Let $E$ be a measurable subset of the real line of finite measure and $f_n$ a sequence of integrable functions on $E$. Show that if $$\sum_{n=1}^\infty \left|f_n(x)\right|\leq \mathcal{C}$$ for almost all $x\in E$, then $$\int_E f=\sum_{n=1}^\infty \int_E f_n.$$

We know that $$f(x)=\sum_{n=1}^\infty f_n(x)$$ converges almost everywhere on $E$. So $$\int_E f(x)=\int_E \sum_{n=1}^\infty f_n(x)=\sum_{n=1}^\infty \int_E f_n(x).$$

  • $\begingroup$ There are some problems here. I think you mean that $E$ has finite measure, not that $E$ is finite. Also, you should define $f = \sum_{n = 1}^{\infty} f_n$. Finally, the statement is clearly wrong. $\endgroup$
    – user296602
    Apr 11, 2016 at 0:57
  • 4
    $\begingroup$ This is a one-line application of dominated convergence. $\endgroup$
    – user296602
    Apr 11, 2016 at 1:00
  • $\begingroup$ Simply saying "converges almost everywhere" is not enough. Use Bongers' hint. $\endgroup$
    – GEdgar
    Apr 11, 2016 at 19:55

1 Answer 1


As pointed out by T. Bongers, we can use the dominated convergence theorem defining $g_N(x):=\sum_{n=1}^Nf_n(x)$. Its absolute value is dominated by a constant independent of $N$, which is integrable over the finite measure set $E$.


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