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Let $X=\{x_1,x_2,x_3,\dots\}$ be countable. Choose $p_n\geq0$ so that $\sum_n p_n=1$. For each subset $S$ of $X$, put $\mu(S)=\sum\limits_{x_n\in S}p_n$. Prove that $\mu$ is countably additive.

Let $\{S_n\}$ be a countable family of disjoint subsets of $X$, and $S=\bigcup_n S_n$. The case $S$ is finite is trivial. So we assume $S$ is infinite. Since $\sum p_n$ converges absolutely, every rearrangement of it converges to the same limit. If each $S_n$ is finite, then we can choose $0=K_0<K_1<K_2<K_3<\cdots$ and $k_1<k_2<k_3<\dots$ such that $x_{k_j} \in S_n$ if $K_{n-1}+1\leq j\leq K_n$. Then $$\mu(S)=\sum_{x_j\in S}p_j=\sum^\infty_{j=1}p_{k_j}=\sum^\infty_{n=1}\sum^{K_n}_{j=K_{n-1}}p_{k_j}=\sum^\infty_{n=1}\mu(S_n)\text{.}$$

The case that exactly one $S_n$ is infinite is similar. How to deal with the case two or more of the $S_n$ are infinite?

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    $\begingroup$ All $p_j$'s are nonnegative, so just feel free to rearrange all the sumations. $\endgroup$
    – Kervyn
    Jun 23 at 9:03
  • $\begingroup$ @Kervyn I think we need to rearrange as a double sequence. Can you state such a theorem and proof? $\endgroup$
    – user912011
    Jun 23 at 9:48
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    $\begingroup$ That's not a big problem. You can relabel the double sequence into a new sequence and show the way you relabel doesn't matter. It's actually some discrete version of Fubini theorem. $\endgroup$
    – Kervyn
    Jun 23 at 10:26

1 Answer 1

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Suppose $\sum a_n$ is an absolute convergent series of complex number.

THEOREM 1 Every rearrangement of $\sum a_n$ converges to the same sum.

THEOREM 2 If $\{n_k\}$ is a sequence of positive integers such that $1=n_1<n_2<n_3<\cdots$, then $$\sum^\infty_{k=1}\sum^{n_{k+1}-1}_{n=n_k}a_n=\sum^\infty_{n=1} a_n\text{.}$$

PROOF Partial sums the former series is a subsequence of the latter.

THEOREM 3 If $\{a_n\}$ is arranged in a double sequence $\{b_{jk}\}$ as follows: $$\begin{array}{ll} b_{11}&=a_1&b_{12}&=a_{N+1}&b_{13}&=a_{2N+1}&\cdots\\ b_{21}&=a_2&b_{22}&=a_{N+2}&b_{23}&=a_{2N+2}&\cdots\\ &\vdots&&\vdots&&\vdots&\\ b_{N1}&=a_N&b_{N2}&=a_{2N}&b_{N3}&=a_{3N}&\cdots \end{array}$$ then $\sum_{k} b_{jk}$ converges for $j=1,\dots,N$, and $\sum_j\sum_kb_{jk}$ converges to $\sum_na_n$.

PROOF The convergence of $\sum_kb_{jk}$ follows from the absolute convergence of $\sum a_n$ and Cauchy criterion. For each $K$, $$ \sum^N_{j=1}\sum^K_{k=1} b_{jk}=\sum^K_{k=1}\sum^N_{j=1}b_{jk}=\sum^K_{k=1}\sum^{kN}_{j=1+(k-1)N}a_j\text{.}$$ Letting $K\rightarrow\infty$, by THEOREM 2, we obtain $\sum^N_{j=1}\sum^\infty_{k=1}b_{jk}=\sum^\infty_{k=1}a_k$.

THEOREM 4 If $\{a_n\}$ is arranged in a double sequence $\{b_{jk}\}$ as follows: $$\begin{array}{ll} T_1:&b_{11}&=a_1&b_{12}&=a_{3}&b_{13}&=a_{6}&\cdots\\ T_2:&b_{21}&=a_2&b_{22}&=a_{5}&b_{23}&=a_{9}&\cdots\\ T_3:&b_{31}&=a_4&b_{32}&=a_{8}&b_{33}&=a_{13}&\cdots\\ &&\vdots&&\vdots&&\vdots& \end{array}$$ then $\sum_{k} b_{jk}$ converges for each $j$, and $\sum_j\sum_kb_{jk}$ converges to $\sum_na_n$.

PROOF Put $a=\sum a_n$. Let $\epsilon>0$. Choose $N$ so that $\sum^\infty_{n=N}|a_n|<\epsilon$. Plainly, it is possible to choose $K$ such that $\{a_1,\dots,a_{N-1}\}\subset T_1\cup\cdots\cup T_K$. Fix $k>K$. By THEOREM 1 and 3, there exists a subsequence $\{c_m\}$ of some rearrangement of $\{a_n\}$ such that $\sum^k_{m=1}\sum^\infty_{n=1}b_{mn}=\sum c_m$. Since $\{c_m\}$ contains $a_n$ for $n<N-1$, $$|\sum c_n-a|\leq \sum^\infty_{n=N}|a_n|<\epsilon\text{.}$$ This completes the proof.

ANSWER TO THE QUESTION By theorem 1 and 4, we can arrange $\{p_n\}$ in a double sequence $\{q_{jk}\}$ so that, for each $j$, $\mu(S_j)=\sum_k q_{jk}$. It follows then that $$\sum p_n=\sum_j\sum_k q_{jk}=\sum_j \mu(S_j)\text{.}$$

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