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I would like to find exemples to show and demonstrate that each of the statements of the definition of:
-measure
$\mu\left(\emptyset \right)=0$
$\mu \left( \bigcup A_n\right)=\sum \mu \left( A_n\right)$ $\mu$ 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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    I think you need to write down here the "statements of the definition" to be used.2012-10-07

2 Answers 2

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$\mu(\emptyset)=0$ 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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    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 you2012-10-16
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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.2012-10-07