Additivity of Lebesgue Integrals on Subintervals of General Intervals

# Additivity of Lebesgue Integrals on Subintervals of General Intervals

On the Additivity of Integrals of Upper Functions on Subintervals of General Intervals page we looked at two very important theorems regarding the additivity of the integrals of upper functions on split domains. We will now see that analogous theorems hold for Lebesgue integrals.

 Theorem 1: Let $f$ be a Lebesgue integrable function on $I$ and let $I_1, I_2 \subset I$ be subintervals of $I$ such that $I_1$ and $I_2$ intersect in at most one point and such that $I = I_1 \cup I_2$. Then $f$ is a Lebesgue integrable function on both $I_1$ and $I_2$ and furthermore, $\displaystyle{\int_I f(x) \: dx = \int_{I_1} f(x) \: dx + \int_{I_2} f(x) \: dx}$.
• Proof: Let $f$ be a Lebesgue integrable function on $I$. Then there exists upper functions $u$ and $v$ on $I$ such that:
(1)
\begin{align} \quad f = u - v \end{align}
• Now $u = u^+ - u^-$ and $v = v^+ - v^-$ and so:
(2)
\begin{align} \quad f = (u^+ - u^-) - (v^+ - v^-) \end{align}
• Each of the functions $u^+$, $u^-$, $v^+$, and $v^-$ are all nonnegative functions and are Lebesgue integrable on $I$, and so by one of the theorems on the page referenced above we have that:
(3)
\begin{align} \int_I u^+(x) \: dx = \int_{I_1} u^+(x) \: dx + \int_{I_2} u^+(x) \: dx \: (*) \quad \quad \int_I u^-(x) \: dx = \int_{I_1} u^- (x) \: dx + \int_{I_2} u^-(x) \: dx \: (**) \end{align}
(4)
\begin{align} \quad \int_I v^+(x) \: dx = \int_{I_1} v^+(x) \: dx + \int_{I_2} v^+(x) \: dx \: (***) \quad \quad \int_I v^-(x) \: dx = \int_{I_1} v^-(x) \: dx + \int_{I_2} v^-(x) \: dx \: (****) \end{align}
• Combining $(*)$, $(**)$, $(***)$, and $(****)$ we see that:
(5)
\begin{align} \quad \int_I f(x) \: dx &= \int_I u^+(x) \: dx - \int_I u^-(x) \: dx - \int_I v^+(x) \: dx + \int_I v^-(x) \: dx \\ \quad &= \left ( \int_{I_1} u^+(x) \: dx + \int_{I_2} u^+(x) \: dx \right ) - \left ( \int_{I_1} u^- (x) \: dx + \int_{I_2} u^-(x) \: dx \right ) \\ \quad & - \left ( \int_{I_1} v^+(x) \: dx + \int_{I_2} v^+(x) \: dx \right ) + \left ( \int_{I_1} v^-(x) \: dx + \int_{I_2} v^-(x) \: dx \right ) \\ \quad &= \left ( \int_{I_1} [(u^+(x) - u^-(x)) - (v^+(x) - v^-(x))] \right ) + \left ( \int_{I_2} [(u^+(x) - u^-(x)) - (v^+(x) - v^-(x))] \right ) \\ \quad &= \int_{I_1} [u(x) - v(x)] \: dx + \int_{I_2} [u(x) - v(x)] \: dx \\ \quad &= \int_{I_1} f(x) \: dx + \int_{I_2} f(x) \: dx \quad \blacksquare \end{align}
 Theorem 2: Let $I$ be an interval and let $I_1, I_2 \subset I$ be intervals such that $I_1$ and $I_2$ intersects in at most one point and such that $I = I_1 \cup I_2$. If $g$ is a Lebesgue integrable function defined on $I_1$ and $h$ is a Lebesgue integrable function defined on $I_2$ then the function $f(x) = \left\{\begin{matrix} g(x) & x \in I_1\\ h(x) & x \in I_2 \setminus (I_1 \cap I_2) \end{matrix}\right.$ is a Lebesgue integrable function on the entire $I$ and $\displaystyle{\int_I f(x) \: dx = \int_{I_1} g(x) \: dx + \int_{I_2} h(x) \: dx}$.
• Proof: Let $g$ be a Lebesgue integrable function on $I_1$ and let $h$ be a Lebesgue integrable function on $I_2$. Then there exists upper functions $u_1$ and $v_1$ on $I_1$ such that $g = u_1 - v_1$. Similarly, there exists upper functions $u_2$ and $v_2$ on $I_2$ such that $h = u_2 - v_2$.
• Define the functions $u$ and $v$ for all $x \in I$ as:
(6)
\begin{align} \quad u(x) = \left\{\begin{matrix} u_1(x) & \mathrm{if} \: x \in I_1 \\ u_2(x) & \mathrm{if} \: x \in I_2 \setminus (I_1 \cap I_2) \end{matrix}\right. \quad \mathrm{and} \quad v(x) = \left\{\begin{matrix} v_1(x) & \mathrm{if} \: x \in I_1 \\ v_2(x) & \mathrm{if} \: x \in I_2 \setminus (I_1 \cap I_2) \end{matrix}\right. \end{align}
• Then by one of the theorems referenced on the page above, we have that $u$ is an upper function on all of $I$ and $v$ is an upper function on all of $I$, so:
(7)
\begin{align} \quad \int_I u(x) \: dx = \int_{I_1} u_1(x) \: dx + \int_{I_2} u_2(x) \: dx \quad \mathrm{and} \quad \int_I v(x) \: dx = \int_{I_1} v_1(x) \: dx + \int_{I_2} v_2(x) \: dx \end{align}
• Furthermore, for all $x \in I$ we see that:
(8)
\begin{align} \quad u(x) - v(x) = \left\{\begin{matrix} u_1(x) - v_1(x) & \mathrm{if} \: x \in I_1 \\ u_2(x) - v_2(x) & \mathrm{if} \: x \in I_2 \setminus (I_1 \cap I_2) \end{matrix}\right. = \left \{\begin{matrix} g(x) & \mathrm{if} \: x \in I_1 \\ h(x) & \mathrm{if} \: x \in I_2 \setminus (I_1 \cap I_2) \end{matrix}\right. = f(x) \end{align}
• So $f$ is Lebesgue integrable on $I$ and furthermore:
(9)
\begin{align} \quad \int_I f(x) \: dx &= \int_I [u(x) - v(x)] \: dx \\ \quad &= \left ( \int_{I_1} u_1(x) \: dx + \int_{I_2} u_2(x) \: dx \right ) - \left ( \int_{I_1} v_1(x) \: dx + \int_{I_2} v_2(x) \: dx \right ) \\ \quad &= \left ( \int_{I_1} [u_1(x) - v_1(x)] \: dx \right ) + \left ( \int_{I_2} [u_2(x) - v_2(x)] \: dx \right ) \\ \quad &= \int_{I_1} g(x) \: dx + \int_{I_2} h(x) \: dx \quad \blacksquare \end{align}