Let $E\subset\mathbb{R}$ a measurable subset, $f\in L^1(E)$ and $\{E_n\}$ a disjoint countable union of measurables sets such that $\bigcup E_n=E$. Show that $$ \int_Ef=\sum_{n=1}^\infty\int_{E_n} f$$

MY ATTEMPT (using a hint of the teacher):

Let $f_n=f\chi_{A_n}$, where $A_n=\bigcup_{n=1}^{\infty}E_n$. As, $f\in L^1(E)\Rightarrow|f|\in L^1(E)$. We have that $|f_n|\leq|f|$ and $$ \lim_{n\rightarrow\infty}f_n=f\lim_{n\rightarrow\infty}\chi_{A_n}=f\chi_E=f $$ By the Dominated Convergence Theorem (learn more here: http://en.wikipedia.org/wiki/Dominated_convergence_theorem), $$ \lim_{n\rightarrow\infty}\int_Ef_n=\int_Ef $$ Now is my doubt $$ \lim_{n\rightarrow\infty}\int_Ef_n{=}^*\sum_{n=1}^\infty\int_{E_n}f_n $$ I don't know if I can affirm this last equality. Can someone explain this to me?

  • 2
    $\begingroup$ can you please tell how does the way you defined $A_{n} $ in 2 nd line of your attempt makes sense? $\endgroup$
    – user775699
    Jun 24, 2020 at 10:45

1 Answer 1


You can affirm that (there is no "n" in the member of the right in your question, makes confusion):

The $E_n$ are disjoint sets, so $$\forall N \in \mathbb{N}, \ \sum_{n=1}^N \int_{E_n} f=\int_{\bigcup\limits_{n=1}^N E_n} f = \int_{A_N}f = \int_E f_N$$

The you take the limit when $N$ goes to $\infty$, and i think you are done!


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