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An additive measure $\mu$ on $R$ has the property that for pairwise disjunct $A_i \in R$ with $\left(\bigcup_{n=1}^\infty A_i\right)\in R$: $$ \mu\left(\bigcup_{n=1}^\infty A_i\right) \geq \sum_{n=1}^\infty \mu(A_n)$$

Referring to this proof: http://www.mathe.wiwi.uni-sb.de/skripte/mit/skriptmi_Kap1-2-Anhang.pdf (p.23)

What's wrong in reversing the argument like this:

$$ \mu\left(\bigcup_{i=1}^m A_i\right) = \sum_{i=1}^m \mu(A_i) \leq \sum_{i=1}^\infty \mu(A_i) \Longrightarrow_\infty \mu\left(\bigcup_{n=1}^\infty A_i\right)\leq \sum_{i=1}^\infty \mu(A_i)$$

The first $=$ comes from additivity, the $\leq$ from $\mu \geq0$ and the implication from taking the limit.

I guess it's the limit but I don't know why...

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  • $\begingroup$ You seem to assume that $\lim\limits_{m\to\infty}\mu\left(\bigcup\limits_{i=1}^m A_i\right) = \mu\left(\bigcup\limits_{i=1}^\infty A_i\right)$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Oct 21 '14 at 17:22
  • $\begingroup$ You refer several times to $\bigcup\limits_{n=1}^{\text{something}} A_i$, with $n$ as the index running from $1$ to something and $i$ as the subscript in $A_i$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Oct 21 '14 at 17:23
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Counterexample:

Prescribe $\mu:\wp\left(\mathbb{N}\right)\rightarrow\left[0,\infty\right]$ by $A\mapsto0$ if $A$ is finite and $A\mapsto\infty$ otherwise.

Then $\mu$ is additive with: $$\mu\left(\bigcup_{n=1}^{\infty}\left\{ n\right\} \right)=\mu\left(\mathbb{N}\right)=\infty>0=\sum_{n=1}^{\infty}\mu\left(\left\{ n\right\} \right)$$

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and the implication from taking the limit.

This assumes that $$\mu\left(\bigcup_{n=1}^{\infty} A_n\right)$$ is the limit when $m\to\infty$ of $$\mu\left(\bigcup_{n=1}^m A_n\right),$$ which is kind of the whole point.

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  • $\begingroup$ So why does the other one work and this is wrong? $\endgroup$ – Greg P. Oct 21 '14 at 16:52
  • $\begingroup$ @GregP. For the other one, the inequality $$\mu \left( \bigcup_{n \geq 1} A_n \right) \geq \mu \left( \bigcup_{n=1}^m A_n \right)$$ is used (which holds for any (sub)additive measure). In contrast, for this one, we want to estimate $\mu (\bigcup_{n \geq 1} A_n)$ from above and therefore the mentioned inequality is useless. $\endgroup$ – saz Oct 21 '14 at 17:19
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The valid argument uses a limit of real numbers, not of sets. The limit of the measure behaves as you'd expect since the measure is a real number, and from ordinary real analysis we have the definition of an infinite sum:

$$\sum_{i=0}^\infty a_i = \lim_{m\rightarrow \infty}\sum_{i=0}^m a_i;$$

$$\sum_{i=0}^\infty \mu(A_i) = \lim_{m\rightarrow \infty}\sum_{i=0}^m \mu(A_i).$$

But with a "limit" of sets you need to be more careful.

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