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?
Best Answer
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!