I know this was asked but I want a proof of this without using Fubini theorem. Anyway the first part of the problem can't be concluded using Fubini. I don't know how to do it :/

Let $f_k:\mathbb R \to \mathbb R$ be a sequence lebesgue measurable and integrable functions such that:

$$\sum{\int{|f_k|}dm}<\infty $$

Prove that $\sum{f_k}$ converges almost surely and:

$$\sum{\int{|f_k|}}<\infty= \int{\sum{|f_k|}} $$

My proof:

Given $\varepsilon >0$ there exist $N$ such that

$$ \sum_{k=N}^{\infty}{\int{|f_k|dm}}<\varepsilon $$


$$ \sum_{k=1}^{\infty}{\int{|f_k|dm}}= \sum_{k=1}^{N-1}{\int{|f_k|dm}}+\sum_{k=N}^{\infty}{\int{|f_k|dm}}=\int{\sum_{k=1}^{N-1}{|f_k|dm}}+\sum_{k=N}^{\infty}{\int{|f_k|dm}} \le \int{\sum_{k=1}^{\infty}{|f_k|dm}}+\sum_{k=N}^{\infty}{\int{|f_k|dm}} \le \int{\sum_{k=1}^{\infty}{|f_k|dm}}+ \varepsilon $$

This holds for each $\varepsilon >0$ therefore

$$ \sum_{k=1}^{\infty}{\int{|f_k|dm}} \le \int{\sum_{k=1}^{\infty}{|f_k|dm}} $$


$$ \int{\sum_{k=1}^{\infty}{|f_k|}}= \int{\sum_{k=1}^{N-1}{|f_k|}}+\int{\sum_{k=N}^{\infty}{|f_k|}}=\sum_{k=1}^{N-1}{\int{|f_k|}}+\int{\sum_{k=N}^{\infty}{|f_k|}}$$

If I prove this I'm done:

$$\lim_{N\to \infty }\int{\sum_{k=N}^{\infty}{|f_k|}}=0$$

Therefore it's enough to prove that $\sum_{k=1}^{\infty}{|f_k|}$ converges almost surely.

  • $\begingroup$ You can cite a proof of your problem in a well known book of Walter Rudin: Real and Complex analysis (1987). You read Theorem 1.38, page 29. $\endgroup$
    – Cao
    Oct 6, 2014 at 2:04

1 Answer 1


Hint: Consider $f\colon {\bf R}\times {\bf N}\to {\bf R}$ given by $f(x,k):=f_k(x)$.


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