# A very important measure.

Let $$f\in L^1(X,\mathcal{A},\mu)$$; we define an application $$\nu\colon\mathcal{A}\to \mathbb{R}$$ placing $$\large \nu(E):=\int_E f\;d\mu\quad (E\in\mathcal{A}).$$

Let $$\{E_k\}_k\subseteq \mathcal{A}$$ be a sequence of disjoint sets and define $$E=\bigcup_{k=1}^\infty E_k$$.

Claim. $$\large \nu(E)=\sum_{k=1}^\infty\nu(E_k)$$

For definition, $$\nu(E)=\nu_+(E)-\nu_-(E)$$, where $$\nu_\pm\colon\mathcal{A}\to\mathbb{R}_+$$ and $$\large \nu_\pm=\int_E f_\pm\; d\mu$$, since $$\nu_\pm$$ are measures we have

\begin{aligned} \large \nu(E)=\nu_+(E)-\nu_-(E)&=\left(\sum_{k=1}^\infty\nu_+(E_k)\right)-\left(\sum_{k=1}^\infty\nu_-(E_k)\right)\\ &\color{RED}{=}\sum_{k=1}^\infty \left(\nu_+(E_k)-\nu_-(E_k)\right)\\ &=\sum_{k=1}^\infty \nu(E_k) \end{aligned}

I know that the red equality derives from the fact that the series $$\large \sum_{k=1}^\infty \nu_\pm(E_k)=\int_E f_\pm\,d\mu<\infty.$$

My text also states that in addition to series convergence, known theorems on absolutely convergent series have been used. However, I do not understand where these theorems have been used. Could you give me some explanation about it? Thanks.

Actually for emphasis and logical flow, I'd re-order and place the red equality as: \begin{align}\tag{1}\sum_{k\ge1}\nu(E_k)&=\sum_{k\ge1}(\nu_+(E_k)-\nu_-(E_k))\\\tag{2}&\color{red}{=}\sum_{k\ge1}\nu_+(E_k)-\sum_{k\ge1}\nu_-(E_k)\\&=\nu_+(E)-\nu_-(E)\\&=\nu(E)\end{align}Since, given convergence of $$2)$$, convergence and equality with $$1)$$ is automatic from basic limit properties, but given convergence of $$1)$$ it is not always possible to obtain convergence and/or equality with $$2)$$.
The classic example of this is $$\nu_{\pm}(E_k):=k$$ for all $$k$$. Then the sum in $$1)$$ converges to $$0$$ as all its summands are $$0$$, but sum $$2)$$ is ill-defined. However, since the integrability of $$f$$ implies both series in $$2)$$ are convergent, you get equality with $$1)$$ by limit properties.
I don't think the transition $$1)\to2)$$ requires absolute convergence of the series, but comments about absolute convergence are often referring to the beautiful phenomenon described by the Riemann series theorem where conditionally-but-not-absolutely convergent series can be made to attain any arbitrary extended real value (!) via reordering of the terms.
N.B. The series $$1)$$ is absolutely convergent anyway, since: $$\sum_{k\ge1}|\nu_+(E_k)-\nu_-(E_k)|=\sum_{k\ge1}|\nu(E_k)|\le\int_E|f|\,\mathrm{d}\mu\lt\infty$$