**Problem**

> 
> Let $(\Omega,\mathscr F,\mu)$ be a measure space where $f$ is a measurable, integrable function and where $\{A\_n\}\_{n=1}^\infty\subseteq\mathscr F$ is a sequence of pairwise disjoint sets with $\bigcup\_{n=1}^\infty A\_n=B\in\mathscr F$, then
> $$\sum\_{n=1}^\infty\int\_{A\_n}f\,d\mu=\int\_B f\,d\mu.$$
> 
> 
> 

**Attempt**

Proving this for non-negative functions requires a fairly straightforward application of the Monotone Convergence Theorem as follows:

> 
> If we impose the further constraint that $f$ is non-negative, then $$ \mu(f \mathbb 1 \_A) = \space \mu (f \sum \_{i=1}^{\infty} \mathbb 1\_{A\_i}) = \space \sum \_{i=1}^{\infty} \mu(f \mathbb 1\_{A\_i}) = \space \sum \_{n=1}^{\infty} \int \_{A\_n} f \space d \mu $$
> by using the Monotone Convergence Theorem to take the sum outside of the integral.
> 
> 
> 

However, generalising this result to all integrable functions is something that I am less sure about. I believe that the correct approach is to apply dominated convergence theorem in some way by using the fact that $\mu (|f|) < \infty$ by the fact that $f$ is integrable. However, it is unclear exactly how this helps to arrive at the desired result.

If we let $f\_n = f \sum \_{j=1}^{n}\mathbb 1 \_{A\_j}$ then clearly $f\_n \uparrow f \mathbb 1\_B$ and DCT tells us that $\mu(f\_n) \rightarrow \mu(f \mathbb 1 \_B)$, however, I'm unsure as to whether or not this is along the right lines in order to prove the desired result.

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Do you agree with my approach above?

Thank you very much for your time.