Additivity of Lebesgue integral over union of disjoint sets 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.

Do you agree with my approach above?
Thank you very much for your time.
 A: Let $f^{+}(x) = f(x)$ when $f(x) \ge 0$ and otherwise $0$, and let $f^{-}(x) = -f(x)$ when $f(x) \le 0$ and otherwise $0$. Then you can use your result for non-negative functions on $f^{+}$ and $f^{-}$ like so:
$$
\sum_{i}^{\infty} \int_{A_i} f \space = \space  \sum_{i} ^{\infty} \big{(}\int_{A_i} f^{+} - \int_{A_i} f^{-}\big{)} $$
by decomposing $f$ as described above and separating out the integral by linearity.
$$ = \sum_{i} ^{\infty} \big{(}\int_{A_i} f^{+}\big{)}  - \sum_{i} ^{\infty} \big{(}\int_{A_i} f^{-}\big{)} = \int_{B} f^{+} - \int_{B} f^{-}
$$
by applying the result for non-negative functions (since $f^+$ and $f^-$ are non-negative).
$$ = \int _B f \space d \mu $$
by linearity. This is the desired result and so the proof is complete.
A: Applying on nonnegative function $\left|f\right|$ we find: $$\sum_{n=1}^{\infty}\int_{A_{n}}\left|f\right|d\mu=\int_{B}\left|f\right|d\mu<\infty\tag1$$
If we define $f_{n}:=\sum_{i=1}^{n}f1_{A_{i}}$ then: $$\int\left|f-f_{n}\right|d\mu\leq\sum_{i=n+1}^{\infty}\int_{A_{i}}\left|f\right|d\mu\tag2$$
Then combining $(1)$ and $(2)$ it can be concluded that: $$\lim_{n\to\infty}\int\left|f-f_{n}\right|d\mu=0$$
This with: $$\left|\int fd\mu-\int f_{n}d\mu\right|\leq\int\left|f-f_{n}\right|d\mu$$
so that we can conclude that: $$\lim_{n\to\infty}\int f_{n}d\mu=\int fd\mu$$or equivalently:$$\sum_{n=1}^{\infty}\int_{A_{n}}fd\mu=\int_{B}fd\mu$$
