I'v seen a proposition saying that:
Let $f: \Omega \rightarrow \mathcal{X}$ be a Bochner integrable function and $\mu$ a $\sigma$-finite measure. Let $\left(A_{n}\right)_{n \in \mathbb{N}}$ be a sequence of disjoint measurable sets and $A:=\bigcup_{n=1}^{\infty} A_{n}$, then $$ \int_{A} f \mathrm{~d} \mu=\sum_{n=1}^{\infty} \int_{A_{n}} f \mathrm{~d} \mu $$ and the sum is absolutely convergent.
I think the condition '$\mu$ is $\sigma$-finite is unnecessary', what's wrong with my proof?
My proof: $$ \sum_{n=1}^{\infty}\left\|\int_{A_{n}} f \mathrm{~d} \mu\right\| \leqslant \sum_{n=1}^{\infty} \int_{A_{n}}\|f(\omega)\| \mathrm{d} \mu=\int_{A}\|f(\omega)\| \mathrm{d} \mu<\infty $$ thus the sum absolutely converges. Then we just need to verify $$ \left\|\int_{A} f \mathrm{~d} \mu - \sum_{n=1}^{m} \int_{A_{n}} f \mathrm{~d} \mu \right\|\rightarrow 0 (m\rightarrow 0) $$ Obviously, $$ \left\|\int_{A} f \mathrm{~d} \mu - \sum_{n=1}^{m} \int_{A_{n}} f \mathrm{~d} \mu \right\| = \left\| \int_{\bigcup_{n=m+1}^{\infty} A_{n} } f \mathrm{~d} \mu\right\| \leq \int_{\bigcup_{n=m+1}^{\infty} A_{n} } \| f \mathrm\|{~d}\mu = \sum_{n=m+ 1}^{\infty} \int_{A_{n}}\|f\| \mathrm{d} \mu \rightarrow 0 $$ which completes the proof.