I've seen various version of the DCT prove that if $f$ is a real valued, or extended real valued, or complex, integrable function, and if $\{E_n\}_n$ is a sequence of disjoint measurable subsets, then:
$$\int_{\bigcup_n E_n}f\ d\lambda=\sum_n\int_{E_n}f\ d \lambda$$
If you didn't know the DCT, but only the MCT, but also knew that if $f$ was real or extended real, then $f=f^+-f^-$, $\int_Kfd\lambda=\int_Kf^+d\lambda-\int_Kf^-d\lambda$.
And in the complex case you have that $\int_Kfd\lambda=\int_K(\Re(f)^+d\lambda-\int_K(\Re(f)^-d\lambda+i\int_K(\Im(f)^+d\lambda-i\int_K(\Im(f)^-d\lambda$.
Can we then prove the proposition above with the MCT on each part instead of using the DCT, or will there be a problem in this case? Can something go wrong here?
First assume $f$ is nonnegative. Define $E=\bigcup_{n=1}^\infty E_n$, then as $f\chi_{E_n}\uparrow f\chi_E$ so by monotone convergence, \begin{align} \sum_{n=1}^\infty \int_{E_n} f\ \mathsf d\lambda &= \lim_{n\to\infty}\sum_{j=1}^n\int_{E_j} f\ \mathsf d\lambda \\ &=\lim_{n\to\infty} \int_{\bigcup_{j=1}^n E_j} f\ \mathsf d\lambda \\ &=\lim_{n\to\infty} \int f\chi_{\bigcup_{j=1}^n E_j}\ \mathsf d\lambda\\ &= \int_E f\mathsf \ d\lambda. \end{align} The cases where $f$ can take negative/complex values follow by linearity (as well as the assumption that $\int |f|\ \mathsf d\lambda < \infty$).