Exercise 1.3.22 (Finite additivity of Lebesgue integral of absolutely integrable functions)

Question

If $E,F$ are disjoint Lebesgue measurable subsets of $\mathbb{R}^d$, and $f:E \cup F	o\mathbb{C}$ is absolutely integrable, show that
$$\int_{E} f = \int_{E \cup F} f  \cdot 1_E$$
and
$$\int_{E \cup F} f = \int_{E} f + \int_{F} f.$$

Elu Thingol · Accepted Answer

For any two disjoint sets $E,F$ we have $1_{E\cup F} = 1_E + 1_F$.
\begin{align*}
	\int_{E \cup F} f &= \int_{\mathbb{R}^d} f \cdot 1_{E \cup F} = \int_{\mathbb{R}^d} f \cdot (1_E + 1_F) = \int_{\mathbb{R}^d} f \cdot 1_E + f \cdot 1_F
\end{align*}
Since $f \cdot 1_E$ and $f \cdot 1_F$ are absolutely integrable, by \href{./exercise_1-3-19}{Exercise 1.3.19} we have
\begin{align*}
&= \int_{\mathbb{R}^d} f \cdot 1_E + \int_{\mathbb{R}^d} f \cdot 1_F = \int_{E} f + \int_{F} f.
\end{align*}

Exercise 1.3.22 (Finite additivity of Lebesgue integral of absolutely integrable functions)

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Exercise 1.3.22 (Finite additivity of Lebesgue integral of absolutely integrable functions)

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