Showing that $\int_{E\cup F}f=\int_E f+\int _F f$, where $E\cap F=\emptyset$

Question

I would like to show that $\int_{E\cup F}f=\int_E f+\int _F f$, where $E\cap F=\emptyset$ and $E,F$ are Lebesgue measurable sets.

Attempt:

First I tried to show that in general I can write $\int (f+g)=\int f+\int g$ like so:

Proof.

Suppose that there are two sequences $(\varphi_n)$ and $(\psi_n)$ such that $\lim_{n\to\infty}\varphi_n=f$ and $\lim_{n\to\infty}\psi_n=g$, where these sequences are made up of non-negative, integrable, simple functions. Then applying Lebesgue's Monotone Convergence Theorem (that is, if we have $f_i$ non-negative and measurable: $f_1\leq\ldots\leq f_n\leq\ldots$, then $\lim_{n\to\infty}\int f_n=\int\lim_{n\to\infty}f_n$), we can get the following:

\begin{align} &\int (f+g)\\ &=\lim_{n\to\infty} \int(\varphi_n+\psi_n)\\ &=\lim_{n\to\infty}\int \varphi_n+\lim_{n\to\infty}\int \psi_n\\ &=\int f+\int g \end{align}

So I (think) I've proved that $\int_T (f+g)=\int_T f+\int_T g$, but this is for a common domain $T$. I'm not sure how to extend this result to split the integral with different domains. (Perhaps there is an easier way to show this instead?)

Any help would be appreciated. Thanks.

Answer 1

Consider any simple function, $\phi = \sum a_k\chi_{E_k}$. Then, \begin{align} \int_{E\cup F} \phi = \int\phi\cdot\chi_{E\cup F} &= \sum_{k=1}^na_k\mu(E_k\cap(E\cup F))\\ &= \sum_{k=1}^na_k\mu(E_k\cap E) +\sum_{k=1}^na_k\mu(E_k\cap F)\\ &= \int_{E} \phi + \int_{F} \phi. \end{align} Can you continue from here?

Answer 2

I realize this is an old question, but here's a simple proof based on indicator functions, requiring no thought about sequences or simple functions.

By definition, if $A$ is measurable, then $\int_A f\, d\lambda = \int f \, \mathbf 1_A \, d\lambda$. We also know that since $E \cap F = \emptyset$, $\mathbf 1_{E \cup F} = \mathbf 1_E + \mathbf 1_F$. Using linearity, we have

\begin{align*} &\int_{E \cup F} f \, d\lambda = \int f \, \mathbf 1_{E \cup F} \, d\lambda = \int f (\mathbf 1_E + \mathbf 1_F) \, d\lambda = \int f \, \mathbf 1_E \, d\lambda + \int f \, \mathbf 1_F \, d\lambda\\ &= \int_E f \, d\lambda + \int_F f \, d\lambda \end{align*}