Let $f_n$ be integrable function and $f_{n}^{+},f_{n}^{-}$ be its positive and negative parts. Are these steps correct? \begin{align} \int {\sum {f_n}} &=\int{\sum{(f_{n}^{+}-f_{n}^{-})}}\\ &=\int{(\sum{f_{n}^{+}}-\sum{f_{n}^{-}})}\\ &=\int{\sum{f_{n}^{+}}}-\int{\sum{f_{n}^{-}}}\\ &=\sum{\int{f_{n}^{+}}}-\sum{\int{f_{n}^{-}}}\\ &=\sum{(\int{f_{n}^{+}}-\int{f_{n}^{-}})}\\ &=\sum{\int{(f_{n}^{+}}-f_{n}^{-})}\\ &=\sum{\int{f_n}} \end{align} Thank you.

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    $\begingroup$ The passing from your third line to the fourth one has to be justified, and I don't think it is true in general: further conditions on the sequence of functions must be imposed. $\endgroup$ – DonAntonio Jul 14 '13 at 10:24
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    $\begingroup$ @DonAntonio The functions $f_n^+$ and $f_n^-$ are non-negative, so that step is justified. The problem is passing from the first to the second line, as it involves rearranging a series that may be conditionally convergent. $\endgroup$ – Ayman Hourieh Jul 14 '13 at 10:26
  • $\begingroup$ Also that, @AymanHourieh, yet I can't see how the functions being non-negative allows the interchange of summatory and integral... $\endgroup$ – DonAntonio Jul 14 '13 at 10:27
  • $\begingroup$ @DonAntonio It's a direct consequence of the monotone convergence theorem. $\endgroup$ – Ayman Hourieh Jul 14 '13 at 10:29
  • $\begingroup$ If $f_n$ non-negative using Monotone Convergence Theorem we have $\int {\sum {f_n}}=\sum {\int {f_n}}$. $\endgroup$ – Deco Jul 14 '13 at 10:32

The steps are correct if both of

$$\sum \int f_n^+ \quad\text{and } \sum \int f_n^- $$

are finite.

If both are finite, the partial sums are dominated by the integrable function $\sum (f_n^+ + f_n^-)$, and except on a null-set you have (unconditional) convergence.

If only one is finite, without loss of generality $\sum\int f_n^- < \infty = \sum\int f_n^+$, then $\sum \int f_n = \infty$, and $g = \sum f_n$ is a measurable function with $\int g^- < \infty = \int g^+$, and with appropriate interpretation of arithmetic involving one infinite operand, it's also valid.

If both sums are infinite, the step from the first line to the second is invalid, and in the (third and) fourth line you have the indeterminate form $\infty - \infty$, which makes the entire chain invalid. ($\int \sum f_n = \sum \int f_n$ may still involve only well-defined entities and hold.)


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