Can somebody tell me why the counting measure (so, if $S=P(X)$, then $\mu(A)$=infinity if $A$ isn't finite and $\mu(A)=$#$A$ if $A$ is finite) is a measure? (The second property of a measure isn't clear for me).
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Which one is the second property? Countable additivity? – Daniel Fischer Jun 26 '14 at 10:32
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sigma-addivity @DanielFischer – Roos Jansen Jun 26 '14 at 10:33
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3If $(A_k)_{k\in\mathbb{N}}$ is a pairwise disjoint family of sets, either only finitely many of the $A_k$ are nonempty, or infinitely many. If infinitely many are nonempty, $\bigcup A_k$ is an infinite set, and $\mu(A_k)\geqslant1$ for infinitely many $k$, so $\mu\left(\bigcup A_k\right)=\infty=\sum\mu(A_k)$. If only finitely many $A_k$ are nonempty, either at least one of the $A_k$ is infinite, in which case the union is infinite too, and the sum is infinite, since at least one of the terms is $\infty$. If all $A_k$ are finite, and only finitely many nonempty, almost all terms in the sum $=0$ – Daniel Fischer Jun 26 '14 at 10:40
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What do you mean by your last sentence? If all A_k are finite and only finitely many nonempty then the sum union is simply the sum of the terms in the different nonempty A_k's? – Roos Jansen Jun 26 '14 at 10:52
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2Comment length limit was reached ;) If all $A_k$ are finite, and only finitely many nonempty, the union equals the union of the finitely many nonempty $A_k$, and the number of its elements is the sum of the number of elements of the nonempty $A_k$. Hence also in this case $\mu\left(\bigcup A_k\right) = \sum \mu(A_k)$ holds. – Daniel Fischer Jun 26 '14 at 10:56
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@DanielFischer Why not posting your comments as an answer, so that the question does not remain unanswered? – Martin Sleziak Jun 26 '14 at 15:36
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To see the $\sigma$-additivity of the counting measure, consider a sequence $(A_k)_{k\in\mathbb{N}}$ of pairwise disjoint sets. If one of the $A_k$ is infinite, say $A_i$, then $\bigcup A_k \supset A_i$ is also infinite, and
$$\sum_{k\in\mathbb{N}} \mu(A_k) = \underbrace{\sum_{k < i}\mu(A_k)}_{0 \leqslant \sum \leqslant \infty} + \underbrace{\mu(A_i)}_{\infty} + \underbrace{\sum_{k > i} \mu(A_k)}_{0\leqslant \sum \leqslant \infty} = \infty = \mu\left(\bigcup_{k\in\mathbb{N}} A_k\right).$$
If infinitely many of the $A_k$ are nonempty, since they are disjoint, the union $\bigcup A_k$ is infinite, and
$$\sum_{k\in\mathbb{N}} \mu(A_k) = \sum_{\substack{k\in\mathbb{N} \\ A_k\neq\varnothing}} \mu(A_k) \geqslant \sum_{\substack{k\in\mathbb{N} \\ A_k\neq\varnothing}} 1 = \infty = \mu\left(\bigcup_{k\in\mathbb{N}} A_k\right).$$
If only finitely many $A_k$ are nonempty and all $A_k$ are finite, then the union is also finite, and the number of its elements is the sum of the number of elements of the nonempty $A_k$ by disjointness, hence
$$\sum_{k\in\mathbb{N}} \mu(A_k) = \sum_{\substack{k\in\mathbb{N} \\ A_k\neq\varnothing}} \mu(A_k) = \mu\left(\bigcup_{\substack{k\in\mathbb{N} \\ A_k\neq\varnothing}} A_k\right)= \mu\left(\bigcup_{k\in\mathbb{N}} A_k\right)$$
holds in that last case too.
Daniel Fischer
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