I've been asked to find the proof for the following proposition:

Let $(f_{n})_{n \geq1}$ be a sequence of Lebesgue measure functions defined on a Lebesgue measurable set $A$ with values in either $R$ or $C$ such that $\sum_{n = 1}^{\infty}\int_{A}|f_{n}(x)|dx < \infty$. Then let the series defined by $f(x) = \sum_{n = 1}^{\infty}f_{n}(x)$ be absolutely convergent almost everywhere for $x \in A, f \in L^{1}(A)$. Prove that: $$ \int_{A}f(x)dx = \sum_{n = 1}^{\infty}\int_{A}f_{n}(x)dx $$

I'm not really sure how to go about solving this problem and appreciate the help in solving it. Many thanks in advance!

  • 1
    $\begingroup$ This is false. You need to remove the absolute values in the conclusion. As a hint: what very well-known measure theoretic result allows you to interchange integrals and limits (series are limits of sums)? $\endgroup$
    – J. De Ro
    Nov 6, 2019 at 14:37
  • $\begingroup$ Yep that was my mistake and I've changed that now. I know using the monotone and dominated convergence theorem will help here but I'm just not sure how to make my argument combining these two theorems. $\endgroup$ Nov 6, 2019 at 14:47

1 Answer 1


This is for $\int f_1 + \int f_2 = \int (f_1 + f_2)$ case, the extension follows naturally.

For any function $f_1$ there exists simple function $\phi_n$ with $0 \leq \phi_n \leq f_1$ such that $\phi_1 \leq \phi_2 \leq \cdots$ and $\phi_x(x) \underset{n \to \infty}{\to} f_1(x)$ for all x.

Similarly, there exists $\psi_n$ such that $0 \leq \psi_n \leq f_2$ with $\psi_1 \leq \psi_2 \leq \cdots$ and $\psi_n(x) \to f_2(x)$.

Now, $\phi_n + \psi_n$ is simple, and increasing in n so $\phi_n(x) + \psi_n(x) \to f_1(x) + f_2(x)$.

$$ \begin{aligned} \int(f_1 + f_2) &\underset{MCT}{=} \lim_{n \to \infty} \int (\phi_n + \psi_n))\\ &= \lim_{n \to \infty} \int \phi_n + \lim_{n \to \infty} \int \psi_n\\ &\underset{MCT}{\longrightarrow} \int f_1 + \int f_2 \end{aligned} $$


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