Additivity of Lebesgue integral over union of disjoint sets

Question

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.

Answer 1

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.

Answer 2

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:$$\sum_{n=1}^{\infty}\int_{A_{n}}fd\mu=\int_{B}fd\mu$$