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I would like to find exemples to show and demonstrate that each of the statements of the definition of:
$\mu\left(\emptyset \right)=0$
$\mu \left( \bigcup A_n\right)=\sum \mu \left( A_n\right)$ $\mu$ defined from a subset of partition of a given set to $\left[0;+\infty\right]$
are not reduntant.

Edit1: I mean Im looking for "applications" which can fit the finite additivity but not that associates the empty set to zero. Or the opposite.

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Take a look at measure theretic probability theory? – Michael Greinecker Oct 7 '12 at 15:00
I edited the questions 'cause I noticed that it was not clear at all.Sorry – Laura Oct 7 '12 at 15:04
I think you need to write down here the "statements of the definition" to be used. – GEdgar Oct 7 '12 at 15:12
up vote 0 down vote accepted

$\mu(\emptyset)=0$ is not redundant because we can have a trivial "measure" which is identically infinite. It is redundant in that it is equivalent to $\exists x \mu(x)<\infty$ (because then $\mu(x)=\mu(\emptyset)+\mu(x)$ and we can subtract $\mu(x)$ to obtain $\mu(\emptyset)=0$).

Countable additivity is not redundant, either. For example, consider the example of a measure $\mu$ defined for all subsets of $\bf R$: $\mu(A)=\infty$ if $0\in \operatorname {cl} A$, $\mu(A)=0$ otherwise. You can see that it is finitely additive (because closure of a finite union is the union of closures), but not countably additive (because the union of singletons $\{1/n\}$ has $0$ in its closure).

For a finite example, see this question.

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The two measures you seggested still works to prove the needing of the other conditions.If I modify the equality of the second condition in a $\leq$ and add the third condition who leads to the outer measure: $ B \subset A \Rightarrow \mu \left(B\right) \leq \mu \left(A\right)$ is still there a measure to show that this isn't redundant? – Laura Oct 16 '12 at 15:45
@Laura: I don't understand. Don't both examples satisfy this condition? Or are you asking if there's a "measure" which satisfies the two conditions and is not monotone? In this case, the answer is no, because if $A\supseteq B$, then $\mu(A)=\mu(B)+\mu(A\setminus B)\geq \mu(B)$ (so any finitely additive nonnegative set function is necessarily monotone). – tomasz Oct 16 '12 at 15:54
Yes, I noticed that the measures satisfy the conditions but I was still wondering if it there was a measure (yet trivial o paradoxal) not monotone.But you've just answered.Thank you – Laura Oct 16 '12 at 15:59

The existence of Lebesgue measure yields elegance and economy of thought that informs much of modern analysis. The technique of its construction is somewhat arcane, but the consequences are huge and important.

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I'm sorry may have explained it wrongly.I'm going to edit the questions in order to be clear. – Laura Oct 7 '12 at 15:00

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