I have to show that \begin{equation} m(\bigcup_{n=1}^{+\infty} A_n)=\sum_{n=1}^{+\infty} m(A_{n}) \end{equation}
where $\{A_{n}\}$ are disjoint and Peano-Jordan measurable, also $\bigcup_{n=1}^{+\infty} A_{n}$ is Peano-Jordan measurable.
The first inequality is clear
\begin{equation} \sum_{n=1}^{+\infty}m(A_{n})\ge m(\bigcup_{n=1}^{+\infty}A_{n}). \end{equation}
Now let $A=\bigcup_{n=1}^{+\infty} A_{n}$ and we consider clousure of $A$, $\bar{A}$. Observe that $\bar{A}$ is compact.
Far all $n\in\mathbb{N}$ let $\tilde{P_{n}}\supset A_{n}$ such that \begin{equation} m(\tilde{P_{n}})<m(A_{n})+\frac{\epsilon}{2^{n+1}}. \end{equation} where $\tilde{P_{n}}$ is a plurinterval.
Now we take a open plurinterval $P_{n}$ such that $P_{n}\supset \overline{\tilde{P_n}}$ such that
\begin{equation} m(\tilde{P_{n}})<m(P_{n})<m(\tilde{P_{n}})+\frac{\epsilon}{2^{n+1}}. \end{equation}
The question is: how can I prove
\begin{equation} \bigcup_{n=1}^{+\infty} P_{n} \supset \bar{A}. \end{equation}
Thanks.
I don't think what you seek to show is true. Consider $A_n = \{\frac{1}{n}\}$ for each $n$. Then $\overline{A_n} = \{\frac{1}{n}\}$ for each $n$, and $\overline{A} = \{0,1,\frac{1}{2},\frac{1}{3},\dots\}$. Let $\tilde{P_n} = (\frac{1}{n}-\frac{1}{2^{n+1}},\frac{1}{n}+\frac{1}{2^{n+1}})$ and $P_n = (\frac{1}{n}-\frac{1}{2^n},\frac{1}{n}+\frac{1}{2^n})$. The point is that these $\tilde{P_n}$'s and $P_n$'s are small enough that they do not include $0$.