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.


1 Answer 1


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: $$\sum_{k\ge1}|\nu_+(E_k)-\nu_-(E_k)|=\sum_{k\ge1}|\nu(E_k)|\le\int_E|f|\,\mathrm{d}\mu\lt\infty$$

  • $\begingroup$ Excuse me, but I have a doubt. If the series absolutely converges, then we can't apply Riemann Dini's theorem, so how does it end? $\endgroup$
    – NatMath
    May 16, 2022 at 10:06
  • $\begingroup$ @NatMath We don’t “apply” the Riemann series theorem. I mention it in passing since it is interesting. All that matters here is that each individual series in the positive and negative measure is convergent, so we can add and subtract them without fear $\endgroup$
    – FShrike
    May 16, 2022 at 10:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .