The Additivity Over Domains of Int. Prop. of Nonneg. Meas. Functs.
The Additivity Over Domains of Integration Property of Nonnegative Measurable Functions
| Theorem 1 (The Finite Additivity Over Domains of Integration): Let $(X, \mathcal A, \mu)$ be a complete measure space and let $f$ be a nonnegative measurable function defined on the measurable sets $A$ and $B$ where $A$ and $B$ are mutually disjoint. Then $\displaystyle{\int_{A \cup B} f(x) \: d \mu = \int_A f(x) \: d \mu + \int_B f(x) \: d \mu}$. |
- Proof: Let $f$ be a nonnegative measurable function defined on the measurable sets $A$ and $B$.
- By the The Simple Function Approximation Lemma and Theorem for General Measurable Spaces page there exists pointwise increasing sequences $(\varphi_n(x))_{n=1}^{\infty}$ and $(\varphi_n'(x))_{n=1}^{\infty}$ of nonnegative simple functions that converge to $f(x)$ on $A$ (and $0$ on $B$) and $f(x)$ on $B$ (and $0$ to $A$) respectively.
- Then $(\varphi_n(x) + \varphi_n'(x))_{n=1}^{\infty}$ is a pointwise increasing sequence of nonnegative simple functions that converge to $f(x)$ on the measurable set $A \cup B$. So by The Monotone Convergence Theorem for Nonnegative Measurable Functions we have that:
\begin{align} \quad \lim_{n \to \infty} \int_{A \cup B} [\varphi_n(x) + \varphi_n'(x)] \: d \mu &= \int_{A \cup B} f(x) \: d \mu \\ \quad \lim_{n \to \infty} \int_{A \cup B} \varphi_n(x) \: d \mu + \lim_{n \to \infty} \int_{A \cup B} \varphi_n'(x) \: d \mu &= \int_{A \cup B} f(x) \: d \mu \\ \quad \lim_{n \to \infty} \int_A \varphi_n \: d \mu + \lim_{n \to\infty} \int_B \varphi_n'(x) \: d \mu &= \int_{A \cup B} f(x) \: d \mu \\ \quad \int_A f(x) \: d \mu + \int_B f(x) \: d \mu &= \int_{A \cup B} f(x) \: d \mu \quad \blacksquare \end{align}
| Theorem 2 (The Countable Additivity Over Domains of Integration): Let $(X, \mathcal A, \mu)$ be a complete measure space and let $f$ be a nonnegative measurable function defined on the measurable sets $(E_n)_{n=1}^{\infty}$ where $(E_n)_{n=1}^{\infty}$ is a collection of mutually disjoint sets. Then $\displaystyle{\int_{\bigcup_{n=1}^{\infty} E_n} f(x) \: d \mu = \sum_{n=1}^{\infty} \int_{E_n} f(x) \: d \mu}$. |