Prove $\mu$ is probability measure

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 $$0=K_0 and $$k_1 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?

• All $p_j$'s are nonnegative, so just feel free to rearrange all the sumations. Jun 23 at 9:03
• @Kervyn I think we need to rearrange as a double sequence. Can you state such a theorem and proof? Jun 23 at 9:48
• 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. Jun 23 at 10:26

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 $$1=n_1, then $$\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 $$n, $$|\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 $$\sum p_n=\sum_j\sum_k q_{jk}=\sum_j \mu(S_j)\text{.}$$