Let $A = \cup_n A_n$ be a finite or countable union of pairwise disjoint sets $A_n$, and suppose $f$ is integrable on each $A_n$, and satisfies the condition $$\sum_n \int_{A_n} |f(x)| d\mu < \infty $$

I have to prove that $f$ is integrable on $A$.

I can't prove this statement, I have tried first that If $f$ is simple, with values $y_1,y_2,\ldots,y_n,\ldots$ let the sets $B_k$ and $B_{nk}$ be $$B_k= \{x | x \in A \ , \ f(x)=y_k \} $$ and $$ B_{nk} = \{x | x\in A_n \ , \ f(x)=y_k \}$$ and then how can I prove that

$$ \int_{A_n} |f(x)| d\mu = \int_k |y_k| \mu(B_{nk})$$

Proving this with the absolute convergence given implies the convergence of $$ \sum_n \sum_k |y_k| \mu(B_{nk}) = \sum_k |y_k| \sum_n \mu(B_{nk})= \sum_k |y_k| \mu(B_{k}) $$ and hence the integrability of $f$ on $A$. Is it correct?

If someone could help me please. Thanks for your time and help.


This is a monotone convergence problem in disguise.

$f_N (x) = |f(x)| {\bf 1}_{\cup_{n\le N} A_n}(x)$.

Now $0\le f_N\nearrow f$, because $A=\cup A_n$.


$$\sum_{n\le N} \int_{A_n} |f| d\mu = \int |f_n|d\mu.$$

Finish it off with monotone convergence.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.