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Let $v : \mathcal {P}(\mathbb R) \to [0,\infty]$ a set-function such that $\displaystyle{ v(F \cup E) \leq v(E) + v(F) \quad \forall E,F \subset \mathbb R}$.

Is always true that: $ \displaystyle{ v\left(\bigcup_{n=1}^{\infty} A_n\right) \leq \sum_{n=1}^{\infty} v(A_n)}$ ?

I think this is not but I can't find a counterexample. Any help?

Thank's in advance!

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Thank you both for your answers! –  passenger May 11 '12 at 19:33
    
However if $v$ is a measure then it is true. –  leo May 15 '12 at 0:56
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3 Answers

up vote 14 down vote accepted

Let $\nu\colon \mathcal{P}(\mathbb{R})\to [0,\infty]$ be the function $\nu(A) = 0$ if $A$ is finite and $\nu(A) = \infty$ if $A$ is infinite. Then $\nu(\mathbb{N}) = \infty$ but $\sum_{n\in \mathbb{N}}\nu(\{n\}) = 0$.

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Heh; we're very close. (-: –  Arturo Magidin May 11 '12 at 19:30
    
You're not kidding! –  froggie May 11 '12 at 19:31
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Define $v$ to be $0$ if the set is bounded, and $\infty$ if the set is unbounded. If $F$ and $E$ are bounded, then so is $F\cup E$, so $v(F\cup E)\leq v(E)+v(F)$ is satisfied.

Now let $A_n = [-n,n]$. Then $\bigcup\limits_{n=1}^{\infty}A_n = \mathbb{R}$ is unbounded, and $v(A_n) = 0$ for all $n$, so $$v\left(\bigcup_{n=1}^{\infty}A_n\right) = \infty \gt 0 = \sum_{n=1}^{\infty}v(A_n).$$

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The purpose of that question is to show that if $\nu$ subadditive it isn't implies that $\nu$ $\sigma$-additive. The answers above shows that fact.

It's well known that, if $\nu$ is a measure, then:

$$ \nu \ \sigma\text{-additive} \ {\Longrightarrow} \ \nu \ \sigma\text{-subadditive} \ {\Longrightarrow} \ \nu \ \text{subadditive} $$

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