Integral of sum of sequence of integrable functions

Question

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.

Answer 1

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.)