 20:28
The answer to that question contains the answer to your question. – José Carlos Santos Nov 07 '19 at 20:28
@JoseCarlosSantos How are additivity and subadditivity same though? – mavavilj Nov 07 '19 at 20:30
They aren't! The person who asked di other question used the wrong expression. – José Carlos Santos Nov 07 '19 at 20:32
@JoséCarlosSantos What do you mean? – mavavilj Nov 07 '19 at 20:35
Read the answer! – José Carlos Santos Nov 07 '19 at 20:36
@JoséCarlosSantos $\sigma$-additivity is countable addivity, whereas I try to prove subadditivity. – mavavilj Nov 07 '19 at 20:38
I have nothing more to say about this. – José Carlos Santos Nov 07 '19 at 20:42

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