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Let $ (X,\mathcal{M},\mu)$ be a measure space and let $ f:X\to[-\infty,+\infty]$ be an integrable function (i.e. at least one of $ f_+ $ and $ f_-$ is integrable). I want to prove that $ \lambda:\mathcal{M}\to[-\infty,+\infty]$ defined as

$$ \lambda(E)=\int_{E}f d\mu $$

is a signed measure.

When I prove the countable additivity, how can I prove that the series $$ \sum_{i=1}^{+\infty}\int_{E_i}f d\mu$$ may only converge absolutely or diverge?

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  • $\begingroup$ Can it not help if you define $\lambda_{+}\left(E\right):=\int f_{+}d\mu$, $\lambda_{-}\left(E\right):=\int f_{-}d\mu$ and finally $\lambda\left(E\right):=\lambda_{+}\left(E\right)-\lambda_{-}\left(E\right)$? Then $\lambda_{+}$and $\lambda_{-}$ are both measures and at least one of them is finite. $\endgroup$
    – drhab
    Commented Mar 17, 2014 at 18:40
  • $\begingroup$ Then how can I prove that these series cannot oscillate? $\endgroup$
    – avati91
    Commented Mar 17, 2014 at 19:22
  • $\begingroup$ Are the sets $E_i$ disjoint? $\endgroup$
    – drhab
    Commented Mar 17, 2014 at 19:28

1 Answer 1

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Define for every $E\in\mathcal M$:

$\lambda_{+}\left(E\right):=\int_{E}f_{+}d\mu$, $\quad\lambda_{-}\left(E\right):=\int_{E}f_{-}d\mu$ and finally $\lambda\left(E\right):=\lambda_{+}\left(E\right)-\lambda_{-}\left(E\right)$.

Let $E=\cup_{i}E_{i}$ where the $E_{i}$ are measurable and disjoint. Then $\sum_{i}\left|\lambda\left(E_{i}\right)\right|=\sum_{i}\left[\lambda_{+}\left(E_{i}\right)+\lambda_{-}\left(E_{i}\right)\right]=\lambda_{+}\left(E\right)+\lambda_{-}\left(E\right)$. If $\lambda_{+}\left(E\right)+\lambda_{-}\left(E\right)<\infty$ then the sequence converges absolutely. If $\lambda_{+}\left(E\right)+\lambda_{-}\left(E\right)=\infty$ then $\lambda_{+}\left(E\right)=\infty\wedge\lambda_{-}\left(E\right)<\infty$ or $\lambda_{+}\left(E\right)<\infty\wedge\lambda_{-}\left(E\right)=\infty$. In both cases $\sum_{i}\lambda\left(E_{i}\right)$ diverges. In the first case to $+\infty$ and in the second to $-\infty$. The option $\lambda_{+}\left(E\right)=\infty=\lambda_{-}\left(E\right)$ is not there since it contradicts that $f$ is integrable.

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  • $\begingroup$ All clear except for the last passage. In which way the fact that $\ f $ is integrable does imply that the series cannot oscillate? $\endgroup$
    – avati91
    Commented Mar 17, 2014 at 19:53
  • $\begingroup$ $f$ integrable means that at least one of $\int f_{+}d\mu$ and $\int f_{-}d\mu$ is finite. Then $\lambda_{+}\left(E\right)\leq\lambda_{+}\left(X\right)=\int f_{+}d\mu<\infty$ or $\lambda_{+}\left(E\right)\leq\lambda_{-}\left(X\right)=\int f_{-}d\mu<\infty$. $\endgroup$
    – drhab
    Commented Mar 17, 2014 at 19:57
  • $\begingroup$ If it is still unclear then tell me what you mean exactly by oscillate. $\endgroup$
    – drhab
    Commented Mar 17, 2014 at 20:00
  • $\begingroup$ Why $\ \sum_{i=1}^{+\infty}\int_{E_i}f_+ d\mu $ and $\ \sum_{i=1}^{+\infty}\int_{E_i}f_- d\mu $ have to exist? $\endgroup$
    – avati91
    Commented Mar 17, 2014 at 20:00
  • $\begingroup$ Both are summations of nonnegative terms and both are allowed to diverge to $+\infty$. Such 'things' always exist. The only thing is that they cannot both diverge to $+\infty$ (that would contradict the integrability of $f$). $\endgroup$
    – drhab
    Commented Mar 17, 2014 at 20:06

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