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I have a little question about measure theory.

Suposse that I have a measurable space $(X,\mathcal{M})$. Moreover, suppose that there exists a partition of $X$ in subsets $\{X_n\}_{n\in \omega}$ and finite measures $\mu_n$ over $X_n$ . To me is natural to define a "measure" over $X$ by $\mu:=\sum_{n\in \omega}\mu_n$ but I have doubts about if this really define a positive measure over $X$. My problem appears with the $\sigma$-aditivity. More precisely, if $\{A_k\}_{k\in\omega}$ is a sequence of disjoint sets on $\mathcal{M}$ I ever can ensure that $$ \sum_{n\in\omega}\sum_{k\in\omega}\mu_n(A_k\cap X_n)=\sum_{k\in\omega}\sum_{n\in\omega}\mu_n(A_k\cap X_n) $$ i,e, under which conditions can I change the order of the sum?

YCB
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1 Answers1

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Positivity and Interchange of Summation and https://en.wikipedia.org/wiki/Monotone_convergence_theorem

The result. If $a_{nk}\ge 0$ is increasing in $n$, then $\lim_n \sum_k a_{nk} = \sum_k \lim_n a_{nk}$ (even if infinite)

Proof. Of course, $\sum_k a_{nk} \le \sum_k \lim_n a_{nk}$, which provides one inequality. But the other inequality is provided by Fatou's lemma, $$ \sum_k\liminf_n a_{nk} \le\liminf_n\sum_ka_{nk}$$ since in both cases $\liminf = \lim$.

By setting $a_{nk} = \sum_{m =0}^n b_{mk}$ and then reversing the order of summation, we get the equality in the question.

Calvin Khor
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